Deck 10: Categorical Data Analysis

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Question
Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table.  RESPONSE  NUMBER  Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50\begin{array} { l | l } \text { RESPONSE } & \text { NUMBER } \\\hline \text { Agree Strongly } & 60 \\\text { Agree Somewhat } & 110 \\\text { Disagree Somewhat } & 80 \\\text { Disagree Strongly } & 50\end{array}
In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. As part of that analysis, a 90% confidence interval for the multinomial probability associated with the "Disagree Somewhat" response was desired. Which of the following confidence intervals should be used?

A) (0.206, 0.327)
B) (0.201, 0.332)
C) (0.225, 0.309)
D) (0.216, 0.317)
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Question
A drug company developed a honey-based liquid medicine designed to calm a child's cough at night. To test the drug, 105 children who were ill with an upper respiratory tract infection were randomly selected to participate in a clinical trial. The children were randomly divided into three groups - one group was given a dosage of the honey drug, the second was given a dosage of liquid DM (an over-the-counter cough medicine), and the third (control group) received a liquid placebo (no dosage at all). After administering the medicine to their coughing child, parents rated their children's cough diagnosis as either better or worse. The results are shown in the table below:
 Diagnosis  Treatment  Better  Worse  Total  Control 43337 DM 122133 Honey 241135 Total 4065105\begin{array}{l}\begin{array}{lccc}&\text { Diagnosis }\\\text { Treatment } & \text { Better } & \text { Worse } & \text { Total } \\\text { Control } & 4 & 33 & 37 \\\text { DM } & 12 & 21 & 33 \\\text { Honey } & 24 & 11 & 35 \\\text { Total } & 40 & 65 & 105\end{array}\end{array}

In order to determine whether the treatment group is independent of the coughing diagnosis, a two-way chi-square test was conducted. Use the chi-square distribution to determine the rejection region for this test when testing at α=0.025\alpha = 0.025 .

A) Reject H0\mathrm { H } _ { 0 } if χ2>5.99147\chi ^ { 2 } > 5.99147
B) Reject H0\mathrm { H } _ { 0 } if χ2>7.81473\chi ^ { 2 } > 7.81473
C) Reject H0\mathrm { H } _ { 0 } if χ2>7.37776\chi ^ { 2 } > 7.37776
D) Reject H0\mathrm { H } _ { 0 } if χ2>9.34840\chi ^ { 2 } > 9.34840
Question
Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table.

 RESPONSE  NUMBER  Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50\begin{array}{l|l}\text { RESPONSE } & \text { NUMBER } \\\hline \text { Agree Strongly } & 60 \\\text { Agree Somewhat } & 110 \\\text { Disagree Somewhat } & 80 \\\text { Disagree Strongly } & 50\end{array}

Specify the null hypothesis for testing whether the true proportions of Internet users in each response category are equal.

A) H0:p1=p2=p3=p4=0.25\mathrm { H } _ { 0 } : \mathrm { p } _ { 1 } = \mathrm { p } _ { 2 } = \mathrm { p } _ { 3 } = \mathrm { p } _ { 4 } = 0.25 , where pi\mathrm { p } _ { \mathrm { i } } represents the proportion of Internet users in one of the four response categories
B) H0:p1=60,p2=110,p3=80,p4=50\mathrm { H } _ { 0 } : \mathrm { p } _ { 1 } = 60 , \mathrm { p } _ { 2 } = 110 , \mathrm { p } _ { 3 } = 80 , \mathrm { p } _ { 4 } = 50 , where pi\mathrm { p } _ { \mathrm { i } } represents the proportion of Internet users in one of the four response categories
C) H0:μ1=μ2=μ3=μ4\mathrm { H } _ { 0 } : \mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } = \mu _ { 4 } , where μi\mu _ { \mathrm { i } } represents the average number of Internet users in one of the found response categories
D) H0\mathrm { H } _ { 0 } : Internet users and Government scanning are independent
Question
The data below show the age and favorite type of music of 779 randomly selected people.
Test the claim that age and preferred music type are independent. Use α=0.05\alpha = 0.05  Age  Country  Rock  Pop  Classical 152121459033213068554248304065473157405060392553\begin{array} { l | c c c c } \text { Age } & \text { Country } & \text { Rock } & \text { Pop } & \text { Classical } \\\hline 15 - 21 & 21 & 45 & 90 & 33 \\21 - 30 & 68 & 55 & 42 & 48 \\30 - 40 & 65 & 47 & 31 & 57 \\40 - 50 & 60 & 39 & 25 & 53\end{array}
Question
A business professor conducted a campus survey to estimate demand among all students for a protein supplement for smoothies and other nutritional drinks. Each of 113 students, randomly selected from all students on campus, provided the following information:
(1) How health conscious are you? (Very, Moderately, Slightly, Not very)
(2) Do you prefer protein supplements in your smoothies? (Yes, No)
As part of his analysis, the professor claims that whether or not the student prefers a protein supplement in smoothies is independent of health consciousness level (Very, Moderate, Slightly, or not very). Use the chi-square distribution to determine the rejection region for this test when
Testing at α=0.05\alpha = 0.05

A) Reject H0\mathrm { H } _ { 0 } if χ2>7.81473\chi ^ { 2 } > 7.81473
B) Reject H0\mathrm { H } _ { 0 } if χ2>5.99147\chi ^ { 2 } > 5.99147
C) Reject H0\mathrm { H } _ { 0 } if χ2>9.34840\chi ^ { 2 } > 9.34840
D) Reject H0\mathrm { H } _ { 0 } if χ2>9.48773\chi ^ { 2 } > 9.48773
Question
Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table.

 RESPONSE  NUMBER  Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50\begin{array}{l|l}\text { RESPONSE } & \text { NUMBER } \\\hline \text { Agree Strongly } & 60 \\\text { Agree Somewhat } & 110 \\\text { Disagree Somewhat } & 80 \\\text { Disagree Strongly } & 50\end{array}

In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. Use the chi-square distribution to determine the rejection region when testing at α=.05\alpha = .05 .

A) Reject H0\mathrm { H } _ { 0 } if χ2>9.48773\chi ^ { 2 } > 9.48773
B) Reject H0\mathrm { H } _ { 0 } if χ2>0.351846\chi ^ { 2 } > 0.351846
C) Reject H0\mathrm { H } _ { 0 } if χ2>7.81473\chi ^ { 2 } > 7.81473
D) Reject H0\mathrm { H } _ { 0 } if χ2>0.710721\chi ^ { 2 } > 0.710721
Question
A survey of entrepreneurs focused on their job characteristics, work habits, social activities, leisure time, etc. One question put to each entrepreneur was, "What make of car (U.S., Europe, or Japan)
Do you drive?" The responses (number in each category) for a sample of 100 entrepreneurs are summarized below. The goal of the analysis is to determine if the proportions of entrepreneurs who drive American, European, and Japanese cars differ.

 U.S.  Europe  Japan 403525\begin{array}{ccc}\text { U.S. } & \text { Europe } & \text { Japan } \\\hline 40 & 35 & 25 \\\hline\end{array}

In order to determine whether the true proportions in each response category differ, a one-way chi-square analysis should be conducted. Use the chi-square distribution to determine the rejection region when testing at α=0.025\alpha = 0.025 .

A) Reject H0\mathrm { H } _ { 0 } if χ2>11.1433\chi ^ { 2 } > 11.1433
B) Reject H0\mathrm { H } _ { 0 } if χ2>9.21034\chi ^ { 2 } > 9.21034
C) Reject H0\mathrm { H } _ { 0 } if χ2>9.34840\chi ^ { 2 } > 9.34840
D) Reject H0\mathrm { H } _ { 0 } if χ2>7.37776\chi ^ { 2 } > 7.37776
Question
Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table.  RESPONSE  NUMBER  Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50\begin{array} { l | l } \text { RESPONSE } & \text { NUMBER } \\\hline \text { Agree Strongly } & 60 \\\text { Agree Somewhat } & 110 \\\text { Disagree Somewhat } & 80 \\\text { Disagree Strongly } & 50\end{array}

In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. Which of the following statements is not necessary for the analysis to be valid?

A) The expected cell counts all must be 30 or more.
B) The probabilities for the four response outcomes remain the same from one internet user to the next.
C) The 300 internet users sampled are independent from one another.
Question
A survey of entrepreneurs focused on their job characteristics, work habits, social activities, leisure time, etc. One question put to each entrepreneur was, "What make of car (U.S., Europe, or Japan) do you drive?" The responses (number in each category) for a sample of 100 entrepreneurs are
Summarized below. The goal of the analysis is to determine if the proportions of entrepreneur who drive American, European, and Japanese cars differ.

 U.S.  Europe  Japan 403525\begin{array}{ccc}\text { U.S. } & \text { Europe } & \text { Japan } \\\hline 40 & 35 & 25 \\\hline\end{array}

In order to determine whether the true proportions in each response category differ, a one-way chi-square analysis should be conducted. When calculating the test statistic, what values for the expected counts should be used in the calculation?

A) E1=33.33,E2=33.33,E3=33.33\mathrm { E } _ { 1 } = 33.33 , \mathrm { E } _ { 2 } = 33.33 , \mathrm { E } _ { 3 } = 33.33
B) E1=46,E2=44,E3=9\mathrm { E } _ { 1 } = 46 , \mathrm { E } _ { 2 } = 44 , \mathrm { E } _ { 3 } = 9
C) E1=100,E2=100,E3=100\mathrm { E } _ { 1 } = 100 , \mathrm { E } _ { 2 } = 100 , \mathrm { E } _ { 3 } = 100
D) E1=0.45,E2=0.46,E3=0.09\mathrm { E } _ { 1 } = 0.45 , \mathrm { E } _ { 2 } = 0.46 , \mathrm { E } _ { 3 } = 0.09
Question
Economists at USF are researching the problem of absenteeism at U.S. firms. A random sample of 100 U.S. organizations was selected to participate in a 1-year study. As part of the study, the economists had collected data on the following two variables for each company: shiftwork available (Yes or No), and union-management relationship (Good or Poor). As part of their analyses, the economists wanted to determine whether or not a company makes shiftwork available depends on the relationship between union and management. The collected data are
Shown below:

 Relation  Shiftwork  Good  Bad  Total  No 112233 Yes 254267 Total 3664100\begin{array}{l}\begin{array}{lccc}&\text { Relation }\\\text { Shiftwork } & \text { Good } & \text { Bad } & \text { Total } \\\text { No } & 11 & 22 & 33 \\\text { Yes } & 25 & 42 & 67 \\\text { Total } & 36 & 64 & 100\end{array}\end{array}

In order to determine whether the shiftwork responses depend on the relationship responses, a two-way chi-square analysis should be conducted. Calculate the value of the test statistic for the desired analysis.

A) χ2=0.70\chi ^ { 2 } = 0.70
B) χ2=0.15\chi ^ { 2 } = 0.15
C) χ2=11.88\chi ^ { 2 } = 11.88
D) χ2=0.07\chi ^ { 2 } = 0.07
Question
Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table.  RESPONSE  NUMBER  Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50\begin{array} { l | l } \text { RESPONSE } & \text { NUMBER } \\\hline \text { Agree Strongly } & 60 \\\text { Agree Somewhat } & 110 \\\text { Disagree Somewhat } & 80 \\\text { Disagree Strongly } & 50\end{array} In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. As part of that analysis, a 95% confidence interval for the multinomial probability associated with the "Agree Strongly" response was desired. Which of the following confidence intervals should be used?

A) (0.141, 0.259)
B) (0.162, 0.238)
C) (0.145, 0.255)
D) (0.155, 0.245)
Question
A survey of entrepreneurs focused on their job characteristics, work habits, social activities, leisure time, etc. One question put to each entrepreneur was, "What make of car (U.S., Europe, or Japan) do you drive?" The responses (number in each category) for a sample of 100 entrepreneurs are summarized below. The goal of the analysis is to determine if the proportions of entrepreneurs who drive American, European, and Japanese cars differ.

 U.S.  Europe  Japan 403525\begin{array} { c c c } \text { U.S. } & \text { Europe } & \text { Japan } \\\hline 40 & 35 & 25 \\\hline\end{array}
In order to determine whether the true proportions in each response category differ, a one-way chi-square analysis should be conducted. As part of that analysis, a 95% confidence interval for the multinomial probability associated with the "Europe" response was desired. Which of the following
Confidence intervals should be used?

A) (0.257, 0.443)
B) (0.271, 0.428)
C) (0.265, 0.440)
D) (0.227, 0.473)
Question
An adverse drug effect (ADE) is an unintended injury caused by prescribed medication. The table summarizes the proximal cause of 95 ADEs that resulted from a dosing error at a Boston hospital. An adverse drug effect (ADE) is an unintended injury caused by prescribed medication. The table summarizes the proximal cause of 95 ADEs that resulted from a dosing error at a Boston hospital.  <div style=padding-top: 35px>
Question
Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table.
 RESPONSE  NUMBER  Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50\begin{array}{l|l}\text { RESPONSE } & \text { NUMBER } \\\hline \text { Agree Strongly } & 60 \\\text { Agree Somewhat } & 110 \\\text { Disagree Somewhat } & 80 \\\text { Disagree Strongly } & 50\end{array}

In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. Calculate the value of the test statistic for the desired analysis.

A) χ2=28.0\chi ^ { 2 } = 28.0
B) χ2=0.25\chi ^ { 2 } = 0.25
C) χ2=75\chi ^ { 2 } = 75
D) χ2=22.54\chi ^ { 2 } = 22.54
Question
A survey of entrepreneurs focused on their job characteristics, work habits, social activities, leisure time, etc. One question put to each entrepreneur was, "What make of car (U.S., Europe, or Japan)do you drive?" The responses (number in each category) for a sample of 100 entrepreneurs are summarized below. The goal of the analysis is to determine if the proportions of entrepreneurs Who drive American, European, and Japanese cars differ.
 U.S.  Europe  Japan 403525\begin{array}{ccc}\text { U.S. } & \text { Europe } & \text { Japan } \\\hline 40 & 35 & 25 \\\hline\end{array}

In order to determine whether the true proportions in each response category differ, a one-way chi-square analysis should be conducted. Suppose the p-value for the test was calculated to be p=0.1738\mathrm { p } = 0.1738 . What is the appropriate conclusion to make when testing at α=0.10\alpha = 0.10 ?

A) There is sufficient evidence to indicate the proportion of entrepreneurs driving Japanese cars is less than the proportion driving U.S. cars.
B) There is sufficient evidence to indicate the proportion of entrepreneurs driving the three makes of car differ.
C) There is sufficient evidence to indicate the proportion of entrepreneurs driving the three makes of car are equal.
D) There is insufficient evidence to indicate the proportion of entrepreneurs driving the three makes of car differ.
Question
A drug company developed a honey-based liquid medicine designed to calm a child's cough at night. To test the drug, 105 children who were ill with an upper respiratory tract infection were randomly selected to participate in a clinical trial. The children were randomly divided into three groups - one group was given a dosage of the honey drug, the second was given a dosage of liquid DM (an over-the-counter cough medicine), and the third (control group) received a liquid placebo (no dosage at all). After administering the medicine to their coughing child, parents rated their children's cough diagnosis as either better or worse. The results are shown in the table below:
 Diagnosis Treatment  Better  Worse  Total  Control 43337 DM 122133 Honey 241135 Total 4065105\begin{array} { l c c c } &\quad\quad\text { \quad Diagnosis}\\\text { Treatment } & \text { Better } & \text { Worse } & \text { Total } \\ \text { Control } & 4 & 33 & 37 \\ \text { DM } & 12 & 21 & 33 \\ \text { Honey } & 24 & 11 & 35 \\ \text { Total } & 40 & 65 & 105 \end{array}
In order to determine whether the treatment group is independent of the coughing diagnosis, a two-way chi-square test was conducted. Suppose the pp -value for the test was calculated to be p=p = 0.00160.0016 . What is the appropriate conclusion to make when testing at α=0.05\alpha = 0.05 ?

A) There is insufficient evidence to indicate the treatment group is dependent on the coughing diagnosis.
B) There is sufficient evidence to indicate the treatment group is independent of the coughing diagnosis.
C) There is insufficient evidence to indicate the treatment group is independent of the coughing diagnosis.
D) There is sufficient evidence to indicate the treatment group is dependent on the coughing diagnosis.
Question
A business professor conducted a campus survey to estimate demand among all students for a protein supplement for smoothies and other nutritional drinks. Each of 113 students, randomly
Selected from all students on campus, provided the following information:
(1) How health conscious are you? (Very, Moderately, Slightly, Not very)
(2) Do you prefer protein supplements in your smoothies? (Yes, No)
As part of his analysis, the professor claims that whether or not the student prefers a protein
Supplement in smoothies is independent of health consciousness level (Very, Moderate, Slightly, or
Not very). Identify the appropriate alternative hypothesis that the professor should use in the test
Of hypothesis he desires.

A) HA\mathrm { H } _ { \mathrm { A } } : Preference and Health Consciousness level are independent variables.
B) HA\mathrm { H } _ { \mathrm { A } } : There is interaction between the Preference and Health Consciousness variables.
C) HA\mathrm { H } _ { \mathrm { A } } : Preference and Health Consciousness level are dependent variables.
D) HA\mathrm { H } _ { \mathrm { A } } : Preference and Health Consciousness level are mutually exclusive variables.
Question
Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table.
 RESPONSE  NUMBER  Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50\begin{array}{l|l}\text { RESPONSE } & \text { NUMBER } \\\hline \text { Agree Strongly } & 60 \\\text { Agree Somewhat } & 110 \\\text { Disagree Somewhat } & 80 \\\text { Disagree Strongly } & 50\end{array}

In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. When calculating the test statistic, what values for the expected counts should be used in the calculation?

A) E1=75,E2=75,E3=75,E4=75\mathrm { E } _ { 1 } = 75 , \mathrm { E } _ { 2 } = 75 , \mathrm { E } _ { 3 } = 75 , \mathrm { E } _ { 4 } = 75
B) E1=300,E2=300,E3=300,E4=300\mathrm { E } _ { 1 } = 300 , \mathrm { E } _ { 2 } = 300 , \mathrm { E } _ { 3 } = 300 , \mathrm { E } _ { 4 } = 300
C) E1=0.25,E2=0.25,E3=0.25,E4=0.25\mathrm { E } _ { 1 } = 0.25 , \mathrm { E } _ { 2 } = 0.25 , \mathrm { E } _ { 3 } = 0.25 , \mathrm { E } _ { 4 } = 0.25
D) E1=60,E2=110,E3=80,E4=50\mathrm { E } _ { 1 } = 60 , \mathrm { E } _ { 2 } = 110 , \mathrm { E } _ { 3 } = 80 , \mathrm { E } _ { 4 } = 50
Question
Economists at USF are researching the problem of absenteeism at U.S. firms. A random sample of 100 U.S. organizations was selected to participate in a 1-year study. As part of the study, the economists had collected data on the following two variables for each company: shiftwork available (Yes or No), and union-management relationship (Good or Poor). As part of their analyses, the economists wanted to determine whether or not a company makes shiftwork available depends on the relationship between union and management. The collected data are
Shown below:


 Relation Shiftwork  Good  Bad  Total  No 112233 Yes 254267 Total 3664100\begin{array} { l c c c }&\text { Relation}\\ \text { Shiftwork } & \text { Good } & \text { Bad } & \text { Total } \\ \text { No } & 11 & 22 & 33 \\ \text { Yes } & 25 & 42 & 67 \\ \text { Total } & 36 & 64 & 100 \end{array}

In order to determine whether the shiftwork responses depend on the relationship responses, a two-way chi-square analysis should be conducted. When calculating the test statistic, what values for the expected counts should be used in the calculation?

A) E11=36,E12=33,E21=64,E22=67\mathrm { E } _ { 11 } = 36 , \mathrm { E } _ { 12 } = 33 , \mathrm { E } _ { 21 } = 64 , \mathrm { E } _ { 22 } = 67
B) E11=0.065,E12=0.036,E21=0.032,E22=0.018\mathrm { E } _ { 11 } = 0.065 , \mathrm { E } _ { 12 } = 0.036 , \mathrm { E } _ { 21 } = 0.032 , \mathrm { E } _ { 22 } = 0.018
C) E11=11.88,E12=21.12,E21=24.12,E22=42.88\mathrm { E } _ { 11 } = 11.88 , \mathrm { E } _ { 12 } = 21.12 , \mathrm { E } _ { 21 } = 24.12 , \mathrm { E } _ { 22 } = 42.88
D) E11=11,E12=22,E21=25,E22=42\mathrm { E } _ { 11 } = 11 , \mathrm { E } _ { 12 } = 22 , \mathrm { E } _ { 21 } = 25 , \mathrm { E } _ { 22 } = 42
Question
Economists at USF are researching the problem of absenteeism at U.S. firms. A random sample of 100 U.S. organizations was selected to participate in a 1-year study. As part of the study, the economists had collected data on the following two variables for each company: shiftwork available (Yes or No), and union-management relationship (Good or Poor). As part of their analyses, the economists wanted to determine whether or not a company makes shiftwork available depends on the relationship between union and management. The collected data are
Shown below:

 Relation  Shiftwork  Good  Bad  Total  No 112233 Yes 254267 Total 3664100\begin{array}{lccc} & {\text { Relation }} & \\\text { Shiftwork } & \text { Good } & \text { Bad } & \text { Total } \\\text { No } & 11 & 22 & 33 \\\text { Yes } & 25 & 42 & 67 \\\text { Total } & 36 & 64 & 100\end{array}

Use the chi-square distribution to determine the rejection region for this test when testing at α=\alpha = 0.050.05 .

A) Reject H0\mathrm { H } _ { 0 } if χ2>7.81473\chi ^ { 2 } > 7.81473
B) Reject H0\mathrm { H } _ { 0 } if χ2>3.84146\chi ^ { 2 } > 3.84146
C) Reject H0\mathrm { H } _ { 0 } if χ2>5.02389\chi ^ { 2 } > 5.02389
D) Reject H0\mathrm { H } _ { 0 } if χ2>5.99147\chi ^ { 2 } > 5.99147
Question
The contingency table below shows the results of a random sample of 200 state representatives that was conducted to see whether their opinions on a bill are related to their party affiliations.
 Opinion  Party  Approve  Disapprove  No Opinion  Republican 422014 Democrat 502418 Independent 10166 Test the claim of independence. Use α=.05\begin{array}{l}\begin{array} { l | c c c } & { \text { Opinion } } \\\hline \text { Party } & \text { Approve } & \text { Disapprove } & \text { No Opinion } \\\text { Republican } & 42 & 20 & 14 \\\text { Democrat } & 50 & 24 & 18 \\\text { Independent } & 10 & 16 & 6\end{array}\\\text { Test the claim of independence. Use } \alpha = .05 \text {. }\end{array}
Question
A sports researcher is interested in determining if there is a relationship between the number of home team and visiting team wins and different sports. A random sample of 526 games is selected and the results are given below. Find the rejection region used to test the claim that the number of home team and visiting team wins is independent of the sport. Use α=0.01\alpha = 0.01

 Football  Basketball  Soccer  Baseball  Home team wins 391562583 Visiting team wins 31981975\begin{array}{l|cccc} & \text { Football } & \text { Basketball } & \text { Soccer } & \text { Baseball } \\\hline \text { Home team wins } & 39 & 156 & 25 & 83 \\\text { Visiting team wins } & 31 & 98 & 19 & 75\end{array}

A) ?2>12.838? ^ { 2 } > 12.838
B) ?2>9.348? ^ { 2 } > 9.348
C) ?2>11.345? ^ { 2 } > 11.345
D) ?2>7.815? ^ { 2 } > 7.815
Question
A professor chose a random sample of 50 recent graduates of an MBA program and recorded the gender of each graduate (M or F) and whether the graduate chose to complete his or her degree requirements by completing a research project (RP) or by taking comprehensive exams (CE). The results are shown below.

 M, CE  M, RP  F, RP  M, CE  M, CE  F, CE  F, RP,  M, CE  M, RP  F, RP  F, CE  M, RP  F, RP  F, CE  M, RP  F, RP  M, RP  F, CE  M, CE  M, CE  M, RP  F, CE  F, RP  M, CE  M, CE  F, CE  M, RP  F, RP  M,CE  M, CE  F, CE  F, RP,  M, CE  M, RP  F, RP  M, CE  M, RP  F, RP  F, CE  M, RP  F, RP  M, RP  M, CE  M,CE  M, CE  M, RP  F, CE  F, RP  M,CE  F, CE \begin{array} { l l l l l } \text { M, CE } & \text { M, RP } & \text { F, RP } & \text { M, CE } & \text { M, CE } \\ \text { F, CE } & \text { F, RP, } & \text { M, CE } & \text { M, RP } & \text { F, RP } \\ \text { F, CE } & \text { M, RP } & \text { F, RP } & \text { F, CE } & \text { M, RP } \\ \text { F, RP } & \text { M, RP } & \text { F, CE } & \text { M, CE } & \text { M, CE } \\ \text { M, RP } & \text { F, CE } & \text { F, RP } & \text { M, CE } & \text { M, CE } \\ \text { F, CE } & \text { M, RP } & \text { F, RP } & \text { M,CE } & \text { M, CE } \\ \text { F, CE } & \text { F, RP, } & \text { M, CE } & \text { M, RP } & \text { F, RP } \\ \text { M, CE } & \text { M, RP } & \text { F, RP } & \text { F, CE } & \text { M, RP } \\ \text { F, RP } & \text { M, RP } & \text { M, CE } & \text { M,CE } & \text { M, CE } \\ \text { M, RP } & \text { F, CE } & \text { F, RP } & \text { M,CE } & \text { F, CE } \end{array}

a. Create a contingency table for the data.
b. Perform a χ2\chi ^ { 2 } -test to determine if there is any evidence that gender and choice of research project or comprehensive exams are not independent. Use α=0.05\alpha = 0.05 .
Question
A new coffeehouse wishes to see whether customers have any preference among 5 different brands of coffee. A sample of 200 customers provided the data below. Test the claim that the probabilities show no preference. Us α=0.01\alpha = 0.01  Brand 12345 Customers 1832556530\begin{array} { l | c c c c c } \text { Brand } & 1 & 2 & 3 & 4 & 5 \\\hline \text { Customers } & 18 & 32 & 55 & 65 & 30\end{array}
Question
What is categorical data?
Question
A sports researcher is interested in determining if there is a relationship between the number of home team and visiting team wins and different sports. A random sample of 526 games is selected and the results are given below. Assuming the row and column classifications are independent, find an estimate for the expected cell count of cell E22E _ { 22 }
 Football  Basketball  Soccer  Baseball  Home team wins 381542784 Visiting team wins 31982272\begin{array}{l|llll} & \text { Football } & \text { Basketball } & \text { Soccer } & \text { Baseball } \\\hline \text { Home team wins } & 38 & 154 & 27 & 84 \\\text { Visiting team wins } & 31 & 98 & 22 & 72\end{array}

A) 106.8106.8
B) 20.820.8
C) 28.228.2
D) 145.2145.2
Question
  Construct a 99% confidence interval for the multinomial probability associated with cell 2.<div style=padding-top: 35px> Construct a 99% confidence interval for the multinomial probability associated with cell 2.
Question
Consider the accompanying contingency table.  Column  Row 12311315162192611\begin{array} { c | c c c } & { \text { Column } } \\\text { Row } & 1 & 2 & 3 \\\hline 1 & 13 & 15 & 16 \\2 & 19 & 26 & 11\end{array} a. Convert the values in row 1 to percentages by calculating the percentage of each column total falling in row 1.
b. Create a bar graph with row 1 percentage on the vertical axis and column number on the horizontal axis.
c. What pattern do you expect to see if the rows and columns are not independent? Is this pattern present in your graph?
Question
The null hypothesis for a test of data resulting from a multinomial experiment is given as The null hypothesis for a test of data resulting from a multinomial experiment is given as   What is the alternative hypothesis for the test?<div style=padding-top: 35px> What is the alternative hypothesis for the test?
Question
Many track runners believe that they have a better chance of winning if they start in the inside lane that is closest to the field. For the data below, the lane closest to the field is Lane 1, the next lane is Lane 2, and so on until the outermost lane, Lane 6. The table displays the starting positions for the winners of 240 competitions. Test the claim that the probability of winning is the same regardless of starting position. Use α=0.05\alpha = 0.05 . The results are based on 240 wins.
 Starting Position 123456 Number of Wins 363344503245\begin{array} { l | c c c c c c } \text { Starting Position } & 1 & 2 & 3 & 4 & 5 & 6 \\\hline \text { Number of Wins } & 36 & 33 & 44 & 50 & 32 & 45\end{array}
Question
When using any procedure to perform a hypothesis test, the user should always be certain that the
experiment satisfies the assumptions given with the procedure.
Question
A random sample of 160 car accidents are selected and categorized by the age of the driver determined to be at fault. The results are listed below. The age distribution of drivers for the given categories is 18% for the under 26 group, 39% for the 26-45 group, 31% for the 46-65 group, and 12% for the group over 65. Test the claim that all ages have crash rates proportional to their number of drivers. Use α=0.05\alpha = 0.05  Age  Under 26 26454665 Over 65 Drivers 66392530\begin{array} { l | c c c c } \text { Age } & \text { Under 26 } & 26 - 45 & 46 - 65 & \text { Over } 65 \\\hline \text { Drivers } & 66 & 39 & 25 & 30\end{array}
Question
 <div style=padding-top: 35px>
Question
 <div style=padding-top: 35px>
Question
 <div style=padding-top: 35px>
Question
 <div style=padding-top: 35px>
Question
Describe probabilities of the k outcomes of the multinomial experiment trials.
Question
A teacher finds that final grades in the statistics department are distributed as: A, 25%; B, 25%; C, 40%; D, 5%; F, 5%. At the end of a randomly selected semester, the following grades were recorded. Determine if the grade distribution for the department is different than expected. Use α=0.01\alpha = 0.01  Grade  A  B  C  D  F  Number 364260148\begin{array} { l | c c c c c } \text { Grade } & \text { A } & \text { B } & \text { C } & \text { D } & \text { F } \\\hline \text { Number } & 36 & 42 & 60 & 14 & 8\end{array}
Question
A sports researcher is interested in determining if there is a relationship between the
number of home team and visiting team wins and different sports. A random sample of
526 games is selected and the results are given below. Test the claim that the number of
home team and visiting team wins is independent of the sport. Use A sports researcher is interested in determining if there is a relationship between the number of home team and visiting team wins and different sports. A random sample of 526 games is selected and the results are given below. Test the claim that the number of home team and visiting team wins is independent of the sport. Use    <div style=padding-top: 35px> A sports researcher is interested in determining if there is a relationship between the number of home team and visiting team wins and different sports. A random sample of 526 games is selected and the results are given below. Test the claim that the number of home team and visiting team wins is independent of the sport. Use    <div style=padding-top: 35px>
Question
What are characteristics of the trials in a multinomial experiment?
Question
<strong> </strong> A) 48.91 B) 37.45 C) 55.63 D) 45.91 <div style=padding-top: 35px>

A) 48.91
B) 37.45
C) 55.63
D) 45.91
Question
 <div style=padding-top: 35px>
Question
The contingency table below shows the results of a random sample of 200 state representatives that was conducted to see whether their opinions on a bill are related to their party affiliations.
Assuming the row and column classifications are independent, find an estimate for the expected Cell count E22E _ { 22 }  Opinion  Party  Approve  Disapprove  No Opinion  Republican 422014 Democrat 502418 Independent 10166\begin{array} { l | c c c } & { \text { Opinion } } \\\hline \text { Party } & \text { Approve } & \text { Disapprove } & \text { No Opinion } \\\text { Republican } & 42 & 20 & 14 \\\text { Democrat } & 50 & 24 & 18 \\\text { Independent } & 10 & 16 & 6\end{array}

A) 17.48
B) 27.6
C) 22.8
D) 46.92
Question
The contingency table below shows the results of a random sample of 200 state representatives that was conducted to see whether their opinions on a bill are related to their party affiliation. <strong>The contingency table below shows the results of a random sample of 200 state representatives that was conducted to see whether their opinions on a bill are related to their party affiliation.  </strong> A) 7.662 B) 11.765 C) 9.483 D) 8.030 <div style=padding-top: 35px>

A) 7.662
B) 11.765
C) 9.483
D) 8.030
Question
A random sample of 160 car accidents are selected and categorized by the age of the driver determined to be at fault. The results are listed below. The age distribution of drivers for the giver categories is 18%18 \% for the under 26 group, 39%39 \% for the 264526 - 45 group, 31%31 \% for the 45 - 65 group, and 12%12 \% for the group over 65 . Find the rejection region used to test the claim that all ages have crash rates proportional to their number of drivers. Use α=0.05\alpha = 0.05 .

 Age  Under 26 26454665 Over 65 Drivers 66392530\begin{array}{l|cccc}\text { Age } & \text { Under 26 } & 26-45 & 46-65 & \text { Over } 65 \\\hline \text { Drivers } & 66 & 39 & 25 & 30\end{array}

A) x2>7.815x ^ { 2 } > 7.815
B) χ2>11.143\chi ^ { 2 } > 11.143
C) x2>6.251x ^ { 2 } > 6.251
D) χ2>9.348\chi ^ { 2 } > 9.348
Question
Use the appropriate table to find the following probabili <strong>Use the appropriate table to find the following probabili  </strong> A) 0.995 B) 0.005 C) 0.010 D) 0.990 <div style=padding-top: 35px>

A) 0.995
B) 0.005
C) 0.010
D) 0.990
Question
<strong> </strong> A) 2.919 B) 4.192 C) 3.290 D) 5.391 <div style=padding-top: 35px>

A) 2.919
B) 4.192
C) 3.290
D) 5.391
Question
Many track runners believe that they have a better chance of winning if they start in the inside lane that is closest to the field. For the data below, the lane closest to the field is Lane 1 , the next lane is Lane 2, and so on until the outermost lane, Lane 6. The table displays the starting positions for the winners of 240 competitions. Find the rejection region used to test the claim that the probability of winning is the same regardless of starting position. Use α=0.05\alpha = 0.05 . The results are based on 240 wins.
 Starting Position 123456 Number of Wins 453236445033\begin{array}{l|cccccc}\text { Starting Position } & 1 & 2 & 3 & 4 & 5 & 6 \\\hline \text { Number of Wins } & 45 & 32 & 36 & 44 & 50 & 33\end{array}

A) χ2>15.086\chi ^ { 2 } > 15.086
B) χ2>11.070\chi ^ { 2 } > 11.070
C) χ2>12.833\chi ^ { 2 } > 12.833
D) χ2>9.236\chi ^ { 2 } > 9.236
Question
 <div style=padding-top: 35px>
Question
The contingency table below shows the results of a random sample of 200 state representatives that was conducted to see whether their opinions on a bill are related to their party affiliations. Use α=\alpha = 0.050.05 .

 Opinion  Party  Approve  Disapprove  No Opinion  Republican 422014 Democrat 502418 Independent 10166\begin{array}{l|ccc} &{\text { Opinion }} \\\hline \text { Party } & \text { Approve } & \text { Disapprove } & \text { No Opinion } \\\text { Republican } & 42 & 20 & 14 \\\text { Democrat } & 50 & 24 & 18 \\\text { Independent } & 10 & 16 & 6\end{array}

Find the rejection region used to test the claim of independence.

A) χ2>7.779\chi ^ { 2 } > 7.779
B) χ2>11.143\chi ^ { 2 } > 11.143
C) χ2>13.277\chi ^ { 2 } > 13.277
D) χ2>9.488\chi ^ { 2 } > 9.488
Question
A teacher finds that final grades in the statistics department are distributed as: A, 25\%; B, 25\%; C, 40%;D,5%;F,5%40 \% ; \mathrm { D } , 5 \% ; \mathrm { F } , 5 \% . At the end of a randomly selected semester, the following grades were recorded. Calculate the chi-square test statistic χ2\chi ^ { 2 } used to determine if the grade distribution for the department is different than expected. Use α=0.01\alpha = 0.01 .
 Grade  A  B  C  D  F  Number 423660814\begin{array}{l|lllll}\text { Grade } & \text { A } & \text { B } & \text { C } & \text { D } & \text { F } \\\hline \text { Number } & 42 & 36 & 60 & 8 & 14\end{array}

A) 6.87
B) 5.25
C) 3.41
D) 4.82
Question
In a test of independence, it is safe to conclude that the events are independent when the value of ?2? ^ { 2 } is very small.
Question
A multinomial experiment with k=4k = 4 cells and n=400n = 400 produced the data shown in the following table.
 Cell 1234ni5021310433\begin{array} { r | r c c r } &\text { Cell }\\& 1 & 2 & 3 & 4 \\\hline n _ { i } & 50 & 213 & 104 & 33\end{array}

Previous studies in this area have shown that p1=p2=p3=p4=.25p _ { 1 } = p _ { 2 } = p _ { 3 } = p _ { 4 } = .25 . Construct a 95%95 \% confidence interval for the multinomial probability associated with cell 2

A) (0.491, 0.574)
B) (0.490, 0.575)
C) (0.093, 0.157)
D) (0.484, 0.581)
Question
The ?2? ^ { 2 } test for independence is a useful tool for establishing a causal relationship between two factors.
Question
Use the appropriate table to find the following  chi-square value: χ.052 for df=3\text { chi-square value: } \chi _ { .05 } ^ { 2 } \text { for } \mathrm { df } = 3

A) 7.815
B) 5.991
C) 9.488
D) 0.352
Question
 <div style=padding-top: 35px>
Question
The sampling distribution for ?² works well when expected counts are very small.
Question
A random sample of 160 car accidents are selected and categorized by the age of the driver determined to be at fault. The results are listed below. The age distribution of drivers for the given categories is 18%18 \% for the under 26 group, 39%39 \% for the 26-45 group, 31%31 \% for the 456545 - 65 group, and 12%12 \% for the group over 65 . Calculate the chi-square test statistic χ2\chi ^ { 2 } used to test the claim that all ages have crash rates proportional to their driving rates.

 Age  Under 26 26454665 Over 65 Drivers 66392530\begin{array}{l|cccc}\text { Age } & \text { Under 26 } & 26-45 & 46-65 & \text { Over } 65 \\\hline \text { Drivers } & 66 & 39 & 25 & 30\end{array}

A) 85.123
B) 95.431
C) 75.101
D) 101.324
Question
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Deck 10: Categorical Data Analysis
1
Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table.  RESPONSE  NUMBER  Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50\begin{array} { l | l } \text { RESPONSE } & \text { NUMBER } \\\hline \text { Agree Strongly } & 60 \\\text { Agree Somewhat } & 110 \\\text { Disagree Somewhat } & 80 \\\text { Disagree Strongly } & 50\end{array}
In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. As part of that analysis, a 90% confidence interval for the multinomial probability associated with the "Disagree Somewhat" response was desired. Which of the following confidence intervals should be used?

A) (0.206, 0.327)
B) (0.201, 0.332)
C) (0.225, 0.309)
D) (0.216, 0.317)
(0.225, 0.309)
2
A drug company developed a honey-based liquid medicine designed to calm a child's cough at night. To test the drug, 105 children who were ill with an upper respiratory tract infection were randomly selected to participate in a clinical trial. The children were randomly divided into three groups - one group was given a dosage of the honey drug, the second was given a dosage of liquid DM (an over-the-counter cough medicine), and the third (control group) received a liquid placebo (no dosage at all). After administering the medicine to their coughing child, parents rated their children's cough diagnosis as either better or worse. The results are shown in the table below:
 Diagnosis  Treatment  Better  Worse  Total  Control 43337 DM 122133 Honey 241135 Total 4065105\begin{array}{l}\begin{array}{lccc}&\text { Diagnosis }\\\text { Treatment } & \text { Better } & \text { Worse } & \text { Total } \\\text { Control } & 4 & 33 & 37 \\\text { DM } & 12 & 21 & 33 \\\text { Honey } & 24 & 11 & 35 \\\text { Total } & 40 & 65 & 105\end{array}\end{array}

In order to determine whether the treatment group is independent of the coughing diagnosis, a two-way chi-square test was conducted. Use the chi-square distribution to determine the rejection region for this test when testing at α=0.025\alpha = 0.025 .

A) Reject H0\mathrm { H } _ { 0 } if χ2>5.99147\chi ^ { 2 } > 5.99147
B) Reject H0\mathrm { H } _ { 0 } if χ2>7.81473\chi ^ { 2 } > 7.81473
C) Reject H0\mathrm { H } _ { 0 } if χ2>7.37776\chi ^ { 2 } > 7.37776
D) Reject H0\mathrm { H } _ { 0 } if χ2>9.34840\chi ^ { 2 } > 9.34840
Reject H0\mathrm { H } _ { 0 } if χ2>7.37776\chi ^ { 2 } > 7.37776
3
Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table.

 RESPONSE  NUMBER  Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50\begin{array}{l|l}\text { RESPONSE } & \text { NUMBER } \\\hline \text { Agree Strongly } & 60 \\\text { Agree Somewhat } & 110 \\\text { Disagree Somewhat } & 80 \\\text { Disagree Strongly } & 50\end{array}

Specify the null hypothesis for testing whether the true proportions of Internet users in each response category are equal.

A) H0:p1=p2=p3=p4=0.25\mathrm { H } _ { 0 } : \mathrm { p } _ { 1 } = \mathrm { p } _ { 2 } = \mathrm { p } _ { 3 } = \mathrm { p } _ { 4 } = 0.25 , where pi\mathrm { p } _ { \mathrm { i } } represents the proportion of Internet users in one of the four response categories
B) H0:p1=60,p2=110,p3=80,p4=50\mathrm { H } _ { 0 } : \mathrm { p } _ { 1 } = 60 , \mathrm { p } _ { 2 } = 110 , \mathrm { p } _ { 3 } = 80 , \mathrm { p } _ { 4 } = 50 , where pi\mathrm { p } _ { \mathrm { i } } represents the proportion of Internet users in one of the four response categories
C) H0:μ1=μ2=μ3=μ4\mathrm { H } _ { 0 } : \mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } = \mu _ { 4 } , where μi\mu _ { \mathrm { i } } represents the average number of Internet users in one of the found response categories
D) H0\mathrm { H } _ { 0 } : Internet users and Government scanning are independent
H0:p1=p2=p3=p4=0.25\mathrm { H } _ { 0 } : \mathrm { p } _ { 1 } = \mathrm { p } _ { 2 } = \mathrm { p } _ { 3 } = \mathrm { p } _ { 4 } = 0.25 , where pi\mathrm { p } _ { \mathrm { i } } represents the proportion of Internet users in one of the four response categories
4
The data below show the age and favorite type of music of 779 randomly selected people.
Test the claim that age and preferred music type are independent. Use α=0.05\alpha = 0.05  Age  Country  Rock  Pop  Classical 152121459033213068554248304065473157405060392553\begin{array} { l | c c c c } \text { Age } & \text { Country } & \text { Rock } & \text { Pop } & \text { Classical } \\\hline 15 - 21 & 21 & 45 & 90 & 33 \\21 - 30 & 68 & 55 & 42 & 48 \\30 - 40 & 65 & 47 & 31 & 57 \\40 - 50 & 60 & 39 & 25 & 53\end{array}
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5
A business professor conducted a campus survey to estimate demand among all students for a protein supplement for smoothies and other nutritional drinks. Each of 113 students, randomly selected from all students on campus, provided the following information:
(1) How health conscious are you? (Very, Moderately, Slightly, Not very)
(2) Do you prefer protein supplements in your smoothies? (Yes, No)
As part of his analysis, the professor claims that whether or not the student prefers a protein supplement in smoothies is independent of health consciousness level (Very, Moderate, Slightly, or not very). Use the chi-square distribution to determine the rejection region for this test when
Testing at α=0.05\alpha = 0.05

A) Reject H0\mathrm { H } _ { 0 } if χ2>7.81473\chi ^ { 2 } > 7.81473
B) Reject H0\mathrm { H } _ { 0 } if χ2>5.99147\chi ^ { 2 } > 5.99147
C) Reject H0\mathrm { H } _ { 0 } if χ2>9.34840\chi ^ { 2 } > 9.34840
D) Reject H0\mathrm { H } _ { 0 } if χ2>9.48773\chi ^ { 2 } > 9.48773
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6
Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table.

 RESPONSE  NUMBER  Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50\begin{array}{l|l}\text { RESPONSE } & \text { NUMBER } \\\hline \text { Agree Strongly } & 60 \\\text { Agree Somewhat } & 110 \\\text { Disagree Somewhat } & 80 \\\text { Disagree Strongly } & 50\end{array}

In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. Use the chi-square distribution to determine the rejection region when testing at α=.05\alpha = .05 .

A) Reject H0\mathrm { H } _ { 0 } if χ2>9.48773\chi ^ { 2 } > 9.48773
B) Reject H0\mathrm { H } _ { 0 } if χ2>0.351846\chi ^ { 2 } > 0.351846
C) Reject H0\mathrm { H } _ { 0 } if χ2>7.81473\chi ^ { 2 } > 7.81473
D) Reject H0\mathrm { H } _ { 0 } if χ2>0.710721\chi ^ { 2 } > 0.710721
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7
A survey of entrepreneurs focused on their job characteristics, work habits, social activities, leisure time, etc. One question put to each entrepreneur was, "What make of car (U.S., Europe, or Japan)
Do you drive?" The responses (number in each category) for a sample of 100 entrepreneurs are summarized below. The goal of the analysis is to determine if the proportions of entrepreneurs who drive American, European, and Japanese cars differ.

 U.S.  Europe  Japan 403525\begin{array}{ccc}\text { U.S. } & \text { Europe } & \text { Japan } \\\hline 40 & 35 & 25 \\\hline\end{array}

In order to determine whether the true proportions in each response category differ, a one-way chi-square analysis should be conducted. Use the chi-square distribution to determine the rejection region when testing at α=0.025\alpha = 0.025 .

A) Reject H0\mathrm { H } _ { 0 } if χ2>11.1433\chi ^ { 2 } > 11.1433
B) Reject H0\mathrm { H } _ { 0 } if χ2>9.21034\chi ^ { 2 } > 9.21034
C) Reject H0\mathrm { H } _ { 0 } if χ2>9.34840\chi ^ { 2 } > 9.34840
D) Reject H0\mathrm { H } _ { 0 } if χ2>7.37776\chi ^ { 2 } > 7.37776
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8
Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table.  RESPONSE  NUMBER  Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50\begin{array} { l | l } \text { RESPONSE } & \text { NUMBER } \\\hline \text { Agree Strongly } & 60 \\\text { Agree Somewhat } & 110 \\\text { Disagree Somewhat } & 80 \\\text { Disagree Strongly } & 50\end{array}

In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. Which of the following statements is not necessary for the analysis to be valid?

A) The expected cell counts all must be 30 or more.
B) The probabilities for the four response outcomes remain the same from one internet user to the next.
C) The 300 internet users sampled are independent from one another.
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9
A survey of entrepreneurs focused on their job characteristics, work habits, social activities, leisure time, etc. One question put to each entrepreneur was, "What make of car (U.S., Europe, or Japan) do you drive?" The responses (number in each category) for a sample of 100 entrepreneurs are
Summarized below. The goal of the analysis is to determine if the proportions of entrepreneur who drive American, European, and Japanese cars differ.

 U.S.  Europe  Japan 403525\begin{array}{ccc}\text { U.S. } & \text { Europe } & \text { Japan } \\\hline 40 & 35 & 25 \\\hline\end{array}

In order to determine whether the true proportions in each response category differ, a one-way chi-square analysis should be conducted. When calculating the test statistic, what values for the expected counts should be used in the calculation?

A) E1=33.33,E2=33.33,E3=33.33\mathrm { E } _ { 1 } = 33.33 , \mathrm { E } _ { 2 } = 33.33 , \mathrm { E } _ { 3 } = 33.33
B) E1=46,E2=44,E3=9\mathrm { E } _ { 1 } = 46 , \mathrm { E } _ { 2 } = 44 , \mathrm { E } _ { 3 } = 9
C) E1=100,E2=100,E3=100\mathrm { E } _ { 1 } = 100 , \mathrm { E } _ { 2 } = 100 , \mathrm { E } _ { 3 } = 100
D) E1=0.45,E2=0.46,E3=0.09\mathrm { E } _ { 1 } = 0.45 , \mathrm { E } _ { 2 } = 0.46 , \mathrm { E } _ { 3 } = 0.09
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10
Economists at USF are researching the problem of absenteeism at U.S. firms. A random sample of 100 U.S. organizations was selected to participate in a 1-year study. As part of the study, the economists had collected data on the following two variables for each company: shiftwork available (Yes or No), and union-management relationship (Good or Poor). As part of their analyses, the economists wanted to determine whether or not a company makes shiftwork available depends on the relationship between union and management. The collected data are
Shown below:

 Relation  Shiftwork  Good  Bad  Total  No 112233 Yes 254267 Total 3664100\begin{array}{l}\begin{array}{lccc}&\text { Relation }\\\text { Shiftwork } & \text { Good } & \text { Bad } & \text { Total } \\\text { No } & 11 & 22 & 33 \\\text { Yes } & 25 & 42 & 67 \\\text { Total } & 36 & 64 & 100\end{array}\end{array}

In order to determine whether the shiftwork responses depend on the relationship responses, a two-way chi-square analysis should be conducted. Calculate the value of the test statistic for the desired analysis.

A) χ2=0.70\chi ^ { 2 } = 0.70
B) χ2=0.15\chi ^ { 2 } = 0.15
C) χ2=11.88\chi ^ { 2 } = 11.88
D) χ2=0.07\chi ^ { 2 } = 0.07
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11
Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table.  RESPONSE  NUMBER  Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50\begin{array} { l | l } \text { RESPONSE } & \text { NUMBER } \\\hline \text { Agree Strongly } & 60 \\\text { Agree Somewhat } & 110 \\\text { Disagree Somewhat } & 80 \\\text { Disagree Strongly } & 50\end{array} In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. As part of that analysis, a 95% confidence interval for the multinomial probability associated with the "Agree Strongly" response was desired. Which of the following confidence intervals should be used?

A) (0.141, 0.259)
B) (0.162, 0.238)
C) (0.145, 0.255)
D) (0.155, 0.245)
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12
A survey of entrepreneurs focused on their job characteristics, work habits, social activities, leisure time, etc. One question put to each entrepreneur was, "What make of car (U.S., Europe, or Japan) do you drive?" The responses (number in each category) for a sample of 100 entrepreneurs are summarized below. The goal of the analysis is to determine if the proportions of entrepreneurs who drive American, European, and Japanese cars differ.

 U.S.  Europe  Japan 403525\begin{array} { c c c } \text { U.S. } & \text { Europe } & \text { Japan } \\\hline 40 & 35 & 25 \\\hline\end{array}
In order to determine whether the true proportions in each response category differ, a one-way chi-square analysis should be conducted. As part of that analysis, a 95% confidence interval for the multinomial probability associated with the "Europe" response was desired. Which of the following
Confidence intervals should be used?

A) (0.257, 0.443)
B) (0.271, 0.428)
C) (0.265, 0.440)
D) (0.227, 0.473)
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13
An adverse drug effect (ADE) is an unintended injury caused by prescribed medication. The table summarizes the proximal cause of 95 ADEs that resulted from a dosing error at a Boston hospital. An adverse drug effect (ADE) is an unintended injury caused by prescribed medication. The table summarizes the proximal cause of 95 ADEs that resulted from a dosing error at a Boston hospital.
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14
Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table.
 RESPONSE  NUMBER  Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50\begin{array}{l|l}\text { RESPONSE } & \text { NUMBER } \\\hline \text { Agree Strongly } & 60 \\\text { Agree Somewhat } & 110 \\\text { Disagree Somewhat } & 80 \\\text { Disagree Strongly } & 50\end{array}

In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. Calculate the value of the test statistic for the desired analysis.

A) χ2=28.0\chi ^ { 2 } = 28.0
B) χ2=0.25\chi ^ { 2 } = 0.25
C) χ2=75\chi ^ { 2 } = 75
D) χ2=22.54\chi ^ { 2 } = 22.54
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15
A survey of entrepreneurs focused on their job characteristics, work habits, social activities, leisure time, etc. One question put to each entrepreneur was, "What make of car (U.S., Europe, or Japan)do you drive?" The responses (number in each category) for a sample of 100 entrepreneurs are summarized below. The goal of the analysis is to determine if the proportions of entrepreneurs Who drive American, European, and Japanese cars differ.
 U.S.  Europe  Japan 403525\begin{array}{ccc}\text { U.S. } & \text { Europe } & \text { Japan } \\\hline 40 & 35 & 25 \\\hline\end{array}

In order to determine whether the true proportions in each response category differ, a one-way chi-square analysis should be conducted. Suppose the p-value for the test was calculated to be p=0.1738\mathrm { p } = 0.1738 . What is the appropriate conclusion to make when testing at α=0.10\alpha = 0.10 ?

A) There is sufficient evidence to indicate the proportion of entrepreneurs driving Japanese cars is less than the proportion driving U.S. cars.
B) There is sufficient evidence to indicate the proportion of entrepreneurs driving the three makes of car differ.
C) There is sufficient evidence to indicate the proportion of entrepreneurs driving the three makes of car are equal.
D) There is insufficient evidence to indicate the proportion of entrepreneurs driving the three makes of car differ.
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16
A drug company developed a honey-based liquid medicine designed to calm a child's cough at night. To test the drug, 105 children who were ill with an upper respiratory tract infection were randomly selected to participate in a clinical trial. The children were randomly divided into three groups - one group was given a dosage of the honey drug, the second was given a dosage of liquid DM (an over-the-counter cough medicine), and the third (control group) received a liquid placebo (no dosage at all). After administering the medicine to their coughing child, parents rated their children's cough diagnosis as either better or worse. The results are shown in the table below:
 Diagnosis Treatment  Better  Worse  Total  Control 43337 DM 122133 Honey 241135 Total 4065105\begin{array} { l c c c } &\quad\quad\text { \quad Diagnosis}\\\text { Treatment } & \text { Better } & \text { Worse } & \text { Total } \\ \text { Control } & 4 & 33 & 37 \\ \text { DM } & 12 & 21 & 33 \\ \text { Honey } & 24 & 11 & 35 \\ \text { Total } & 40 & 65 & 105 \end{array}
In order to determine whether the treatment group is independent of the coughing diagnosis, a two-way chi-square test was conducted. Suppose the pp -value for the test was calculated to be p=p = 0.00160.0016 . What is the appropriate conclusion to make when testing at α=0.05\alpha = 0.05 ?

A) There is insufficient evidence to indicate the treatment group is dependent on the coughing diagnosis.
B) There is sufficient evidence to indicate the treatment group is independent of the coughing diagnosis.
C) There is insufficient evidence to indicate the treatment group is independent of the coughing diagnosis.
D) There is sufficient evidence to indicate the treatment group is dependent on the coughing diagnosis.
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17
A business professor conducted a campus survey to estimate demand among all students for a protein supplement for smoothies and other nutritional drinks. Each of 113 students, randomly
Selected from all students on campus, provided the following information:
(1) How health conscious are you? (Very, Moderately, Slightly, Not very)
(2) Do you prefer protein supplements in your smoothies? (Yes, No)
As part of his analysis, the professor claims that whether or not the student prefers a protein
Supplement in smoothies is independent of health consciousness level (Very, Moderate, Slightly, or
Not very). Identify the appropriate alternative hypothesis that the professor should use in the test
Of hypothesis he desires.

A) HA\mathrm { H } _ { \mathrm { A } } : Preference and Health Consciousness level are independent variables.
B) HA\mathrm { H } _ { \mathrm { A } } : There is interaction between the Preference and Health Consciousness variables.
C) HA\mathrm { H } _ { \mathrm { A } } : Preference and Health Consciousness level are dependent variables.
D) HA\mathrm { H } _ { \mathrm { A } } : Preference and Health Consciousness level are mutually exclusive variables.
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18
Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table.
 RESPONSE  NUMBER  Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50\begin{array}{l|l}\text { RESPONSE } & \text { NUMBER } \\\hline \text { Agree Strongly } & 60 \\\text { Agree Somewhat } & 110 \\\text { Disagree Somewhat } & 80 \\\text { Disagree Strongly } & 50\end{array}

In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. When calculating the test statistic, what values for the expected counts should be used in the calculation?

A) E1=75,E2=75,E3=75,E4=75\mathrm { E } _ { 1 } = 75 , \mathrm { E } _ { 2 } = 75 , \mathrm { E } _ { 3 } = 75 , \mathrm { E } _ { 4 } = 75
B) E1=300,E2=300,E3=300,E4=300\mathrm { E } _ { 1 } = 300 , \mathrm { E } _ { 2 } = 300 , \mathrm { E } _ { 3 } = 300 , \mathrm { E } _ { 4 } = 300
C) E1=0.25,E2=0.25,E3=0.25,E4=0.25\mathrm { E } _ { 1 } = 0.25 , \mathrm { E } _ { 2 } = 0.25 , \mathrm { E } _ { 3 } = 0.25 , \mathrm { E } _ { 4 } = 0.25
D) E1=60,E2=110,E3=80,E4=50\mathrm { E } _ { 1 } = 60 , \mathrm { E } _ { 2 } = 110 , \mathrm { E } _ { 3 } = 80 , \mathrm { E } _ { 4 } = 50
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19
Economists at USF are researching the problem of absenteeism at U.S. firms. A random sample of 100 U.S. organizations was selected to participate in a 1-year study. As part of the study, the economists had collected data on the following two variables for each company: shiftwork available (Yes or No), and union-management relationship (Good or Poor). As part of their analyses, the economists wanted to determine whether or not a company makes shiftwork available depends on the relationship between union and management. The collected data are
Shown below:


 Relation Shiftwork  Good  Bad  Total  No 112233 Yes 254267 Total 3664100\begin{array} { l c c c }&\text { Relation}\\ \text { Shiftwork } & \text { Good } & \text { Bad } & \text { Total } \\ \text { No } & 11 & 22 & 33 \\ \text { Yes } & 25 & 42 & 67 \\ \text { Total } & 36 & 64 & 100 \end{array}

In order to determine whether the shiftwork responses depend on the relationship responses, a two-way chi-square analysis should be conducted. When calculating the test statistic, what values for the expected counts should be used in the calculation?

A) E11=36,E12=33,E21=64,E22=67\mathrm { E } _ { 11 } = 36 , \mathrm { E } _ { 12 } = 33 , \mathrm { E } _ { 21 } = 64 , \mathrm { E } _ { 22 } = 67
B) E11=0.065,E12=0.036,E21=0.032,E22=0.018\mathrm { E } _ { 11 } = 0.065 , \mathrm { E } _ { 12 } = 0.036 , \mathrm { E } _ { 21 } = 0.032 , \mathrm { E } _ { 22 } = 0.018
C) E11=11.88,E12=21.12,E21=24.12,E22=42.88\mathrm { E } _ { 11 } = 11.88 , \mathrm { E } _ { 12 } = 21.12 , \mathrm { E } _ { 21 } = 24.12 , \mathrm { E } _ { 22 } = 42.88
D) E11=11,E12=22,E21=25,E22=42\mathrm { E } _ { 11 } = 11 , \mathrm { E } _ { 12 } = 22 , \mathrm { E } _ { 21 } = 25 , \mathrm { E } _ { 22 } = 42
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20
Economists at USF are researching the problem of absenteeism at U.S. firms. A random sample of 100 U.S. organizations was selected to participate in a 1-year study. As part of the study, the economists had collected data on the following two variables for each company: shiftwork available (Yes or No), and union-management relationship (Good or Poor). As part of their analyses, the economists wanted to determine whether or not a company makes shiftwork available depends on the relationship between union and management. The collected data are
Shown below:

 Relation  Shiftwork  Good  Bad  Total  No 112233 Yes 254267 Total 3664100\begin{array}{lccc} & {\text { Relation }} & \\\text { Shiftwork } & \text { Good } & \text { Bad } & \text { Total } \\\text { No } & 11 & 22 & 33 \\\text { Yes } & 25 & 42 & 67 \\\text { Total } & 36 & 64 & 100\end{array}

Use the chi-square distribution to determine the rejection region for this test when testing at α=\alpha = 0.050.05 .

A) Reject H0\mathrm { H } _ { 0 } if χ2>7.81473\chi ^ { 2 } > 7.81473
B) Reject H0\mathrm { H } _ { 0 } if χ2>3.84146\chi ^ { 2 } > 3.84146
C) Reject H0\mathrm { H } _ { 0 } if χ2>5.02389\chi ^ { 2 } > 5.02389
D) Reject H0\mathrm { H } _ { 0 } if χ2>5.99147\chi ^ { 2 } > 5.99147
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21
The contingency table below shows the results of a random sample of 200 state representatives that was conducted to see whether their opinions on a bill are related to their party affiliations.
 Opinion  Party  Approve  Disapprove  No Opinion  Republican 422014 Democrat 502418 Independent 10166 Test the claim of independence. Use α=.05\begin{array}{l}\begin{array} { l | c c c } & { \text { Opinion } } \\\hline \text { Party } & \text { Approve } & \text { Disapprove } & \text { No Opinion } \\\text { Republican } & 42 & 20 & 14 \\\text { Democrat } & 50 & 24 & 18 \\\text { Independent } & 10 & 16 & 6\end{array}\\\text { Test the claim of independence. Use } \alpha = .05 \text {. }\end{array}
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22
A sports researcher is interested in determining if there is a relationship between the number of home team and visiting team wins and different sports. A random sample of 526 games is selected and the results are given below. Find the rejection region used to test the claim that the number of home team and visiting team wins is independent of the sport. Use α=0.01\alpha = 0.01

 Football  Basketball  Soccer  Baseball  Home team wins 391562583 Visiting team wins 31981975\begin{array}{l|cccc} & \text { Football } & \text { Basketball } & \text { Soccer } & \text { Baseball } \\\hline \text { Home team wins } & 39 & 156 & 25 & 83 \\\text { Visiting team wins } & 31 & 98 & 19 & 75\end{array}

A) ?2>12.838? ^ { 2 } > 12.838
B) ?2>9.348? ^ { 2 } > 9.348
C) ?2>11.345? ^ { 2 } > 11.345
D) ?2>7.815? ^ { 2 } > 7.815
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23
A professor chose a random sample of 50 recent graduates of an MBA program and recorded the gender of each graduate (M or F) and whether the graduate chose to complete his or her degree requirements by completing a research project (RP) or by taking comprehensive exams (CE). The results are shown below.

 M, CE  M, RP  F, RP  M, CE  M, CE  F, CE  F, RP,  M, CE  M, RP  F, RP  F, CE  M, RP  F, RP  F, CE  M, RP  F, RP  M, RP  F, CE  M, CE  M, CE  M, RP  F, CE  F, RP  M, CE  M, CE  F, CE  M, RP  F, RP  M,CE  M, CE  F, CE  F, RP,  M, CE  M, RP  F, RP  M, CE  M, RP  F, RP  F, CE  M, RP  F, RP  M, RP  M, CE  M,CE  M, CE  M, RP  F, CE  F, RP  M,CE  F, CE \begin{array} { l l l l l } \text { M, CE } & \text { M, RP } & \text { F, RP } & \text { M, CE } & \text { M, CE } \\ \text { F, CE } & \text { F, RP, } & \text { M, CE } & \text { M, RP } & \text { F, RP } \\ \text { F, CE } & \text { M, RP } & \text { F, RP } & \text { F, CE } & \text { M, RP } \\ \text { F, RP } & \text { M, RP } & \text { F, CE } & \text { M, CE } & \text { M, CE } \\ \text { M, RP } & \text { F, CE } & \text { F, RP } & \text { M, CE } & \text { M, CE } \\ \text { F, CE } & \text { M, RP } & \text { F, RP } & \text { M,CE } & \text { M, CE } \\ \text { F, CE } & \text { F, RP, } & \text { M, CE } & \text { M, RP } & \text { F, RP } \\ \text { M, CE } & \text { M, RP } & \text { F, RP } & \text { F, CE } & \text { M, RP } \\ \text { F, RP } & \text { M, RP } & \text { M, CE } & \text { M,CE } & \text { M, CE } \\ \text { M, RP } & \text { F, CE } & \text { F, RP } & \text { M,CE } & \text { F, CE } \end{array}

a. Create a contingency table for the data.
b. Perform a χ2\chi ^ { 2 } -test to determine if there is any evidence that gender and choice of research project or comprehensive exams are not independent. Use α=0.05\alpha = 0.05 .
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24
A new coffeehouse wishes to see whether customers have any preference among 5 different brands of coffee. A sample of 200 customers provided the data below. Test the claim that the probabilities show no preference. Us α=0.01\alpha = 0.01  Brand 12345 Customers 1832556530\begin{array} { l | c c c c c } \text { Brand } & 1 & 2 & 3 & 4 & 5 \\\hline \text { Customers } & 18 & 32 & 55 & 65 & 30\end{array}
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25
What is categorical data?
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26
A sports researcher is interested in determining if there is a relationship between the number of home team and visiting team wins and different sports. A random sample of 526 games is selected and the results are given below. Assuming the row and column classifications are independent, find an estimate for the expected cell count of cell E22E _ { 22 }
 Football  Basketball  Soccer  Baseball  Home team wins 381542784 Visiting team wins 31982272\begin{array}{l|llll} & \text { Football } & \text { Basketball } & \text { Soccer } & \text { Baseball } \\\hline \text { Home team wins } & 38 & 154 & 27 & 84 \\\text { Visiting team wins } & 31 & 98 & 22 & 72\end{array}

A) 106.8106.8
B) 20.820.8
C) 28.228.2
D) 145.2145.2
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27
  Construct a 99% confidence interval for the multinomial probability associated with cell 2. Construct a 99% confidence interval for the multinomial probability associated with cell 2.
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28
Consider the accompanying contingency table.  Column  Row 12311315162192611\begin{array} { c | c c c } & { \text { Column } } \\\text { Row } & 1 & 2 & 3 \\\hline 1 & 13 & 15 & 16 \\2 & 19 & 26 & 11\end{array} a. Convert the values in row 1 to percentages by calculating the percentage of each column total falling in row 1.
b. Create a bar graph with row 1 percentage on the vertical axis and column number on the horizontal axis.
c. What pattern do you expect to see if the rows and columns are not independent? Is this pattern present in your graph?
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29
The null hypothesis for a test of data resulting from a multinomial experiment is given as The null hypothesis for a test of data resulting from a multinomial experiment is given as   What is the alternative hypothesis for the test? What is the alternative hypothesis for the test?
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30
Many track runners believe that they have a better chance of winning if they start in the inside lane that is closest to the field. For the data below, the lane closest to the field is Lane 1, the next lane is Lane 2, and so on until the outermost lane, Lane 6. The table displays the starting positions for the winners of 240 competitions. Test the claim that the probability of winning is the same regardless of starting position. Use α=0.05\alpha = 0.05 . The results are based on 240 wins.
 Starting Position 123456 Number of Wins 363344503245\begin{array} { l | c c c c c c } \text { Starting Position } & 1 & 2 & 3 & 4 & 5 & 6 \\\hline \text { Number of Wins } & 36 & 33 & 44 & 50 & 32 & 45\end{array}
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31
When using any procedure to perform a hypothesis test, the user should always be certain that the
experiment satisfies the assumptions given with the procedure.
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32
A random sample of 160 car accidents are selected and categorized by the age of the driver determined to be at fault. The results are listed below. The age distribution of drivers for the given categories is 18% for the under 26 group, 39% for the 26-45 group, 31% for the 46-65 group, and 12% for the group over 65. Test the claim that all ages have crash rates proportional to their number of drivers. Use α=0.05\alpha = 0.05  Age  Under 26 26454665 Over 65 Drivers 66392530\begin{array} { l | c c c c } \text { Age } & \text { Under 26 } & 26 - 45 & 46 - 65 & \text { Over } 65 \\\hline \text { Drivers } & 66 & 39 & 25 & 30\end{array}
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33
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34
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35
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36
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37
Describe probabilities of the k outcomes of the multinomial experiment trials.
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38
A teacher finds that final grades in the statistics department are distributed as: A, 25%; B, 25%; C, 40%; D, 5%; F, 5%. At the end of a randomly selected semester, the following grades were recorded. Determine if the grade distribution for the department is different than expected. Use α=0.01\alpha = 0.01  Grade  A  B  C  D  F  Number 364260148\begin{array} { l | c c c c c } \text { Grade } & \text { A } & \text { B } & \text { C } & \text { D } & \text { F } \\\hline \text { Number } & 36 & 42 & 60 & 14 & 8\end{array}
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39
A sports researcher is interested in determining if there is a relationship between the
number of home team and visiting team wins and different sports. A random sample of
526 games is selected and the results are given below. Test the claim that the number of
home team and visiting team wins is independent of the sport. Use A sports researcher is interested in determining if there is a relationship between the number of home team and visiting team wins and different sports. A random sample of 526 games is selected and the results are given below. Test the claim that the number of home team and visiting team wins is independent of the sport. Use    A sports researcher is interested in determining if there is a relationship between the number of home team and visiting team wins and different sports. A random sample of 526 games is selected and the results are given below. Test the claim that the number of home team and visiting team wins is independent of the sport. Use
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40
What are characteristics of the trials in a multinomial experiment?
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41
<strong> </strong> A) 48.91 B) 37.45 C) 55.63 D) 45.91

A) 48.91
B) 37.45
C) 55.63
D) 45.91
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42
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43
The contingency table below shows the results of a random sample of 200 state representatives that was conducted to see whether their opinions on a bill are related to their party affiliations.
Assuming the row and column classifications are independent, find an estimate for the expected Cell count E22E _ { 22 }  Opinion  Party  Approve  Disapprove  No Opinion  Republican 422014 Democrat 502418 Independent 10166\begin{array} { l | c c c } & { \text { Opinion } } \\\hline \text { Party } & \text { Approve } & \text { Disapprove } & \text { No Opinion } \\\text { Republican } & 42 & 20 & 14 \\\text { Democrat } & 50 & 24 & 18 \\\text { Independent } & 10 & 16 & 6\end{array}

A) 17.48
B) 27.6
C) 22.8
D) 46.92
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44
The contingency table below shows the results of a random sample of 200 state representatives that was conducted to see whether their opinions on a bill are related to their party affiliation. <strong>The contingency table below shows the results of a random sample of 200 state representatives that was conducted to see whether their opinions on a bill are related to their party affiliation.  </strong> A) 7.662 B) 11.765 C) 9.483 D) 8.030

A) 7.662
B) 11.765
C) 9.483
D) 8.030
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45
A random sample of 160 car accidents are selected and categorized by the age of the driver determined to be at fault. The results are listed below. The age distribution of drivers for the giver categories is 18%18 \% for the under 26 group, 39%39 \% for the 264526 - 45 group, 31%31 \% for the 45 - 65 group, and 12%12 \% for the group over 65 . Find the rejection region used to test the claim that all ages have crash rates proportional to their number of drivers. Use α=0.05\alpha = 0.05 .

 Age  Under 26 26454665 Over 65 Drivers 66392530\begin{array}{l|cccc}\text { Age } & \text { Under 26 } & 26-45 & 46-65 & \text { Over } 65 \\\hline \text { Drivers } & 66 & 39 & 25 & 30\end{array}

A) x2>7.815x ^ { 2 } > 7.815
B) χ2>11.143\chi ^ { 2 } > 11.143
C) x2>6.251x ^ { 2 } > 6.251
D) χ2>9.348\chi ^ { 2 } > 9.348
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46
Use the appropriate table to find the following probabili <strong>Use the appropriate table to find the following probabili  </strong> A) 0.995 B) 0.005 C) 0.010 D) 0.990

A) 0.995
B) 0.005
C) 0.010
D) 0.990
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47
<strong> </strong> A) 2.919 B) 4.192 C) 3.290 D) 5.391

A) 2.919
B) 4.192
C) 3.290
D) 5.391
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48
Many track runners believe that they have a better chance of winning if they start in the inside lane that is closest to the field. For the data below, the lane closest to the field is Lane 1 , the next lane is Lane 2, and so on until the outermost lane, Lane 6. The table displays the starting positions for the winners of 240 competitions. Find the rejection region used to test the claim that the probability of winning is the same regardless of starting position. Use α=0.05\alpha = 0.05 . The results are based on 240 wins.
 Starting Position 123456 Number of Wins 453236445033\begin{array}{l|cccccc}\text { Starting Position } & 1 & 2 & 3 & 4 & 5 & 6 \\\hline \text { Number of Wins } & 45 & 32 & 36 & 44 & 50 & 33\end{array}

A) χ2>15.086\chi ^ { 2 } > 15.086
B) χ2>11.070\chi ^ { 2 } > 11.070
C) χ2>12.833\chi ^ { 2 } > 12.833
D) χ2>9.236\chi ^ { 2 } > 9.236
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49
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50
The contingency table below shows the results of a random sample of 200 state representatives that was conducted to see whether their opinions on a bill are related to their party affiliations. Use α=\alpha = 0.050.05 .

 Opinion  Party  Approve  Disapprove  No Opinion  Republican 422014 Democrat 502418 Independent 10166\begin{array}{l|ccc} &{\text { Opinion }} \\\hline \text { Party } & \text { Approve } & \text { Disapprove } & \text { No Opinion } \\\text { Republican } & 42 & 20 & 14 \\\text { Democrat } & 50 & 24 & 18 \\\text { Independent } & 10 & 16 & 6\end{array}

Find the rejection region used to test the claim of independence.

A) χ2>7.779\chi ^ { 2 } > 7.779
B) χ2>11.143\chi ^ { 2 } > 11.143
C) χ2>13.277\chi ^ { 2 } > 13.277
D) χ2>9.488\chi ^ { 2 } > 9.488
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51
A teacher finds that final grades in the statistics department are distributed as: A, 25\%; B, 25\%; C, 40%;D,5%;F,5%40 \% ; \mathrm { D } , 5 \% ; \mathrm { F } , 5 \% . At the end of a randomly selected semester, the following grades were recorded. Calculate the chi-square test statistic χ2\chi ^ { 2 } used to determine if the grade distribution for the department is different than expected. Use α=0.01\alpha = 0.01 .
 Grade  A  B  C  D  F  Number 423660814\begin{array}{l|lllll}\text { Grade } & \text { A } & \text { B } & \text { C } & \text { D } & \text { F } \\\hline \text { Number } & 42 & 36 & 60 & 8 & 14\end{array}

A) 6.87
B) 5.25
C) 3.41
D) 4.82
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52
In a test of independence, it is safe to conclude that the events are independent when the value of ?2? ^ { 2 } is very small.
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53
A multinomial experiment with k=4k = 4 cells and n=400n = 400 produced the data shown in the following table.
 Cell 1234ni5021310433\begin{array} { r | r c c r } &\text { Cell }\\& 1 & 2 & 3 & 4 \\\hline n _ { i } & 50 & 213 & 104 & 33\end{array}

Previous studies in this area have shown that p1=p2=p3=p4=.25p _ { 1 } = p _ { 2 } = p _ { 3 } = p _ { 4 } = .25 . Construct a 95%95 \% confidence interval for the multinomial probability associated with cell 2

A) (0.491, 0.574)
B) (0.490, 0.575)
C) (0.093, 0.157)
D) (0.484, 0.581)
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54
The ?2? ^ { 2 } test for independence is a useful tool for establishing a causal relationship between two factors.
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55
Use the appropriate table to find the following  chi-square value: χ.052 for df=3\text { chi-square value: } \chi _ { .05 } ^ { 2 } \text { for } \mathrm { df } = 3

A) 7.815
B) 5.991
C) 9.488
D) 0.352
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56
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57
The sampling distribution for ?² works well when expected counts are very small.
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58
A random sample of 160 car accidents are selected and categorized by the age of the driver determined to be at fault. The results are listed below. The age distribution of drivers for the given categories is 18%18 \% for the under 26 group, 39%39 \% for the 26-45 group, 31%31 \% for the 456545 - 65 group, and 12%12 \% for the group over 65 . Calculate the chi-square test statistic χ2\chi ^ { 2 } used to test the claim that all ages have crash rates proportional to their driving rates.

 Age  Under 26 26454665 Over 65 Drivers 66392530\begin{array}{l|cccc}\text { Age } & \text { Under 26 } & 26-45 & 46-65 & \text { Over } 65 \\\hline \text { Drivers } & 66 & 39 & 25 & 30\end{array}

A) 85.123
B) 95.431
C) 75.101
D) 101.324
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59
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