Deck 11: Chi-Square and Analysis of Variance

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Question
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.

-Given the sample data below, test the claim that the populations have the same mean. Use a significance level of 0.05.

 Brand A Brand B Brand C \hlinen=10n=10n=10xˉ=32.9xˉ=31.6xˉ=27.1s2=3.16s2=3.04s2=4.66\begin{array} { l } \text { Brand A}&\text { Brand B}&\text { Brand C }\\\hlinen=10 & n=10 & n=10 \\\bar{x}=32.9 & \bar{x}=31.6 & \bar{x}=27.1 \\s^{2}=3.16 & s^{2}=3.04 & s^{2}=4.66\end{array}
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Question
The test statistic  one-way ANOVA is F= variance between samples  variance within samples \text { one-way } \mathrm { ANOVA } \text { is } \mathrm { F } = \frac { \text { variance between samples } } { \text { variance within samples } } . Describe variance within samples and variance between samples. What relationship between variance within samples and variance between samples would result in the conclusion that the value of F is significant?
Question
A researcher wishes to test the effectiveness of a flu vaccination. 150 people are vaccinated, 180 people are vaccinated with a placebo, and 100 people are not vaccinated. The number in each group who later caught the flu was recorded. The results are shown below.
 Vaccinated  Placebo  Control  Caught the flu 81921 Did not catch the flu 14216179\begin{array} { r | r r r } & \text { Vaccinated } & \text { Placebo } & \text { Control } \\\hline \text { Caught the flu } & 8 & 19 & 21 \\\text { Did not catch the flu } & 142 & 161 & 79\end{array}
Use a 0.05 significance level to test the claim that the proportion of people catching the flu is the same in all three groups.
Question
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.

-A consumer magazine wants to compare the lifetimes of ballpoint pens of three different types. The magazine takes a random sample of pens of each type in the following table.

 Brand A Brand B Brand C 260181238218240257184162241219218213\begin{array} { l } \text { Brand A}&\text { Brand B}&\text { Brand C }\\\hline260&181&238\\218&240&257\\184&162&241\\219&218&213\end{array}
Do the data indicate that there is a difference in mean lifetime for the three brands of ballpoint pens?  Use α=0.01\text { Use } \alpha = 0.01 \text {. }
Question
Define categorical data and give an example.
Question
Use a X² test to test the claim that in the given contingency table, the row variable and the column variable are independent.

-The table below shows the age and favorite type of music of 668 randomly selected people.  Rock  Pop  Classical 152550857325356891603545907477\begin{array} { l | r r r } & \text { Rock } & \text { Pop } & \text { Classical } \\\hline 15 - 25 & 50 & 85 & 73 \\25 - 35 & 68 & 91 & 60 \\35 - 45 & 90 & 74 & 77\end{array}
Use a 5 percent level of significance to test the null hypothesis that age and preferred music type are independent.
Question
The following table shows the number of employees who called in sick at a business for different days of a particular week.  Day  Sun  Mon  Tues  Wed  Thurs  Fri  Sat  Number sick 81271191112\begin{array} { l | c c c c c c c } \text { Day } & \text { Sun } & \text { Mon } & \text { Tues } & \text { Wed } & \text { Thurs } & \text { Fri } & \text { Sat } \\\hline \text { Number sick } & 8 & 12 & 7 & 11 & 9 & 11 & 12\end{array}
i) At the 0.05 level of significance, test the claim that sick days occur with equal frequency on the different days of the week.
ii) Test the claim after changing the frequency for Saturday to 152. Describe the effect of this outlier on the test.
Question
Use the data given below to verify that the t test for independent samples and the ANOVA method are equivalent.
AB101929181127193015181621\begin{array} { c | c } \mathrm { A } & \mathrm { B } \\\hline 10 & 19 \\29 & 18 \\11 & 27 \\19 & 30 \\15 & 18 \\16 & 21\end{array}
i) Use a t test with a 0.05 significance level to test the claim that the two samples come from populations with the same means.
ii) Use the ANOVA method with a 0.05 significance level to test the same claim.
iii) Verify that the squares of the t test statistic and the critical value are equal to the F test statistic and critical value.
Question
At the 0.025 significance level, test the claim that the four brands have the same mean if the following sample results have been obtained.
 Brand A Brand B Brand C  Brand D 1718212220182425212325272225262921262935293637\begin{array} { l } \text { Brand A}&\text { Brand B}&\text { Brand C }&\text { Brand D }\\\hline17&18&21&22\\20&18&24&25\\21&23&25&27\\22&25&26&29\\21&26&29&35\\&&29&36\\&&&37\end{array}
Question
According to Benford's Law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. Test for goodness-of-fit with Benford's Law.  Leading Digit 123456789 Benford’s law:  distribution of  leading digits 30.1%17.6%12.5%9.7%7.9%6.7%5.8%5.1%4.6%\begin{array} { | l | c | c | c | c | c | c | c | c | c | } \hline \text { Leading Digit } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\\hline \begin{array} { l } \text { Benford's law: } \\\text { distribution of } \\\text { leading digits }\end{array} & 30.1 \% & 17.6 \% & 12.5 \% & 9.7 \% & 7.9 \% & 6.7 \% & 5.8 \% & 5.1 \% & 4.6 \% \\\hline\end{array}

-When working for the Brooklyn District Attorney, investigator Robert Burton analyzed the leading digits of the amounts from 784 checks issued by seven suspect companies. The frequencies were found to be 0, 9, 0, 70, 485, 189, 8, 23, and 0, and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. If the observed frequencies are substantially different from the frequencies expected with Benford's Law, the check amounts appear to result from fraud. Use a 0.05 significance level to test for goodness-of-fit with Benford's Law. Does it appear that the checks are the result of fraud?
Question
A researcher wishes to test whether the proportion of college students who smoke is the same in four different colleges. She randomly selects 100 students from each college and records the number that smoke. The results are shown below.
 College A  College B  College C  College D  Smoke 17261134 Don’t smoke 83748966\begin{array} { r | r r r r } & \text { College A } & \text { College B } & \text { College C } & \text { College D } \\\hline \text { Smoke } & 17 & 26 & 11 & 34 \\\text { Don't smoke } & 83 & 74 & 89 & 66\end{array}
Use a 0.01 significance level to test the claim that the proportion of students smoking is the same at all four colleges.
Question
Four independent samples of 100 values each are randomly drawn from populations that are normally distributed with equal variances. You wish to test the claim that μ1=μ2=μ3\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } =μ4= \mu _ { 4 }
i) If you test the individual claims μ1=μ2,μ1=μ3,μ1=μ4,,μ3=μ4\mu _ { 1 } = \mu _ { 2 } , \mu _ { 1 } = \mu _ { 3 } , \mu _ { 1 } = \mu _ { 4 } , \ldots , \mu _ { 3 } = \mu _ { 4 } , how many ways can you pair off the 4 means?
ii) Assume that the tests are independent and that for each test of equality between two means, there is a 0.990.99 probability of not making a type I error. If all possible pairs of means are tested for equality, what is the probability of making no type I errors?
iii) If you use analysis of variance to test the claim that μ1=μ2=μ3=μ4\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } = \mu _ { 4 } at the 0.010.01 level of significance, what is the probability of not making a type I error?
Question
Use a significance level of 0.01 to test the claim that workplace accidents are distributed on workdays as follows: Monday 25%, Tuesday: 15%, Wednesday: 15%, Thursday: 15%, and Friday: 30%.
In a study of 100 workplace accidents, 26 occurred on a Monday, 15 occurred on a Tuesday, 17 occurred on a Wednesday, 17 occurred on a Thursday, and 25 occurred on a Friday.
Question
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.

-The data below represent the weight losses for people on three different exercise programs.
Exercise A  Exercise B Exercise C2.55.84.38.84.96.27.31.15.89.87.88.15.11.27.9\begin{array} { l } \text {Exercise A }& \text { Exercise B}& \text { Exercise C}\\\hline2.5&5.8&4.3\\8.8&4.9&6.2\\7.3&1.1&5.8\\9.8&7.8&8.1\\5.1&1.2&7.9\end{array}

At the 1% significance level, does it appear that a difference exists in the true mean weight loss produced by the three exercise programs?
Question
Suppose you are to test for equality of four different population means, with H0: µA = µB =μC=μD= \mu _ { C } = \mu _ { D } . Write the hypotheses for the paired tests. Use methods of probability to explain why the process of ANOVA has a higher degree of confidence than testing each of the pairs separately.
Question
An observed frequency distribution of exam scores is as follows:
Exam Score under 60 60697079808990100 Frequency30301406040\begin{array} {| l|ccccc| } \hline \text {Exam Score }&\text {under 60 }&60-69&70-79&80-89&90-100\\\hline \text { Frequency}&30&30&140&60&40\\\hline\end{array}


i) Assuming a normal distribution with μ=75\mu = 75 and σ=15\sigma = 15 , find the probability of a randomly selected subject belonging to each class. (Use boundaries of 59.5,69.5,79.5,89.559.5,69.5,79.5,89.5 , 100.)
ii) Using the probabilities found in part (i), find the expected frequency for each category.
iii) Use a 0.050.05 significance level to test the claim that the exam scores were randomly selected from a normally distributed population with μ=75\mu = 75 and σ=15\sigma = 15 .
Question
A survey conducted in a small business yielded the results shown in the table.  Men Women  Health insurance 4122 No health insurance 3424\begin{array} { l | c c } & \text { Men}&\text { Women } \\\hline \text { Health insurance } & 41 & 22 \\\hline \text { No health insurance } & 34 & 24\end{array} i) Test the claim that health care coverage is independent of gender. Use a 0.05
significance level.
ii) Using Yates' correction, replace (OE)2E with (OE0.5)2E\sum \frac{(O-E)^{2}}{E}\text { with } \sum \frac{(|O-E|-0.5)^{2}}{E} and repeat the test.
What effect does Yates' correction have on the value of the test statistic?
Question
Use a χ2\chi ^ { 2 } test to test the claim that in the given contingency table, the row variable and the column variable are independent.

-Responses to a survey question are broken down according to employment status and the sample results are given below. At the 0.10 significance level, test the claim that response and employment status are independent.  Yes  No  Undecided  Employed 30155 Unemployed 202510\begin{array} { r | r r r } & \text { Yes } & \text { No } & \text { Undecided } \\\hline \text { Employed } & 30 & 15 & 5 \\\text { Unemployed } & 20 & 25 & 10\end{array}
Question
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.

-At the 0.025 significance level, test the claim that the three brands have the same mean if the following sample results have been obtained.  Brand ABrand B Brand C 322722342425373332333022362139\begin{array} { l } \text { Brand A}& \text {Brand B }& \text {Brand C }\\\hline32&27&22\\34&24&25\\37&33&32\\33&30&22\\36&&21\\39\end{array}
Question
Use a χ2\chi ^ { 2 } test to test the claim that in the given contingency table, the row variable and the column variable are independent.

-Use the sample data below to test whether car color affects the likelihood of being in an accident. Use a significance level of 0.01.  Red  Blue  White  Car has been in  accident 283336 Car has not been  in accident 232230\begin{array} { c | c c c } & \text { Red } & \text { Blue } & \text { White } \\\hline \text { Car has been in } \\\text { accident } & 28 & 33 & 36 \\\begin{array} { c } \text { Car has not been } \\\text { in accident }\end{array} & 23 & 22 & 30\end{array}
Question
Describe the null hypothesis for the test of independence. List the assumptions for the X² test of independence. What is the major difference between the assumptions for this test and the assumptions for the previous tests we have studied?
Question
When using statistical software packages, the critical value is typically not given. What method is used to determine whether you reject or fail to reject the null hypothesis?
Question
At a high school debate tournament, half of the teams were asked to wear suits and ties and the rest were asked to wear jeans and t-shirts. The results are given in the table below.
Test the hypothesis at the 0.05 level that the proportion of wins is the same for teams wearing suits as for teams wearing jeans.
 Win  Loss  Suit 2228 T-shirt 2822\begin{array} { r | r r } & \text { Win } & \text { Loss } \\\hline \text { Suit } & 22 & 28 \\\text { T-shirt } & 28 & 22\end{array}
Question
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.

-At the 0.025 significance level, test the claim that the four brands have the same mean if the following sample results have been obtained.  Brand A Brand B Brand C Brand D 15202115251722152122201423231923221822202828\begin{array} { l } \text { Brand A } &\text{Brand }\mathrm{B}& \text { Brand } \mathrm{C} &\text { Brand D }\\\hline15&20&21&15\\25&17&22&15\\21&22&20&14\\23&23&19&23\\22&&18&22\\20&&&28\\&&&28\end{array}
Question
Using the data below and a 0.05 significance level, test the claim that the responses occur with percentages of 15%, 20%, 25%, 25%, and 15% respectively.  Response  A  B  C  D  E  Frequency 1215161819\begin{array} { c | r r r r r } \text { Response } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \\\hline \text { Frequency } & 12 & 15 & 16 & 18 & 19\end{array}
Question
Explain the computation of expected values for contingency tables in terms of
probabilities. Refer to the assumptions of the null hypothesis as part of your explanation.
You might give a brief example to illustrate.
Question
You roll a die 48 times with the following results. You roll a die 48 times with the following results.   Use a significance level of 0.05 to test the claim that the die is fair.<div style=padding-top: 35px> Use a significance level of 0.05 to test the claim that the die is fair.
Question
A survey conducted in a small town yielded the results shown in the table.  Men Women  Plan to vote 10587 Do not plan to vote 312246\begin{array} { l | l c } & \text { Men}&\text { Women } \\\hline \text { Plan to vote } & 105 & 87 \\\hline \text { Do not plan to vote } & 312 & 246\end{array}
i) Test the claim that the intention to vote in the next presidential election is independent of the gender of the person being surveyed. Use a 0.05 significance level.
ii) Using Yates' correction, replace (OE)2E with (OE0.5)2E\sum \frac{(\mathrm{O}-\mathrm{E})^{2}}{E}\text { with } \sum \frac{(|O-E|-0.5)^{2}}{E} and repeat the test.
What effect does Yates' correction have on the value of the test statistic?
Question
Among the four northwestern states, Washington has 51% of the total population, Oregon has 30%, Idaho has 11%, and Montana has 8%. A market researcher selects a sample of 1000 subjects, with 450 in Washington, 340 in Oregon, 150 in Idaho, and 60 in Montana. At the 0.05 significance level, test the claim that the sample of 1000 subjects has a distribution that agrees with the distribution of state populations.
Question
Describe the null and alternative hypotheses for one-way ANOVA. Give an example.
Question
A survey conducted in a small business yielded the results shown in the table.  Men Women  Health insurance 5020 No health insurance 3010\begin{array} { l | c c } & \text { Men}&\text { Women } \\\hline \text { Health insurance } & 50 & 20 \\\hline \text { No health insurance } & 30 & 10\end{array} i) Test the claim that health care coverage is independent of gender. Use a 0.05
significance level.
ii) Using Yates' correction, replace (OE)2E with (OE0.5)2E\sum { \frac { ( \mathrm { O } - \mathrm { E } ) ^ { 2 } } { E } } \text { with } \sum { \frac { ( | \mathrm { O } - \mathrm { E } | - 0.5 ) ^ { 2 } } { E } } and repeat the test.
What effect does Yates' correction have on the value of the test statistic?
Question
Use a 0.01 significance level to test the claim that the proportion of men who plan to vote in the next election is the same as the proportion of women who plan to vote. 300 men and 300 women were randomly selected and asked whether they planned to vote in the next election. The results are shown below.
 Men  Women  Plan to vote 170185 Do not plan to vote 130115\begin{array} { r | r r } & \text { Men } & \text { Women } \\\hline \text { Plan to vote } & 170 & 185 \\\text { Do not plan to vote } & 130 & 115\end{array}
Question
In studying the occurrence of genetic characteristics, the following sample data were obtained. At the 0.05 significance level, test the claim that the characteristics occur with the same frequency.
 Characteristic  A  B  C  D  E  F  Frequency 283045483839\begin{array} { c | r r r r r r } \text { Characteristic } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } \\\hline \text { Frequency } & 28 & 30 & 45 & 48 & 38 & 39\end{array}
Question
Define the term "treatment". What other term means the same thing? Give an example.
Question
Draw an example of an F distribution and list the characteristics of the F distribution.
Question
Use a χ2\chi ^ { 2 } test to test the claim that in the given contingency table, the row variable and the column variable are independent.

-Responses to a survey question are broken down according to gender and the sample results are given below. At the 0.05 significance level, test the claim that response and gender are independent.  Yes  No  Undecided  Male 255015 Female 203010\begin{array} { r | r r r } & \text { Yes } & \text { No } & \text { Undecided } \\\hline \text { Male } & 25 & 50 & 15 \\\text { Female } & 20 & 30 & 10\end{array}
Question
Use a χ2\chi ^ { 2 } test to test the claim that in the given contingency table, the row variable and the column variable are independent.

-160 students who were majoring in either math or English were asked a test question, and the researcher recorded whether they answered the question correctly. The sample results are given below. At the 0.10 significance level, test the claim that response and major are independent.
 Correct  Incorrect  Math 2753 English 4337\begin{array} { r | r r } & \text { Correct } & \text { Incorrect } \\\hline \text { Math } & 27 & 53 \\\text { English } & 43 & 37\end{array}
Question
Describe the test of homogeneity. What characteristic distinguishes a test of homogeneity from a test of independence?
Question
Use the data given below to verify that the t test for independent samples and the ANOVA method are equivalent.
AB85748172736591836459\begin{array} { c | c } \mathrm { A } & \mathrm { B } \\\hline 85 & 74 \\81 & 72 \\73 & 65 \\91 & 83 \\64 & 59\end{array}
i) Use a t test with a 0.05 significance level to test the claim that the two samples come from populations with the same means.
ii) Use the ANOVA method with a 0.05 significance level to test the same claim.
iii) Verify that the squares of the t test statistic and the critical value are equal to the F test statistic and critical value.
Question
At the same time each day, a researcher records the temperature in each of three greenhouses. The table shows the temperatures in degrees Fahrenheit recorded for one week.
 Greenhouse #1  Greenhouse #2  Greenhouse #3 737167726963737262667261686560717362727159\begin{array} { c | c | c } \text { Greenhouse \#1 } & \text { Greenhouse \#2 } & \text { Greenhouse \#3 } \\\hline 73 & 71 & 67 \\72 & 69 & 63 \\73 & 72 & 62 \\66 & 72 & 61 \\68 & 65 & 60 \\71 & 73 & 62 \\72 & 71 & 59\end{array}
i) Use a 0.05 significance level to test the claim that the average temperature is the same in each greenhouse.
ii) How are the analysis of variance results affected if the same constant is added to every one of the original sample values?
Question
What can you conclude about the equality of the population means?

 Source  DF  SS  MS  F  p  Factor 313.5004.5005.170.011 Error 1613.9250.870 Total 1927.425\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\ \text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \\ \text { Error } & 16 & 13.925 & 0.870 & & \\ \text { Total } & 19 & 27.425 & & & \end{array}


A) Accept the null hypothesis since the p-value is greater than the significance level.
B) Reject the null hypothesis since the p-value is less than the significance level.
C) Reject the null hypothesis since the p-value is greater than the significance level.
D) Accept the null hypothesis since the p-value is less than the significance level.
Question
In studying the responses to a multiple-choice test question, the following sample data were obtained. At the 0.05 significance level, test the claim that the responses occur with the same frequency.
 Response  A  B  C  D  E  Frequency 1215161819\begin{array} { r | r r r r r } \text { Response } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \\\hline \text { Frequency } & 12 & 15 & 16 & 18 & 19\end{array}
Question
An observed frequency distribution is as follows:  Number of successes 012 Frequency 419366\begin{array}{l|ccc}\text { Number of successes } & 0 & 1 & 2 \\\hline \text { Frequency } & 41 & 93 & 66\end{array}
i) Assuming a binomial distribution with n=2\mathrm { n } = 2 and p=1/2\mathrm { p } = 1 / 2 , use the binomial formula to find the probability corresponding to each category of the table.
ii) Using the probabilities found in part (i), find the expected frequency for each category.
iii) Use a 0.050.05 level of significance to test the claim that the observed frequencies fit a binomial distribution for which n=2\mathrm { n } = 2 and p=1/2\mathrm { p } = 1 / 2 .
Question
An observed frequency distribution is as follows: An observed frequency distribution is as follows:  <div style=padding-top: 35px>
Question
Find the critical value.  Source  DF  SS  MS  F  p  Factor 33010.001.60.264 Error 8506.25 Total 1180\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\\text { Factor } & 3 & 30 & 10.00 & 1.6 & 0.264 \\\text { Error } & 8 & 50 & 6.25 & & \\\text { Total } & 11 & 80 & & &\end{array}

A) 4.07
B) 8.85
C) 7.59
D) 1.6
Question
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.

-Random samples of four different models of cars were selected and the gas mileage of each car was measured. The results are shown below.  Model A Model B Model C Model D 23283025252628262429322526302728\begin{array} { l } \text { Model A}&\text { Model B}&\text { Model C}&\text { Model D }\\\hline23&28&30&25\\25&26&28&26\\24&29&32&25\\26&30&27&28\end{array}
Test the claim that the four different models have the same population mean. Use a significance level of 0.05.
Question
According to Benford's Law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. Test for goodness-of-fit with Benford's Law.  Leading Digit 123456789 Benford’s law:  distribution of  leading digits 30.1%17.6%12.5%9.7%7.9%6.7%5.8%5.1%4.6%\begin{array} { | l | c | c | c | c | c | c | c | c | c | } \hline \text { Leading Digit } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\\hline \begin{array} { l } \text { Benford's law: } \\\text { distribution of } \\\text { leading digits }\end{array} & 30.1 \% & 17.6 \% & 12.5 \% & 9.7 \% & 7.9 \% & 6.7 \% & 5.8 \% & 5.1 \% & 4.6 \% \\\hline\end{array}

-When working for the Brooklyn District Attorney, investigator Robert Burton analyzed the leading digits of the amounts from 784 checks issued by seven suspect companies. The frequencies were found to be 0, 18, 0, 79, 476, 180, 8, 23, and 0, and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. If the observed frequencies are substantially different from the frequencies expected with Benford's Law, the check amounts appear to result from fraud. Use a 0.05 significance level to test for goodness-of-fit with Benford's Law. Does it appear that the checks are the result of fraud?
Question
An observed frequency distribution of exam scores is as follows: An observed frequency distribution of exam scores is as follows:  <div style=padding-top: 35px>
Question
At the same time each day, a researcher records the temperature in each of three greenhouses. The table shows the temperatures in degrees Fahrenheit recorded for one week.
 Greenhouse #1  Greenhouse #2  Greenhouse #3 737167726963737262667261686560717362727159\begin{array} { c | c | c } \text { Greenhouse \#1 } & \text { Greenhouse \#2 } & \text { Greenhouse \#3 } \\\hline 73 & 71 & 67 \\72 & 69 & 63 \\73 & 72 & 62 \\66 & 72 & 61 \\68 & 65 & 60 \\71 & 73 & 62 \\72 & 71 & 59\end{array}
i) Use a 0.05 significance level to test the claim that the average temperature is the same in each greenhouse.
ii) How are the analysis of variance results affected if 8° is added to each temperature listed for greenhouse #3?
Question
According to Benford's Law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. Test for goodness-of-fit with Benford's Law.  Leading Digit 123456789 Benford’s law:  distribution of  leading digits 30.1%17.6%12.5%9.7%7.9%6.7%5.8%5.1%4.6%\begin{array} { | l | c | c | c | c | c | c | c | c | c | } \hline \text { Leading Digit } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\\hline \begin{array} { l } \text { Benford's law: } \\\text { distribution of } \\\text { leading digits }\end{array} & 30.1 \% & 17.6 \% & 12.5 \% & 9.7 \% & 7.9 \% & 6.7 \% & 5.8 \% & 5.1 \% & 4.6 \% \\\hline\end{array}

-When working for the Brooklyn District Attorney, investigator Robert Burton analyzed the leading digits of the amounts from 784 checks issued by seven suspect companies. The frequencies were found to be 0, 12, 0, 73, 482, 186, 8, 23, and 0, and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. If the observed frequencies are substantially different from the frequencies expected with Benford's Law, the check amounts appear to result from fraud. Use a 0.05 significance level to test for goodness-of-fit with Benford's Law. Does it appear that the checks are the result of fraud?
Question
A company manager wishes to test a union leader's claim that absences occur on the different week days with the same frequencies. Test this claim at the 0.05 level of significance if the following sample data have been compiled.  Day  MonTue Wed Thur Fri absences 3715122343\begin{array} { l | } \text { Day } & \text { Mon} & \text {Tue} & \text { Wed} & \text { Thur} & \text { Fri }\\\hline \text {absences }&37&15&12&23&43\\\end{array}
Question
Six independent samples of 100 values each are randomly drawn from populations that are normally distributed with equal variances. You wish to test the claim that μ1=μ2=μ3=\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } = μ4=μ5=μ6\mu _ { 4 } = \mu _ { 5 } = \mu _ { 6 }
i) If you test the individual claims μ1=μ2,μ1=μ3,μ1=μ4,,μ5=μ6\mu _ { 1 } = \mu _ { 2 } , \mu _ { 1 } = \mu _ { 3 } , \mu _ { 1 } = \mu _ { 4 } , \ldots , \mu _ { 5 } = \mu _ { 6 } , how many ways can you pair off the 6 means?
ii) Assume that the tests are independent and that for each test of equality between two means, there is a 0.900.90 probability of not making a type I error. If all possible pairs of means are tested for equality, what is the probability of making no type I errors?
iii) If you use analysis of variance to test the claim that μ1=μ2=μ3=μ4=μ5=μ6\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } = \mu _ { 4 } = \mu _ { 5 } = \mu _ { 6 } at the 0.100.10 level of significance, what is the probability of not making a type I error?
Question
Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean.

-What can you conclude about the equality of the population means?

 Source  DF  SS  MS  F  p  Factor 33010.001.60.264 Error 8506.25 Total 1180\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\ \text { Factor } & 3 & 30 & 10.00 & 1.6 & 0.264 \\ \text { Error } & 8 & 50 & 6.25 & & \\ \text { Total } & 11 & 80 & & & \end{array}


A) Accept the null hypothesis since the p-value is less than the significance level.
B) Reject the null hypothesis since the p-value is less than the significance level.
C) Accept the null hypothesis since the p-value is greater than the significance level.
D) Reject the null hypothesis since the p-value is greater than the significance level.
Question
Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean.

-Identify the p-value.  Source  DF  SS  MS  F  p  Factor 313.5004.5005.170.011 Error 1613.9250.870 Total 1927.425\begin{array} { l r c l c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\\text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \\\text { Error } & 16 & 13.925 & 0.870 & & \\\text { Total } & 19 & 27.425 & & &\end{array}

A) 5.17
B) 0.011
C) 4.500
D) 0.870
Question
Fill in the missing entries in the following partially completed one-way ANOVA table.  Source  df  SS  MS=SS/df  F-statistic  Treatment 28.9 Error 303.5 Total 33\begin{array} { | l | l | l | l | l | } \hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & & 28.9 & & \\\hline \text { Error } & 30 & & 3.5 & \\\hline \text { Total } & 33 & & & \\\hline\end{array}

A)
 Source  df  SS  MS=SS/df  F-stati stic  Treatment 6328.90.46520.20 Error 30105.03.5 Total 33133.9\begin{array}{|l|c|c|c|c|}\hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-stati stic } \\\hline \text { Treatment } & 63 & 28.9 & 0.46 & 520.20 \\\hline \text { Error } & 30 & 105.0 & 3.5 & \\\hline \text { Total } & 33 & 133.9 & & \\\hline\end{array}

B)
 Source  df  SS  MS=SS/df  F-statistic  Treatment 328.99.632.75 Error 30105.03.5 Total 33133.9\begin{array}{|l|c|c|c|c|}\hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & 3 & 28.9 & 9.63 & 2.75 \\\hline \text { Error } & 30 & 105.0 & 3.5 & \\\hline \text { Total } & 33 & 133.9 & & \\\hline\end{array}

C)
 Source dfSSMS=SS/df F-statistic  Treatment 328.99.630.36 Error 30105.03.5 Total 33133.9\begin{array}{|l|c|c|c|c|}\hline \text { Source } & \mathrm{df} & \mathrm{SS} & \mathrm{MS}=\mathrm{SS} / \mathrm{df} & \text { F-statistic } \\\hline \text { Treatment } & 3 & 28.9 & 9.63 & 0.36 \\\hline \text { Error } & 30 & 105.0 & 3.5 & \\\hline \text { Total } & 33 & 133.9 & & \\\hline\end{array}

D)
 Source  df  SS  MS=SS/df  F-statistic  Treatment 328.99.632.75 Error 30105.03.5 Total 3329.02\begin{array}{|l|r|c|c|c|}\hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & 3 & 28.9 & 9.63 & 2.75 \\\hline \text { Error } & 30 & 105.0 & 3.5 & \\\hline \text { Total } & 33 & 29.02 & & \\\hline\end{array}
Question
Use a χ2\chi ^ { 2 } test to test the claim that in the given contingency table, the row variable and the column variable are independent.

-Tests for adverse reactions to a new drug yielded the results given in the table. At the 0.05 significance level, test the claim that the treatment (drug or placebo) is independent of the reaction (whether or not headaches were experienced).  Drug  Placebo  Headaches 117 No headaches 7391\begin{array} { r | r r } & \text { Drug }&\text { Placebo } \\\hline \text { Headaches } & 11 & 7 \\\text { No headaches } & 73 & 91\end{array}
Question
Describe a goodness-of-fit test. What assumptions are made when using a goodness-of-fit test?
Question
List the assumptions for testing hypotheses that three or more means are equivalent.
Question
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.

-Given the sample data below, test the claim that the populations have the same mean. Use a significance level of 0.05.
 Brand A  Brand B  Brand C  Brand D n=16n=16n=16n=16x=2.09x=3.48x=1.86x=2.84 s=0.37 s=0.61 s=0.45 s=0.53\begin{array} { l } \text { Brand A }&\text { Brand B }&\text { Brand C }&\text { Brand D }\\\hline\mathrm{n}=16 & \mathrm{n}=16 & \mathrm{n}=16 & \mathrm{n}=16 \\\overline{\mathrm{x}}=2.09 & \overline{\mathrm{x}}=3.48 & \overline{\mathrm{x}}=1.86 & \overline{\mathrm{x}}=2.84 \\\mathrm{~s}=0.37 & \mathrm{~s}=0.61 & \mathrm{~s}=0.45 & \mathrm{~s}=0.53\end{array}
Question
In the chi-square test of independence, the formula used is χ2=Σ(OE)2E\chi ^ { 2 } = \frac { \Sigma ( \mathrm { O } - \mathrm { E } ) ^ { 2 } } { \mathrm { E } } . Discuss the meaning of O\mathrm { O } and E\mathrm { E } and explain the circumstances under which the χ2\chi ^ { 2 } values will be smaller or larger. What is the relationship between a significant χ2\chi ^ { 2 } value and the values of O and E?
Question
Find the critical value.  Source  DF  SS  MS  F  p  Factor 313.5004.5005.170.011 Error 1613.9250.870 Total 1927.425\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\\text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \\\text { Error } & 16 & 13.925 & 0.870 & & \\\text { Total } & 19 & 27.425 & & &\end{array}

A) 3.06
B) 8.70
C) 3.24
D) 5.42
Question
Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean.

-Identify the value of the test statistic.  Source  DF  SS  MS  F  P  Factor 313.5004.5005.170.011 Error 1613.9250.870 Total 1927.425\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\\text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \\\text { Error } & 16 & 13.925 & 0.870 & & \\\text { Total } & 19 & 27.425 & & &\end{array}

A) 5.17
B) 13.500
C) 4.500
D) 0.011
Question
Identify the p-value.  Source  DF  SS  MS  F  p  Factor 33010.001.60.264 Error 8506.25 Total 1180\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\\text { Factor } & 3 & 30 & 10.00 & 1.6 & 0.264 \\\text { Error } & 8 & 50 & 6.25 & & \\\text { Total } & 11 & 80 & & &\end{array}

A) 0.264
B) 1.6
C) 10.00
D) 6.25
Question
Identify the value of the test statistic.  Source  DF  SS  MS  F  p  Factor 33010.001.60.264 Error 8506.25 Total 1180\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\\text { Factor } & 3 & 30 & 10.00 & 1.6 & 0.264 \\\text { Error } & 8 & 50 & 6.25 & & \\\text { Total } & 11 & 80 & & &\end{array}

A) 0.264
B) 1.6
C) 30
D) 10.00
Question
Fill in the missing entries in the following partially completed one-way ANOVA table.
 Source  df  SS  MS=SS/df  F-statistic  Treatment 311.16 Error 13.720.686 Total \begin{array} { | l | c | c | c | c | } \hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & 3 & & & 11.16 \\\hline \text { Error } & & 13.72 & 0.686 & \\\hline \text { Total } & & & & \\\hline\end{array}

A)
 Source  df SSMS=SS/df F-statistic  Treatment 30.1840.06111.16 Error 2013.720.686 Total 2313.90\begin{array}{|l|c|c|c|c|}\hline \text { Source } & \text { df } & \mathrm{SS} & \mathrm{MS}=\mathrm{SS} / \mathrm{df} & \text { F-statistic } \\\hline \text { Treatment } & 3 & 0.184 & 0.061 & 11.16 \\\hline \text { Error } & 20 & 13.72 & 0.686 & \\\hline \text { Total } & 23 & 13.90 & & \\\hline\end{array}

B)
 Source  df  SS  MS=SS/df  F-statistic  Treatment 32.557.6611.16 Error 2013.720.686 Total 2316.27\begin{array}{|l|c|c|c|c|}\hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & 3 & 2.55 & 7.66 & 11.16 \\\hline \text { Error } & 20 & 13.72 & 0.686 & \\\hline \text { Total } & 23 & 16.27 & & \\\hline\end{array}

C)
 Source  df  SS  MS=SS/df  F-statistic  Treatment 322.977.6611.16 Error 2013.720.686 Total 2336.69\begin{array}{|l|c|c|c|c|}\hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & 3 & 22.97 & 7.66 & 11.16 \\\hline \text { Error } & 20 & 13.72 & 0.686 & \\\hline \text { Total } & 23 & 36.69 & & \\\hline\end{array}

D)
 Source  df  SS  MS=SS/df  F-statistic  Treatment 348.8016.2711.16 Error 2013.720.686 Total 2362.52\begin{array}{|l|c|c|c|c|}\hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & 3 & 48.80 & 16.27 & 11.16 \\\hline \text { Error } & 20 & 13.72 & 0.686 & \\\hline \text { Total } & 23 & 62.52 & & \\\hline\end{array}
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Deck 11: Chi-Square and Analysis of Variance
1
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.

-Given the sample data below, test the claim that the populations have the same mean. Use a significance level of 0.05.

 Brand A Brand B Brand C \hlinen=10n=10n=10xˉ=32.9xˉ=31.6xˉ=27.1s2=3.16s2=3.04s2=4.66\begin{array} { l } \text { Brand A}&\text { Brand B}&\text { Brand C }\\\hlinen=10 & n=10 & n=10 \\\bar{x}=32.9 & \bar{x}=31.6 & \bar{x}=27.1 \\s^{2}=3.16 & s^{2}=3.04 & s^{2}=4.66\end{array}
Test statistic: F=25.589\mathrm { F } = 25.589 . Critical value: F=3.35\mathrm { F } = 3.35 .
Reject the claim of equal means. The different brands do not appear to have the same mean.
2
The test statistic  one-way ANOVA is F= variance between samples  variance within samples \text { one-way } \mathrm { ANOVA } \text { is } \mathrm { F } = \frac { \text { variance between samples } } { \text { variance within samples } } . Describe variance within samples and variance between samples. What relationship between variance within samples and variance between samples would result in the conclusion that the value of F is significant?
Variance between samples measures the variation between the sample means, that is the variation due to the treatment. The variance within the samples depends solely on the sample variances and is a measure of pooled variation. The F ratio compares the two. If the F ratio is relatively close to 1, the two variances are about the same, and we conclude that there are no significant differences among the sample means. When the value of F is excessively large (that is, greater than 1), we conclude that the variation among the samples is not the same and that the means are not equal.
3
A researcher wishes to test the effectiveness of a flu vaccination. 150 people are vaccinated, 180 people are vaccinated with a placebo, and 100 people are not vaccinated. The number in each group who later caught the flu was recorded. The results are shown below.
 Vaccinated  Placebo  Control  Caught the flu 81921 Did not catch the flu 14216179\begin{array} { r | r r r } & \text { Vaccinated } & \text { Placebo } & \text { Control } \\\hline \text { Caught the flu } & 8 & 19 & 21 \\\text { Did not catch the flu } & 142 & 161 & 79\end{array}
Use a 0.05 significance level to test the claim that the proportion of people catching the flu is the same in all three groups.
H0\mathrm { H } _ { 0 } : The proportion of people catching the flu is the same in all three groups.
H1\mathrm { H } _ { 1 } : The proportions are different.
Test statistic: χ2=14.965\chi ^ { 2 } = 14.965 . Critical value: χ2=5.991\chi ^ { 2 } = 5.991 .
Reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that the proportion of people catching the flu is the same in all three groups.
4
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.

-A consumer magazine wants to compare the lifetimes of ballpoint pens of three different types. The magazine takes a random sample of pens of each type in the following table.

 Brand A Brand B Brand C 260181238218240257184162241219218213\begin{array} { l } \text { Brand A}&\text { Brand B}&\text { Brand C }\\\hline260&181&238\\218&240&257\\184&162&241\\219&218&213\end{array}
Do the data indicate that there is a difference in mean lifetime for the three brands of ballpoint pens?  Use α=0.01\text { Use } \alpha = 0.01 \text {. }
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5
Define categorical data and give an example.
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6
Use a X² test to test the claim that in the given contingency table, the row variable and the column variable are independent.

-The table below shows the age and favorite type of music of 668 randomly selected people.  Rock  Pop  Classical 152550857325356891603545907477\begin{array} { l | r r r } & \text { Rock } & \text { Pop } & \text { Classical } \\\hline 15 - 25 & 50 & 85 & 73 \\25 - 35 & 68 & 91 & 60 \\35 - 45 & 90 & 74 & 77\end{array}
Use a 5 percent level of significance to test the null hypothesis that age and preferred music type are independent.
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7
The following table shows the number of employees who called in sick at a business for different days of a particular week.  Day  Sun  Mon  Tues  Wed  Thurs  Fri  Sat  Number sick 81271191112\begin{array} { l | c c c c c c c } \text { Day } & \text { Sun } & \text { Mon } & \text { Tues } & \text { Wed } & \text { Thurs } & \text { Fri } & \text { Sat } \\\hline \text { Number sick } & 8 & 12 & 7 & 11 & 9 & 11 & 12\end{array}
i) At the 0.05 level of significance, test the claim that sick days occur with equal frequency on the different days of the week.
ii) Test the claim after changing the frequency for Saturday to 152. Describe the effect of this outlier on the test.
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8
Use the data given below to verify that the t test for independent samples and the ANOVA method are equivalent.
AB101929181127193015181621\begin{array} { c | c } \mathrm { A } & \mathrm { B } \\\hline 10 & 19 \\29 & 18 \\11 & 27 \\19 & 30 \\15 & 18 \\16 & 21\end{array}
i) Use a t test with a 0.05 significance level to test the claim that the two samples come from populations with the same means.
ii) Use the ANOVA method with a 0.05 significance level to test the same claim.
iii) Verify that the squares of the t test statistic and the critical value are equal to the F test statistic and critical value.
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9
At the 0.025 significance level, test the claim that the four brands have the same mean if the following sample results have been obtained.
 Brand A Brand B Brand C  Brand D 1718212220182425212325272225262921262935293637\begin{array} { l } \text { Brand A}&\text { Brand B}&\text { Brand C }&\text { Brand D }\\\hline17&18&21&22\\20&18&24&25\\21&23&25&27\\22&25&26&29\\21&26&29&35\\&&29&36\\&&&37\end{array}
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10
According to Benford's Law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. Test for goodness-of-fit with Benford's Law.  Leading Digit 123456789 Benford’s law:  distribution of  leading digits 30.1%17.6%12.5%9.7%7.9%6.7%5.8%5.1%4.6%\begin{array} { | l | c | c | c | c | c | c | c | c | c | } \hline \text { Leading Digit } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\\hline \begin{array} { l } \text { Benford's law: } \\\text { distribution of } \\\text { leading digits }\end{array} & 30.1 \% & 17.6 \% & 12.5 \% & 9.7 \% & 7.9 \% & 6.7 \% & 5.8 \% & 5.1 \% & 4.6 \% \\\hline\end{array}

-When working for the Brooklyn District Attorney, investigator Robert Burton analyzed the leading digits of the amounts from 784 checks issued by seven suspect companies. The frequencies were found to be 0, 9, 0, 70, 485, 189, 8, 23, and 0, and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. If the observed frequencies are substantially different from the frequencies expected with Benford's Law, the check amounts appear to result from fraud. Use a 0.05 significance level to test for goodness-of-fit with Benford's Law. Does it appear that the checks are the result of fraud?
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11
A researcher wishes to test whether the proportion of college students who smoke is the same in four different colleges. She randomly selects 100 students from each college and records the number that smoke. The results are shown below.
 College A  College B  College C  College D  Smoke 17261134 Don’t smoke 83748966\begin{array} { r | r r r r } & \text { College A } & \text { College B } & \text { College C } & \text { College D } \\\hline \text { Smoke } & 17 & 26 & 11 & 34 \\\text { Don't smoke } & 83 & 74 & 89 & 66\end{array}
Use a 0.01 significance level to test the claim that the proportion of students smoking is the same at all four colleges.
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12
Four independent samples of 100 values each are randomly drawn from populations that are normally distributed with equal variances. You wish to test the claim that μ1=μ2=μ3\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } =μ4= \mu _ { 4 }
i) If you test the individual claims μ1=μ2,μ1=μ3,μ1=μ4,,μ3=μ4\mu _ { 1 } = \mu _ { 2 } , \mu _ { 1 } = \mu _ { 3 } , \mu _ { 1 } = \mu _ { 4 } , \ldots , \mu _ { 3 } = \mu _ { 4 } , how many ways can you pair off the 4 means?
ii) Assume that the tests are independent and that for each test of equality between two means, there is a 0.990.99 probability of not making a type I error. If all possible pairs of means are tested for equality, what is the probability of making no type I errors?
iii) If you use analysis of variance to test the claim that μ1=μ2=μ3=μ4\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } = \mu _ { 4 } at the 0.010.01 level of significance, what is the probability of not making a type I error?
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13
Use a significance level of 0.01 to test the claim that workplace accidents are distributed on workdays as follows: Monday 25%, Tuesday: 15%, Wednesday: 15%, Thursday: 15%, and Friday: 30%.
In a study of 100 workplace accidents, 26 occurred on a Monday, 15 occurred on a Tuesday, 17 occurred on a Wednesday, 17 occurred on a Thursday, and 25 occurred on a Friday.
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14
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.

-The data below represent the weight losses for people on three different exercise programs.
Exercise A  Exercise B Exercise C2.55.84.38.84.96.27.31.15.89.87.88.15.11.27.9\begin{array} { l } \text {Exercise A }& \text { Exercise B}& \text { Exercise C}\\\hline2.5&5.8&4.3\\8.8&4.9&6.2\\7.3&1.1&5.8\\9.8&7.8&8.1\\5.1&1.2&7.9\end{array}

At the 1% significance level, does it appear that a difference exists in the true mean weight loss produced by the three exercise programs?
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15
Suppose you are to test for equality of four different population means, with H0: µA = µB =μC=μD= \mu _ { C } = \mu _ { D } . Write the hypotheses for the paired tests. Use methods of probability to explain why the process of ANOVA has a higher degree of confidence than testing each of the pairs separately.
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16
An observed frequency distribution of exam scores is as follows:
Exam Score under 60 60697079808990100 Frequency30301406040\begin{array} {| l|ccccc| } \hline \text {Exam Score }&\text {under 60 }&60-69&70-79&80-89&90-100\\\hline \text { Frequency}&30&30&140&60&40\\\hline\end{array}


i) Assuming a normal distribution with μ=75\mu = 75 and σ=15\sigma = 15 , find the probability of a randomly selected subject belonging to each class. (Use boundaries of 59.5,69.5,79.5,89.559.5,69.5,79.5,89.5 , 100.)
ii) Using the probabilities found in part (i), find the expected frequency for each category.
iii) Use a 0.050.05 significance level to test the claim that the exam scores were randomly selected from a normally distributed population with μ=75\mu = 75 and σ=15\sigma = 15 .
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17
A survey conducted in a small business yielded the results shown in the table.  Men Women  Health insurance 4122 No health insurance 3424\begin{array} { l | c c } & \text { Men}&\text { Women } \\\hline \text { Health insurance } & 41 & 22 \\\hline \text { No health insurance } & 34 & 24\end{array} i) Test the claim that health care coverage is independent of gender. Use a 0.05
significance level.
ii) Using Yates' correction, replace (OE)2E with (OE0.5)2E\sum \frac{(O-E)^{2}}{E}\text { with } \sum \frac{(|O-E|-0.5)^{2}}{E} and repeat the test.
What effect does Yates' correction have on the value of the test statistic?
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18
Use a χ2\chi ^ { 2 } test to test the claim that in the given contingency table, the row variable and the column variable are independent.

-Responses to a survey question are broken down according to employment status and the sample results are given below. At the 0.10 significance level, test the claim that response and employment status are independent.  Yes  No  Undecided  Employed 30155 Unemployed 202510\begin{array} { r | r r r } & \text { Yes } & \text { No } & \text { Undecided } \\\hline \text { Employed } & 30 & 15 & 5 \\\text { Unemployed } & 20 & 25 & 10\end{array}
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19
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.

-At the 0.025 significance level, test the claim that the three brands have the same mean if the following sample results have been obtained.  Brand ABrand B Brand C 322722342425373332333022362139\begin{array} { l } \text { Brand A}& \text {Brand B }& \text {Brand C }\\\hline32&27&22\\34&24&25\\37&33&32\\33&30&22\\36&&21\\39\end{array}
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20
Use a χ2\chi ^ { 2 } test to test the claim that in the given contingency table, the row variable and the column variable are independent.

-Use the sample data below to test whether car color affects the likelihood of being in an accident. Use a significance level of 0.01.  Red  Blue  White  Car has been in  accident 283336 Car has not been  in accident 232230\begin{array} { c | c c c } & \text { Red } & \text { Blue } & \text { White } \\\hline \text { Car has been in } \\\text { accident } & 28 & 33 & 36 \\\begin{array} { c } \text { Car has not been } \\\text { in accident }\end{array} & 23 & 22 & 30\end{array}
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21
Describe the null hypothesis for the test of independence. List the assumptions for the X² test of independence. What is the major difference between the assumptions for this test and the assumptions for the previous tests we have studied?
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22
When using statistical software packages, the critical value is typically not given. What method is used to determine whether you reject or fail to reject the null hypothesis?
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23
At a high school debate tournament, half of the teams were asked to wear suits and ties and the rest were asked to wear jeans and t-shirts. The results are given in the table below.
Test the hypothesis at the 0.05 level that the proportion of wins is the same for teams wearing suits as for teams wearing jeans.
 Win  Loss  Suit 2228 T-shirt 2822\begin{array} { r | r r } & \text { Win } & \text { Loss } \\\hline \text { Suit } & 22 & 28 \\\text { T-shirt } & 28 & 22\end{array}
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24
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.

-At the 0.025 significance level, test the claim that the four brands have the same mean if the following sample results have been obtained.  Brand A Brand B Brand C Brand D 15202115251722152122201423231923221822202828\begin{array} { l } \text { Brand A } &\text{Brand }\mathrm{B}& \text { Brand } \mathrm{C} &\text { Brand D }\\\hline15&20&21&15\\25&17&22&15\\21&22&20&14\\23&23&19&23\\22&&18&22\\20&&&28\\&&&28\end{array}
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25
Using the data below and a 0.05 significance level, test the claim that the responses occur with percentages of 15%, 20%, 25%, 25%, and 15% respectively.  Response  A  B  C  D  E  Frequency 1215161819\begin{array} { c | r r r r r } \text { Response } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \\\hline \text { Frequency } & 12 & 15 & 16 & 18 & 19\end{array}
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26
Explain the computation of expected values for contingency tables in terms of
probabilities. Refer to the assumptions of the null hypothesis as part of your explanation.
You might give a brief example to illustrate.
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27
You roll a die 48 times with the following results. You roll a die 48 times with the following results.   Use a significance level of 0.05 to test the claim that the die is fair. Use a significance level of 0.05 to test the claim that the die is fair.
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28
A survey conducted in a small town yielded the results shown in the table.  Men Women  Plan to vote 10587 Do not plan to vote 312246\begin{array} { l | l c } & \text { Men}&\text { Women } \\\hline \text { Plan to vote } & 105 & 87 \\\hline \text { Do not plan to vote } & 312 & 246\end{array}
i) Test the claim that the intention to vote in the next presidential election is independent of the gender of the person being surveyed. Use a 0.05 significance level.
ii) Using Yates' correction, replace (OE)2E with (OE0.5)2E\sum \frac{(\mathrm{O}-\mathrm{E})^{2}}{E}\text { with } \sum \frac{(|O-E|-0.5)^{2}}{E} and repeat the test.
What effect does Yates' correction have on the value of the test statistic?
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29
Among the four northwestern states, Washington has 51% of the total population, Oregon has 30%, Idaho has 11%, and Montana has 8%. A market researcher selects a sample of 1000 subjects, with 450 in Washington, 340 in Oregon, 150 in Idaho, and 60 in Montana. At the 0.05 significance level, test the claim that the sample of 1000 subjects has a distribution that agrees with the distribution of state populations.
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30
Describe the null and alternative hypotheses for one-way ANOVA. Give an example.
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31
A survey conducted in a small business yielded the results shown in the table.  Men Women  Health insurance 5020 No health insurance 3010\begin{array} { l | c c } & \text { Men}&\text { Women } \\\hline \text { Health insurance } & 50 & 20 \\\hline \text { No health insurance } & 30 & 10\end{array} i) Test the claim that health care coverage is independent of gender. Use a 0.05
significance level.
ii) Using Yates' correction, replace (OE)2E with (OE0.5)2E\sum { \frac { ( \mathrm { O } - \mathrm { E } ) ^ { 2 } } { E } } \text { with } \sum { \frac { ( | \mathrm { O } - \mathrm { E } | - 0.5 ) ^ { 2 } } { E } } and repeat the test.
What effect does Yates' correction have on the value of the test statistic?
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32
Use a 0.01 significance level to test the claim that the proportion of men who plan to vote in the next election is the same as the proportion of women who plan to vote. 300 men and 300 women were randomly selected and asked whether they planned to vote in the next election. The results are shown below.
 Men  Women  Plan to vote 170185 Do not plan to vote 130115\begin{array} { r | r r } & \text { Men } & \text { Women } \\\hline \text { Plan to vote } & 170 & 185 \\\text { Do not plan to vote } & 130 & 115\end{array}
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33
In studying the occurrence of genetic characteristics, the following sample data were obtained. At the 0.05 significance level, test the claim that the characteristics occur with the same frequency.
 Characteristic  A  B  C  D  E  F  Frequency 283045483839\begin{array} { c | r r r r r r } \text { Characteristic } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } \\\hline \text { Frequency } & 28 & 30 & 45 & 48 & 38 & 39\end{array}
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34
Define the term "treatment". What other term means the same thing? Give an example.
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35
Draw an example of an F distribution and list the characteristics of the F distribution.
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36
Use a χ2\chi ^ { 2 } test to test the claim that in the given contingency table, the row variable and the column variable are independent.

-Responses to a survey question are broken down according to gender and the sample results are given below. At the 0.05 significance level, test the claim that response and gender are independent.  Yes  No  Undecided  Male 255015 Female 203010\begin{array} { r | r r r } & \text { Yes } & \text { No } & \text { Undecided } \\\hline \text { Male } & 25 & 50 & 15 \\\text { Female } & 20 & 30 & 10\end{array}
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37
Use a χ2\chi ^ { 2 } test to test the claim that in the given contingency table, the row variable and the column variable are independent.

-160 students who were majoring in either math or English were asked a test question, and the researcher recorded whether they answered the question correctly. The sample results are given below. At the 0.10 significance level, test the claim that response and major are independent.
 Correct  Incorrect  Math 2753 English 4337\begin{array} { r | r r } & \text { Correct } & \text { Incorrect } \\\hline \text { Math } & 27 & 53 \\\text { English } & 43 & 37\end{array}
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38
Describe the test of homogeneity. What characteristic distinguishes a test of homogeneity from a test of independence?
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39
Use the data given below to verify that the t test for independent samples and the ANOVA method are equivalent.
AB85748172736591836459\begin{array} { c | c } \mathrm { A } & \mathrm { B } \\\hline 85 & 74 \\81 & 72 \\73 & 65 \\91 & 83 \\64 & 59\end{array}
i) Use a t test with a 0.05 significance level to test the claim that the two samples come from populations with the same means.
ii) Use the ANOVA method with a 0.05 significance level to test the same claim.
iii) Verify that the squares of the t test statistic and the critical value are equal to the F test statistic and critical value.
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40
At the same time each day, a researcher records the temperature in each of three greenhouses. The table shows the temperatures in degrees Fahrenheit recorded for one week.
 Greenhouse #1  Greenhouse #2  Greenhouse #3 737167726963737262667261686560717362727159\begin{array} { c | c | c } \text { Greenhouse \#1 } & \text { Greenhouse \#2 } & \text { Greenhouse \#3 } \\\hline 73 & 71 & 67 \\72 & 69 & 63 \\73 & 72 & 62 \\66 & 72 & 61 \\68 & 65 & 60 \\71 & 73 & 62 \\72 & 71 & 59\end{array}
i) Use a 0.05 significance level to test the claim that the average temperature is the same in each greenhouse.
ii) How are the analysis of variance results affected if the same constant is added to every one of the original sample values?
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41
What can you conclude about the equality of the population means?

 Source  DF  SS  MS  F  p  Factor 313.5004.5005.170.011 Error 1613.9250.870 Total 1927.425\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\ \text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \\ \text { Error } & 16 & 13.925 & 0.870 & & \\ \text { Total } & 19 & 27.425 & & & \end{array}


A) Accept the null hypothesis since the p-value is greater than the significance level.
B) Reject the null hypothesis since the p-value is less than the significance level.
C) Reject the null hypothesis since the p-value is greater than the significance level.
D) Accept the null hypothesis since the p-value is less than the significance level.
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42
In studying the responses to a multiple-choice test question, the following sample data were obtained. At the 0.05 significance level, test the claim that the responses occur with the same frequency.
 Response  A  B  C  D  E  Frequency 1215161819\begin{array} { r | r r r r r } \text { Response } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \\\hline \text { Frequency } & 12 & 15 & 16 & 18 & 19\end{array}
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43
An observed frequency distribution is as follows:  Number of successes 012 Frequency 419366\begin{array}{l|ccc}\text { Number of successes } & 0 & 1 & 2 \\\hline \text { Frequency } & 41 & 93 & 66\end{array}
i) Assuming a binomial distribution with n=2\mathrm { n } = 2 and p=1/2\mathrm { p } = 1 / 2 , use the binomial formula to find the probability corresponding to each category of the table.
ii) Using the probabilities found in part (i), find the expected frequency for each category.
iii) Use a 0.050.05 level of significance to test the claim that the observed frequencies fit a binomial distribution for which n=2\mathrm { n } = 2 and p=1/2\mathrm { p } = 1 / 2 .
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44
An observed frequency distribution is as follows: An observed frequency distribution is as follows:
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45
Find the critical value.  Source  DF  SS  MS  F  p  Factor 33010.001.60.264 Error 8506.25 Total 1180\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\\text { Factor } & 3 & 30 & 10.00 & 1.6 & 0.264 \\\text { Error } & 8 & 50 & 6.25 & & \\\text { Total } & 11 & 80 & & &\end{array}

A) 4.07
B) 8.85
C) 7.59
D) 1.6
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46
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.

-Random samples of four different models of cars were selected and the gas mileage of each car was measured. The results are shown below.  Model A Model B Model C Model D 23283025252628262429322526302728\begin{array} { l } \text { Model A}&\text { Model B}&\text { Model C}&\text { Model D }\\\hline23&28&30&25\\25&26&28&26\\24&29&32&25\\26&30&27&28\end{array}
Test the claim that the four different models have the same population mean. Use a significance level of 0.05.
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47
According to Benford's Law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. Test for goodness-of-fit with Benford's Law.  Leading Digit 123456789 Benford’s law:  distribution of  leading digits 30.1%17.6%12.5%9.7%7.9%6.7%5.8%5.1%4.6%\begin{array} { | l | c | c | c | c | c | c | c | c | c | } \hline \text { Leading Digit } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\\hline \begin{array} { l } \text { Benford's law: } \\\text { distribution of } \\\text { leading digits }\end{array} & 30.1 \% & 17.6 \% & 12.5 \% & 9.7 \% & 7.9 \% & 6.7 \% & 5.8 \% & 5.1 \% & 4.6 \% \\\hline\end{array}

-When working for the Brooklyn District Attorney, investigator Robert Burton analyzed the leading digits of the amounts from 784 checks issued by seven suspect companies. The frequencies were found to be 0, 18, 0, 79, 476, 180, 8, 23, and 0, and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. If the observed frequencies are substantially different from the frequencies expected with Benford's Law, the check amounts appear to result from fraud. Use a 0.05 significance level to test for goodness-of-fit with Benford's Law. Does it appear that the checks are the result of fraud?
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48
An observed frequency distribution of exam scores is as follows: An observed frequency distribution of exam scores is as follows:
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49
At the same time each day, a researcher records the temperature in each of three greenhouses. The table shows the temperatures in degrees Fahrenheit recorded for one week.
 Greenhouse #1  Greenhouse #2  Greenhouse #3 737167726963737262667261686560717362727159\begin{array} { c | c | c } \text { Greenhouse \#1 } & \text { Greenhouse \#2 } & \text { Greenhouse \#3 } \\\hline 73 & 71 & 67 \\72 & 69 & 63 \\73 & 72 & 62 \\66 & 72 & 61 \\68 & 65 & 60 \\71 & 73 & 62 \\72 & 71 & 59\end{array}
i) Use a 0.05 significance level to test the claim that the average temperature is the same in each greenhouse.
ii) How are the analysis of variance results affected if 8° is added to each temperature listed for greenhouse #3?
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50
According to Benford's Law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. Test for goodness-of-fit with Benford's Law.  Leading Digit 123456789 Benford’s law:  distribution of  leading digits 30.1%17.6%12.5%9.7%7.9%6.7%5.8%5.1%4.6%\begin{array} { | l | c | c | c | c | c | c | c | c | c | } \hline \text { Leading Digit } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\\hline \begin{array} { l } \text { Benford's law: } \\\text { distribution of } \\\text { leading digits }\end{array} & 30.1 \% & 17.6 \% & 12.5 \% & 9.7 \% & 7.9 \% & 6.7 \% & 5.8 \% & 5.1 \% & 4.6 \% \\\hline\end{array}

-When working for the Brooklyn District Attorney, investigator Robert Burton analyzed the leading digits of the amounts from 784 checks issued by seven suspect companies. The frequencies were found to be 0, 12, 0, 73, 482, 186, 8, 23, and 0, and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. If the observed frequencies are substantially different from the frequencies expected with Benford's Law, the check amounts appear to result from fraud. Use a 0.05 significance level to test for goodness-of-fit with Benford's Law. Does it appear that the checks are the result of fraud?
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51
A company manager wishes to test a union leader's claim that absences occur on the different week days with the same frequencies. Test this claim at the 0.05 level of significance if the following sample data have been compiled.  Day  MonTue Wed Thur Fri absences 3715122343\begin{array} { l | } \text { Day } & \text { Mon} & \text {Tue} & \text { Wed} & \text { Thur} & \text { Fri }\\\hline \text {absences }&37&15&12&23&43\\\end{array}
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52
Six independent samples of 100 values each are randomly drawn from populations that are normally distributed with equal variances. You wish to test the claim that μ1=μ2=μ3=\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } = μ4=μ5=μ6\mu _ { 4 } = \mu _ { 5 } = \mu _ { 6 }
i) If you test the individual claims μ1=μ2,μ1=μ3,μ1=μ4,,μ5=μ6\mu _ { 1 } = \mu _ { 2 } , \mu _ { 1 } = \mu _ { 3 } , \mu _ { 1 } = \mu _ { 4 } , \ldots , \mu _ { 5 } = \mu _ { 6 } , how many ways can you pair off the 6 means?
ii) Assume that the tests are independent and that for each test of equality between two means, there is a 0.900.90 probability of not making a type I error. If all possible pairs of means are tested for equality, what is the probability of making no type I errors?
iii) If you use analysis of variance to test the claim that μ1=μ2=μ3=μ4=μ5=μ6\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } = \mu _ { 4 } = \mu _ { 5 } = \mu _ { 6 } at the 0.100.10 level of significance, what is the probability of not making a type I error?
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53
Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean.

-What can you conclude about the equality of the population means?

 Source  DF  SS  MS  F  p  Factor 33010.001.60.264 Error 8506.25 Total 1180\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\ \text { Factor } & 3 & 30 & 10.00 & 1.6 & 0.264 \\ \text { Error } & 8 & 50 & 6.25 & & \\ \text { Total } & 11 & 80 & & & \end{array}


A) Accept the null hypothesis since the p-value is less than the significance level.
B) Reject the null hypothesis since the p-value is less than the significance level.
C) Accept the null hypothesis since the p-value is greater than the significance level.
D) Reject the null hypothesis since the p-value is greater than the significance level.
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54
Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean.

-Identify the p-value.  Source  DF  SS  MS  F  p  Factor 313.5004.5005.170.011 Error 1613.9250.870 Total 1927.425\begin{array} { l r c l c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\\text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \\\text { Error } & 16 & 13.925 & 0.870 & & \\\text { Total } & 19 & 27.425 & & &\end{array}

A) 5.17
B) 0.011
C) 4.500
D) 0.870
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55
Fill in the missing entries in the following partially completed one-way ANOVA table.  Source  df  SS  MS=SS/df  F-statistic  Treatment 28.9 Error 303.5 Total 33\begin{array} { | l | l | l | l | l | } \hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & & 28.9 & & \\\hline \text { Error } & 30 & & 3.5 & \\\hline \text { Total } & 33 & & & \\\hline\end{array}

A)
 Source  df  SS  MS=SS/df  F-stati stic  Treatment 6328.90.46520.20 Error 30105.03.5 Total 33133.9\begin{array}{|l|c|c|c|c|}\hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-stati stic } \\\hline \text { Treatment } & 63 & 28.9 & 0.46 & 520.20 \\\hline \text { Error } & 30 & 105.0 & 3.5 & \\\hline \text { Total } & 33 & 133.9 & & \\\hline\end{array}

B)
 Source  df  SS  MS=SS/df  F-statistic  Treatment 328.99.632.75 Error 30105.03.5 Total 33133.9\begin{array}{|l|c|c|c|c|}\hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & 3 & 28.9 & 9.63 & 2.75 \\\hline \text { Error } & 30 & 105.0 & 3.5 & \\\hline \text { Total } & 33 & 133.9 & & \\\hline\end{array}

C)
 Source dfSSMS=SS/df F-statistic  Treatment 328.99.630.36 Error 30105.03.5 Total 33133.9\begin{array}{|l|c|c|c|c|}\hline \text { Source } & \mathrm{df} & \mathrm{SS} & \mathrm{MS}=\mathrm{SS} / \mathrm{df} & \text { F-statistic } \\\hline \text { Treatment } & 3 & 28.9 & 9.63 & 0.36 \\\hline \text { Error } & 30 & 105.0 & 3.5 & \\\hline \text { Total } & 33 & 133.9 & & \\\hline\end{array}

D)
 Source  df  SS  MS=SS/df  F-statistic  Treatment 328.99.632.75 Error 30105.03.5 Total 3329.02\begin{array}{|l|r|c|c|c|}\hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & 3 & 28.9 & 9.63 & 2.75 \\\hline \text { Error } & 30 & 105.0 & 3.5 & \\\hline \text { Total } & 33 & 29.02 & & \\\hline\end{array}
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56
Use a χ2\chi ^ { 2 } test to test the claim that in the given contingency table, the row variable and the column variable are independent.

-Tests for adverse reactions to a new drug yielded the results given in the table. At the 0.05 significance level, test the claim that the treatment (drug or placebo) is independent of the reaction (whether or not headaches were experienced).  Drug  Placebo  Headaches 117 No headaches 7391\begin{array} { r | r r } & \text { Drug }&\text { Placebo } \\\hline \text { Headaches } & 11 & 7 \\\text { No headaches } & 73 & 91\end{array}
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57
Describe a goodness-of-fit test. What assumptions are made when using a goodness-of-fit test?
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58
List the assumptions for testing hypotheses that three or more means are equivalent.
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59
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.

-Given the sample data below, test the claim that the populations have the same mean. Use a significance level of 0.05.
 Brand A  Brand B  Brand C  Brand D n=16n=16n=16n=16x=2.09x=3.48x=1.86x=2.84 s=0.37 s=0.61 s=0.45 s=0.53\begin{array} { l } \text { Brand A }&\text { Brand B }&\text { Brand C }&\text { Brand D }\\\hline\mathrm{n}=16 & \mathrm{n}=16 & \mathrm{n}=16 & \mathrm{n}=16 \\\overline{\mathrm{x}}=2.09 & \overline{\mathrm{x}}=3.48 & \overline{\mathrm{x}}=1.86 & \overline{\mathrm{x}}=2.84 \\\mathrm{~s}=0.37 & \mathrm{~s}=0.61 & \mathrm{~s}=0.45 & \mathrm{~s}=0.53\end{array}
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60
In the chi-square test of independence, the formula used is χ2=Σ(OE)2E\chi ^ { 2 } = \frac { \Sigma ( \mathrm { O } - \mathrm { E } ) ^ { 2 } } { \mathrm { E } } . Discuss the meaning of O\mathrm { O } and E\mathrm { E } and explain the circumstances under which the χ2\chi ^ { 2 } values will be smaller or larger. What is the relationship between a significant χ2\chi ^ { 2 } value and the values of O and E?
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61
Find the critical value.  Source  DF  SS  MS  F  p  Factor 313.5004.5005.170.011 Error 1613.9250.870 Total 1927.425\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\\text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \\\text { Error } & 16 & 13.925 & 0.870 & & \\\text { Total } & 19 & 27.425 & & &\end{array}

A) 3.06
B) 8.70
C) 3.24
D) 5.42
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62
Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean.

-Identify the value of the test statistic.  Source  DF  SS  MS  F  P  Factor 313.5004.5005.170.011 Error 1613.9250.870 Total 1927.425\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\\text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \\\text { Error } & 16 & 13.925 & 0.870 & & \\\text { Total } & 19 & 27.425 & & &\end{array}

A) 5.17
B) 13.500
C) 4.500
D) 0.011
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63
Identify the p-value.  Source  DF  SS  MS  F  p  Factor 33010.001.60.264 Error 8506.25 Total 1180\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\\text { Factor } & 3 & 30 & 10.00 & 1.6 & 0.264 \\\text { Error } & 8 & 50 & 6.25 & & \\\text { Total } & 11 & 80 & & &\end{array}

A) 0.264
B) 1.6
C) 10.00
D) 6.25
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64
Identify the value of the test statistic.  Source  DF  SS  MS  F  p  Factor 33010.001.60.264 Error 8506.25 Total 1180\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\\text { Factor } & 3 & 30 & 10.00 & 1.6 & 0.264 \\\text { Error } & 8 & 50 & 6.25 & & \\\text { Total } & 11 & 80 & & &\end{array}

A) 0.264
B) 1.6
C) 30
D) 10.00
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65
Fill in the missing entries in the following partially completed one-way ANOVA table.
 Source  df  SS  MS=SS/df  F-statistic  Treatment 311.16 Error 13.720.686 Total \begin{array} { | l | c | c | c | c | } \hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & 3 & & & 11.16 \\\hline \text { Error } & & 13.72 & 0.686 & \\\hline \text { Total } & & & & \\\hline\end{array}

A)
 Source  df SSMS=SS/df F-statistic  Treatment 30.1840.06111.16 Error 2013.720.686 Total 2313.90\begin{array}{|l|c|c|c|c|}\hline \text { Source } & \text { df } & \mathrm{SS} & \mathrm{MS}=\mathrm{SS} / \mathrm{df} & \text { F-statistic } \\\hline \text { Treatment } & 3 & 0.184 & 0.061 & 11.16 \\\hline \text { Error } & 20 & 13.72 & 0.686 & \\\hline \text { Total } & 23 & 13.90 & & \\\hline\end{array}

B)
 Source  df  SS  MS=SS/df  F-statistic  Treatment 32.557.6611.16 Error 2013.720.686 Total 2316.27\begin{array}{|l|c|c|c|c|}\hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & 3 & 2.55 & 7.66 & 11.16 \\\hline \text { Error } & 20 & 13.72 & 0.686 & \\\hline \text { Total } & 23 & 16.27 & & \\\hline\end{array}

C)
 Source  df  SS  MS=SS/df  F-statistic  Treatment 322.977.6611.16 Error 2013.720.686 Total 2336.69\begin{array}{|l|c|c|c|c|}\hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & 3 & 22.97 & 7.66 & 11.16 \\\hline \text { Error } & 20 & 13.72 & 0.686 & \\\hline \text { Total } & 23 & 36.69 & & \\\hline\end{array}

D)
 Source  df  SS  MS=SS/df  F-statistic  Treatment 348.8016.2711.16 Error 2013.720.686 Total 2362.52\begin{array}{|l|c|c|c|c|}\hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & 3 & 48.80 & 16.27 & 11.16 \\\hline \text { Error } & 20 & 13.72 & 0.686 & \\\hline \text { Total } & 23 & 62.52 & & \\\hline\end{array}
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