Consumer panel preferences for three proposed Cafes are as follows: \[\begin{array} { | c | c | c | } \hline \text { Cafe A } & \text { Cafe B } & \text { Cafe C } \\ \hline 48 & 62 & 40 \\ \hline \end{array}\] Use 0.05 level of significance and test to see if there is a preference among the three Cafes, according to the data.

H 0 : p 1 = p 2 = p 3 = 1/3 H 1 : At least two propo

NARRBEGIN: Students Absenteeism Student Absenteeism Consider a multinomial experiment involving n = 200 students of a large high school. The attendance department recorded the number of students who were absent during the weekdays. The null hypothesis to be tested is: HS1U1B10S1U1B0: pS1U1B11S1U1B0 = 0.10, pS1U1B12S1U1B0 = 0.25, pS1U1B13S1U1B0 = 0.30, pS1U1B14S1U1B0 = 0.20, pS1U1B15S1U1B0 = 0.15.NARREND -{Student Absenteeism Narrative} Test the hypothesis at the 5% level of significance with the following frequencies: \[\begin{array} { | l | c c c c c | } \hline \text { Day of the Week } & \text { Mon. } & \text { Tues. } & \text { Wed. } & \text { Thurs. } & \text { Fri } \\ \hline \text { Number Absent } & \mathbf { 8 } & 22 & 28 & 24 & 18 \\ \hline \end{array}\]

Rejection region: @#LAT-DLM& 2 > @#LAT-DLM& 2 0.05,4 = 9.488

The number of ATVs sold by three salespersons over a 3-month period is shown below: \[\begin{array}{l} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text {Brand of ATV }\\ \begin{array} { | l | c c c | } \hline \text { Falesparson } & \text { Brand } \mathrm { A } & \text { Brand } \mathrm { B } & \text { Brand } \mathrm { C } \\ \hline \text { Juan } & 7 & 2 & 6 \\ \text { pedro } & 11 & 4 & \mathbf { 8 } \\ \text { Fernando } & \mathbf { 8 } & 5 & 3 \\ \hline \end{array} \end{array}\] Use the 5% level of significance and test for the independence of salesperson and type of product sold.

H 0 : Salesperson and brand of car sold are

The Chairman of a committee has recently circulated pamphlets among the members, attempting to convince them that pension benefits should be the primary issue. A subsequent survey revealed the following breakdown of the members according to the plant at which they worked and the issue that they felt should be supported as the primary one. \[\begin{array}{l} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text { Issues }\\ \begin{array} { | l | c c c | } \hline \text { Plant Location } & \text { Very Interesting } & \text { Fairly Interesting } & \text { Not Interesting } \\ \hline \text { New Mexico } & 60 & 62 & 78 \\ \text { Utah } & 70 & 56 & 74\\\hline \end{array} \end{array}\] Do the data indicate at the 5% significant level that there are differences between the two plants regarding which issue should be the primary one?

H 0 : Plant location and work issues are in

Exam 15: Chi-Squared Tests Optional

Consumer panel preferences for three proposed Cafes are as follows: Cafe A Cafe B Cafe C 48 62 40 Use 0.05 level of significance and test to see if there is a preference among the three Cafes, according to the data.

(Essay)

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Question 141

NARRBEGIN: Students Absenteeism Student Absenteeism Consider a multinomial experiment involving n = 200 students of a large high school. The attendance department recorded the number of students who were absent during the weekdays. The null hypothesis to be tested is: H₀: p₁ = 0.10, p₂ = 0.25, p₃ = 0.30, p₄ = 0.20, p₅ = 0.15.NARREND -{Student Absenteeism Narrative} Test the hypothesis at the 5% level of significance with the following frequencies: Day of the Week Mon. Tues. Wed. Thurs. Fri Number Absent 22 28 24 18

(Essay)

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Question 142

What statistic do we get when we square the value of z?

(Essay)

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Question 143

The number of degrees of freedom in testing for normality is the:

(Multiple Choice)

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Question 144

The number of ATVs sold by three salespersons over a 3-month period is shown below: Brand of ATV Falesparson Brand Brand Brand Juan 7 2 6 pedro 11 4 Fernando 5 3 Use the 5% level of significance and test for the independence of salesperson and type of product sold.

(Essay)

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Question 145

In a goodness-of-fit test, the null hypothesis states that the data came from a normally distributed population. The researcher estimated the population mean and population standard deviation from a sample of 500 observations. In addition, the researcher used 6 standardized intervals to test for normality. Using a 5% level of significance, the critical value for this test is:

(Multiple Choice)

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Question 146

Which of the following statements regarding the chi-squared distribution is true?

(Multiple Choice)

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Question 147

The Chairman of a committee has recently circulated pamphlets among the members, attempting to convince them that pension benefits should be the primary issue. A subsequent survey revealed the following breakdown of the members according to the plant at which they worked and the issue that they felt should be supported as the primary one. Issues Plant Location Very Interesting Fairly Interesting Not Interesting New Mexico 60 62 78 Utah 70 56 74 Do the data indicate at the 5% significant level that there are differences between the two plants regarding which issue should be the primary one?

(Essay)

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Question 148

In 2011, computers of Brand A controlled 25% of the market, Brand B 20%, Brand C 10%, and brand D 45%. In 2015, sample data was collected from many randomly selected stores throughout the country. Of the 1,200 computers sold, 280 were Brand A, 270 were Brand B, 90 were Brand C, and 560 were Brand D. Has the market changed since 2011? Test at the 1% significance level.

(Essay)

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Question 149

If you want to compare two populations that each have two categories, which test(s) can you use?

(Essay)

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Question 150

A test for the differences between two proportions can be performed using the chi-squared distribution.

(True/False)

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Question 151

How do you find the probabilities needed to obtain expected frequencies for a test of a contingency table?

(Multiple Choice)

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Question 152

To describe a population with more than two categories you can only use a chi-squared goodness-of-fit test.

(True/False)

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Question 153

The chi-squared test of a contingency table is based upon:

(Multiple Choice)

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Question 154

In a test of a contingency table, rejecting the null hypothesis concludes the variables are not independent.

(True/False)

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Question 155

In a chi-squared test of a contingency table, the value of the test statistic was $\chi$ ² = 15.652, and the critical value at $\alpha$ = 0.025 was 11.1433. Thus, we must reject the null hypothesis at $\alpha$ = 0.025.

(True/False)

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Question 156

A chi-squared test of a contingency table can be used to infer that differences exist between ____________________ populations of nominal variables.

(Short Answer)

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Question 157

The number of degrees of freedom for a contingency table with 4 rows and 8 columns is

(Multiple Choice)

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Question 158

To determine the critical values in the chi-squared distribution table, you need to know the:

(Multiple Choice)

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Question 159

For comparing two or more populations each having two or more categories, use a(n) test of a(n) .

(Essay)

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Question 160

Showing 141 - 160 of 175

Consumer panel preferences for three proposed Cafes are as follows: Cafe A Cafe B Cafe C 48 62 40 Use 0.05 level of significance and test to see if there is a preference among the three Cafes, according to the data.

What statistic do we get when we square the value of z?

The number of degrees of freedom in testing for normality is the:

The number of ATVs sold by three salespersons over a 3-month period is shown below: Brand of ATV Falesparson Brand Brand Brand Juan 7 2 6 pedro 11 4 Fernando 5 3 Use the 5% level of significance and test for the independence of salesperson and type of product sold.

Which of the following statements regarding the chi-squared distribution is true?

If you want to compare two populations that each have two categories, which test(s) can you use?

A test for the differences between two proportions can be performed using the chi-squared distribution.

How do you find the probabilities needed to obtain expected frequencies for a test of a contingency table?

To describe a population with more than two categories you can only use a chi-squared goodness-of-fit test.

The chi-squared test of a contingency table is based upon:

In a test of a contingency table, rejecting the null hypothesis concludes the variables are not independent.

In a chi-squared test of a contingency table, the value of the test statistic was $\chi$ ² = 15.652, and the critical value at $\alpha$ = 0.025 was 11.1433. Thus, we must reject the null hypothesis at $\alpha$ = 0.025.

A chi-squared test of a contingency table can be used to infer that differences exist between ____________________ populations of nominal variables.

The number of degrees of freedom for a contingency table with 4 rows and 8 columns is

To determine the critical values in the chi-squared distribution table, you need to know the:

For comparing two or more populations each having two or more categories, use a(n) test of a(n) .

What Is Statistics

Graphical Descriptive Techniques I

Graphical Descriptive Techniques II

Numerical Descriptive Techniques

Data Collection and Sampling

Probability

Random Variables and Discrete Probability Distributions

Continuous Probability Distributions

Sampling Distributions

Introduction to Estimation

Introduction to Hypothesis Testing

Inference About a Population

Inference About Comparing Two Populations

Analysis of Variance

Simple Linear Regression and Correlation

Multiple Regression

Filters

Exam 15: Chi-Squared Tests Optional

Consumer panel preferences for three proposed Cafes are as follows: Cafe A Cafe B Cafe C 48 62 40 Use 0.05 level of significance and test to see if there is a preference among the three Cafes, according to the data.

What statistic do we get when we square the value of z?

The number of degrees of freedom in testing for normality is the:

The number of ATVs sold by three salespersons over a 3-month period is shown below: Brand of ATV Falesparson Brand Brand Brand Juan 7 2 6 pedro 11 4 Fernando 5 3 Use the 5% level of significance and test for the independence of salesperson and type of product sold.

Which of the following statements regarding the chi-squared distribution is true?

If you want to compare two populations that each have two categories, which test(s) can you use?

A test for the differences between two proportions can be performed using the chi-squared distribution.

How do you find the probabilities needed to obtain expected frequencies for a test of a contingency table?

To describe a population with more than two categories you can only use a chi-squared goodness-of-fit test.

The chi-squared test of a contingency table is based upon:

In a test of a contingency table, rejecting the null hypothesis concludes the variables are not independent.

In a chi-squared test of a contingency table, the value of the test statistic was χ\chiχ 2 = 15.652, and the critical value at α\alphaα = 0.025 was 11.1433. Thus, we must reject the null hypothesis at α\alphaα = 0.025.

A chi-squared test of a contingency table can be used to infer that differences exist between ____________________ populations of nominal variables.

The number of degrees of freedom for a contingency table with 4 rows and 8 columns is

To determine the critical values in the chi-squared distribution table, you need to know the:

For comparing two or more populations each having two or more categories, use a(n) ____________________ test of a(n) ____________________.

What Is Statistics

Graphical Descriptive Techniques I

Graphical Descriptive Techniques II

Numerical Descriptive Techniques

Data Collection and Sampling

Probability

Random Variables and Discrete Probability Distributions

Continuous Probability Distributions

Sampling Distributions

Introduction to Estimation

Introduction to Hypothesis Testing

Inference About a Population

Inference About Comparing Two Populations

Analysis of Variance

Simple Linear Regression and Correlation

Multiple Regression

Filters

In a chi-squared test of a contingency table, the value of the test statistic was $\chi$ ² = 15.652, and the critical value at $\alpha$ = 0.025 was 11.1433. Thus, we must reject the null hypothesis at $\alpha$ = 0.025.

For comparing two or more populations each having two or more categories, use a(n) test of a(n) .