The personnel manager of a consumer product company asked a random sample of employees how they felt about the work they were doing.The following table gives a breakdown of their responses by gender.Do the data provide sufficient evidence to conclude that the level of job satisfaction is related to gender? Conduct the test at the 0.10 level of significance. \[\begin{array}{l} \text { Response }\\ \begin{array} { | l | c c c | } \hline \text { Gender } & \begin{array} { c } \text { Very } \\ \text { Interesting } \end{array} & \begin{array} { c } \text { Fairly } \\ \text { Interesting } \end{array} & \begin{array} { c } \text { Not } \\ \text { Interesting } \end{array} \\ \hline \text { Male } & 70 & 41 & 9 \\ \text { Female } & 35 & 34 & 11 \\ \hline \end{array} \end{array}\] H0: ___________________________________________________________ H1: ___________________________________________________________ Test statistic = _________________________________ Critical Value = ________________________________ Conclusion: ____________________________________

Gender and level of job satisfaction are

Mean Profit A report stated that the mean profit margin of 1000 publicly held corporations is 7.5 percent and the standard deviation is 5.1 percent.Profit margins appear normally distributed.A financial analyst sampled 100 of these companies to test whether or not the mean and standard deviation are as stated.The sample results were: \[\begin{array} { | l | c | } \hline \text { Category } & \text { Frequency } \\ \hline \text { Under } 2 & 6 \\ 2 - \text { under } 6 & 32 \\ 6 \text {-under } 10 & 34 \\ 10 \text {-under } 14 & 14 \\ 14 \text {-under } 18 & 6 \\ 18 \text { or more } & 8 \\ \hline \end{array}\] -A test for independence is applied to a contingency table with 4 rows and 5 columns for two nominal variables.Calculate the degrees of freedom for this test.

The answer of Mean Profit A report stated that the mean...

Grade distribution A professor would like to test to see if the grade distribution for male and female students are the same.The following table shows the distribution of grades for 1450 students. \[\begin{array} { | c | c | c | c | c | c | c } \hline & \text { A } & \text { B } & \mathbf { C } & \mathbf { D } & \mathbf { F } & \text { Tatal } \\ \hline \text { Male } & 124 & 192 & 212 & \mathbf { 8 7 } & 15 & \mathbf { 6 3 0 } \\ \hline \text { Female } & 195 & 238 & 24 \mathrm { D } & 120 & 27 & \mathbf { 8 2 0 } \\ \hline \text { Tatal } & 319 & 430 & 452 & 207 & 42 & 1450 \\ \hline \end{array}\] -Compute the expected counts for each category.

\[\begin{array} { | c | c | c | c | c |

In a chi-squared test of a contingency table,the value of the test statistic was$\chi$S1U1P12 S1S1P0= 12.678,and the critical value at $\alpha$ = 0.025 was 14.4494.Thus:

A) we fail to reject the null hypothesis at= 0.025. B) we reject the null hypothesis at= 0.025. C) we don't have enough evidence to accept or reject the null hypothesis at= 0.025. D) we should decrease the level of significance in order to reject the null hypothesis. A) we fail to reject the null hypothesis at= 0.025. B) we reject the null hypothesis at= 0.025. C) we don't have enough evidence to accept or reject the null hypothesis at= 0.025. D) we should decrease the level of significance in order to reject the null hypothesis.

Exam 13: Chi-Square Applications

Metro Bus System The management of the Metro Bus System of Houston,TX was faced with the task of improving efficiency in terms of running on time.One of its greatest problems was breakdown of equipment.Between 1994 and 1997,emphasis was placed on maintenance.Random samples of 10,000 bus runs were taken from the records of 1994 and 1997.The percentage of breakdowns in 1994 was 0.4 and in 1997 it was 0.04.Use a chi-square test for equal proportions to analyze these data. -State the conclusion at the 0.01 significance level.

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Question 21

Hypotheses In testing the hypotheses: H₀: $\sigma ^ { 2 } = 0.01$ H₁: $σ ^ { 2 } \neq 0.01$ a random sample of 10 observations was drawn from a normal population,and the sample standard deviation was 0.043. -Determine the rejection region at the 5% significance level.

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Question 22

Gifts Three choices of free gifts are offered to 270 persons who subscribed to a new magazine.The number choosing each gift is shown in the table below.Assume the null hypothesis that the gifts are equally attractive to the subscribers.Use a 0.01 significance level. Gift Number Choosing gift A 110 B 100 C 60 -State the decision rule.

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Question 23

Find the chi-square value for each of the right-tail areas below,given that the degrees of freedom are 7: A)0.95 B)0.01 C)0.025 D)0.05

(Short Answer)

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Question 24

In carrying out a chi-square goodness-of-fit test,what are the "k" and "m" terms in the "df = k - 1 - m" expression and why is each term present?

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Question 25

For the chi-square goodness-of-fit test,the null hypothesis is rejected whenever:

(Multiple Choice)

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Question 26

Gifts Three choices of free gifts are offered to 270 persons who subscribed to a new magazine.The number choosing each gift is shown in the table below.Assume the null hypothesis that the gifts are equally attractive to the subscribers.Use a 0.01 significance level. Gift Number Choosing gift A 110 B 100 C 60 -State the null and alternative hypotheses. H₀: H₁:

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Question 27

In the goodness-of-fit chi-square test,under which assumption are the expected frequencies constructed?

(Multiple Choice)

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Question 28

The personnel manager of a consumer product company asked a random sample of employees how they felt about the work they were doing.The following table gives a breakdown of their responses by gender.Do the data provide sufficient evidence to conclude that the level of job satisfaction is related to gender? Conduct the test at the 0.10 level of significance. Response Gender Very Interesting Fairly Interesting Not Interesting Male 70 41 9 Female 35 34 11 H₀: _________________ H₁: _ Test statistic = _ Critical Value = Conclusion: ____________________

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Question 29

TV Model Three hundred individuals were asked to view the picture quality of four different TVs and asked to choose the one they preferred.The data is shown in the following table: TV Madel Number Wha preferred A 73 D Tatal 300 -Test the equality of proportions at the 0.05 level the hypotheses. State the decision rule in terms of the critical value. What is the conclusion?

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Question 30

Employee Four employees are monitored to determine whether there is any difference in the proportions of acceptable parts produced by the employees.The sample of parts produced is given below. Employee Quality Total Acceptable 265 270 225 180 950 Unacceptable 10 5 15 20 50 Total 275 275 250 200 1000 -These data provide the information for testing the hypothesis that the proportion of unacceptable is the same for the four employees.What is the expected frequency for Acceptable and employee 3?

(Short Answer)

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Question 31

In a goodness-of-fit chi-square test,a large amount of discrepancy between the frequencies that are observed and those that are expected tend to cause us to fail to reject the null hypothesis.

(True/False)

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Question 32

Gifts Three choices of free gifts are offered to 270 persons who subscribed to a new magazine.The number choosing each gift is shown in the table below.Assume the null hypothesis that the gifts are equally attractive to the subscribers.Use a 0.01 significance level. Gift Number Choosing gift A 110 B 100 C 60 -Compute the test statistic.

(Short Answer)

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Question 33

For all practical purposes,the test for equality of proportions is really just a special case for the independence of two variables.

(True/False)

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Question 34

Mean Profit A report stated that the mean profit margin of 1000 publicly held corporations is 7.5 percent and the standard deviation is 5.1 percent.Profit margins appear normally distributed.A financial analyst sampled 100 of these companies to test whether or not the mean and standard deviation are as stated.The sample results were: Category Frequency Under 2 6 2- under 6 32 6 -under 10 34 10 -under 14 14 14 -under 18 6 18 or more 8 -A test for independence is applied to a contingency table with 4 rows and 5 columns for two nominal variables.Calculate the degrees of freedom for this test.

(Short Answer)

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Question 35

Grade distribution A professor would like to test to see if the grade distribution for male and female students are the same.The following table shows the distribution of grades for 1450 students. A B Tatal Male 124 192 212 15 Female 195 238 24 120 27 Tatal 319 430 452 207 42 1450 -Compute the expected counts for each category.

(Essay)

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Question 36

Mean Profit A report stated that the mean profit margin of 1000 publicly held corporations is 7.5 percent and the standard deviation is 5.1 percent.Profit margins appear normally distributed.A financial analyst sampled 100 of these companies to test whether or not the mean and standard deviation are as stated.The sample results were: Category Frequency Under 2 6 2- under 6 32 6 -under 10 34 10 -under 14 14 14 -under 18 6 18 or more 8 -What is the expected frequency of the under 2 category?

(Short Answer)

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Question 37

In a chi-squared test of a contingency table,the value of the test statistic was $\chi$ ²= 12.678,and the critical value at $\alpha$ = 0.025 was 14.4494.Thus:

(Multiple Choice)

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Question 38

In a hypothesis test for the population variance,the alternate hypothesis is the population variance does not equal 17.0.The significance level to be used is 0.05 and the sample size to be taken is 25.The table value(s)to use from the chi-square distribution is (are):

(Multiple Choice)

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Question 39

The number of degrees of freedom in a test of a contingency table with r rows and c columns is _________________________.

(Essay)

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Question 40

Hypotheses In testing the hypotheses: H₀: $\sigma ^ { 2 } = 0.01$ H₁: $σ ^ { 2 } \neq 0.01$ a random sample of 10 observations was drawn from a normal population,and the sample standard deviation was 0.043. -Determine the rejection region at the 5% significance level.

Find the chi-square value for each of the right-tail areas below,given that the degrees of freedom are 7: A)0.95 B)0.01 C)0.025 D)0.05

In carrying out a chi-square goodness-of-fit test,what are the "k" and "m" terms in the "df = k - 1 - m" expression and why is each term present?

For the chi-square goodness-of-fit test,the null hypothesis is rejected whenever:

In the goodness-of-fit chi-square test,under which assumption are the expected frequencies constructed?

In a goodness-of-fit chi-square test,a large amount of discrepancy between the frequencies that are observed and those that are expected tend to cause us to fail to reject the null hypothesis.

For all practical purposes,the test for equality of proportions is really just a special case for the independence of two variables.

In a chi-squared test of a contingency table,the value of the test statistic was $\chi$ ²= 12.678,and the critical value at $\alpha$ = 0.025 was 14.4494.Thus:

In a hypothesis test for the population variance,the alternate hypothesis is the population variance does not equal 17.0.The significance level to be used is 0.05 and the sample size to be taken is 25.The table value(s)to use from the chi-square distribution is (are):

The number of degrees of freedom in a test of a contingency table with r rows and c columns is _________________________.

A Preview of Business Statistics

Visual Description of Data

Statistical Description of Data

Data Collection and Sampling Methods

Probability: Review of Basic Concepts

Discrete Probability Distributions

Continuous Probability Distributions

Sampling Distributions

Estimation From Sample Data

Hypothesis Tests Involving a Sample Mean or Proportion

Hypothesis Tests Involving Two Sample Means

Analysis of Variance Tests

Nonparametric Methods

Simple Linear Regression and Correlation

Multiple Regression and Correlation

Model Building

Models for Time Series and Forecasting

Decision Theory

Total Quality Management

Filters

Exam 13: Chi-Square Applications

Find the chi-square value for each of the right-tail areas below,given that the degrees of freedom are 7: A)0.95 ____________________ B)0.01 ____________________ C)0.025 ____________________ D)0.05 ____________________

In carrying out a chi-square goodness-of-fit test,what are the "k" and "m" terms in the "df = k - 1 - m" expression and why is each term present?

For the chi-square goodness-of-fit test,the null hypothesis is rejected whenever:

In the goodness-of-fit chi-square test,under which assumption are the expected frequencies constructed?

In a goodness-of-fit chi-square test,a large amount of discrepancy between the frequencies that are observed and those that are expected tend to cause us to fail to reject the null hypothesis.

For all practical purposes,the test for equality of proportions is really just a special case for the independence of two variables.

In a chi-squared test of a contingency table,the value of the test statistic was χ\chiχ 2 = 12.678,and the critical value at α\alphaα = 0.025 was 14.4494.Thus:

In a hypothesis test for the population variance,the alternate hypothesis is the population variance does not equal 17.0.The significance level to be used is 0.05 and the sample size to be taken is 25.The table value(s)to use from the chi-square distribution is (are):

The number of degrees of freedom in a test of a contingency table with r rows and c columns is _________________________.

A Preview of Business Statistics

Visual Description of Data

Statistical Description of Data

Data Collection and Sampling Methods

Probability: Review of Basic Concepts

Discrete Probability Distributions

Continuous Probability Distributions

Sampling Distributions

Estimation From Sample Data

Hypothesis Tests Involving a Sample Mean or Proportion

Hypothesis Tests Involving Two Sample Means

Analysis of Variance Tests

Nonparametric Methods

Simple Linear Regression and Correlation

Multiple Regression and Correlation

Model Building

Models for Time Series and Forecasting

Decision Theory

Total Quality Management

Filters

Find the chi-square value for each of the right-tail areas below,given that the degrees of freedom are 7: A)0.95 B)0.01 C)0.025 D)0.05

In a chi-squared test of a contingency table,the value of the test statistic was $\chi$ ²= 12.678,and the critical value at $\alpha$ = 0.025 was 14.4494.Thus: