Exam 15: Chi-Squared Tests

Seat Belts A study was conducted to determine whether the use of seat belts in vehicles depends on whether or not a child was present in the car.A sample of 1,000 people treated for injuries sustained from vehicle accidents was obtained,and each person was classified according to (1)child present (yes/no)and (2)seat belt usage (worn or not worn)during the accident.The data are shown in the table below. -{Seat Belts Narrative} Prepare the table of expected values.

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Question 1

Correct Answer:

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A chi-squared goodness-of-fit test is always conducted as a(n):

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Question 2

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Seat Belts A study was conducted to determine whether the use of seat belts in vehicles depends on whether or not a child was present in the car.A sample of 1,000 people treated for injuries sustained from vehicle accidents was obtained,and each person was classified according to (1)child present (yes/no)and (2)seat belt usage (worn or not worn)during the accident.The data are shown in the table below. -{Seat Belts Narrative} Calculate the value of the test statistic.

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Question 3

Correct Answer:

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χ² = 26.016

The only way the chi-squared test statistic can be zero is if the observed frequencies are all exactly the same as the expected frequencies.

(True/False)

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Question 4

The owner of a consumer products 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 age.Is there sufficient evidence to conclude that the level of job satisfaction is related to age? Use α = .10.

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Question 5

If two events A and B are independent,the P(A and B)= P(A)+ P(B).

(True/False)

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Question 6

A chi-squared test of a contingency table with 10 degrees of freedom results in a test statistic of 17.894.Using the chi-squared table,the most accurate statement that can be made about the p-value for this test is that .05 < p-value < .10.

(True/False)

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Question 7

In the test of a contingency table,the observed cell frequencies must satisfy the rule of 5.

(True/False)

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Question 8

The sampling distribution of the test statistic for a goodness-of-fit test with k categories is a:

(Multiple Choice)

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Question 9

In the test of a contingency table,the expected cell frequencies must satisfy the rule of 5.

(True/False)

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Question 10

Seat Belts A study was conducted to determine whether the use of seat belts in vehicles depends on whether or not a child was present in the car.A sample of 1,000 people treated for injuries sustained from vehicle accidents was obtained,and each person was classified according to (1)child present (yes/no)and (2)seat belt usage (worn or not worn)during the accident.The data are shown in the table below. -{Seat Belts Narrative} State the appropriate null and alternative hypotheses for this experiment.

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Question 11

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

(Multiple Choice)

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Question 12

If we want to perform a two-tail test for differences between two populations of nominal data with exactly two categories,we can employ either the z-test of p₁ − p₂,or the chi-squared test of a contingency table.

(True/False)

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Question 13

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

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Question 14

Suppose that a random sample of 150 observations was drawn from a population.After calculating the mean and standard deviation,each observation was standardized and the number of observations in each of the intervals below was counted.Can we infer at the 5% significance level that the data were drawn from a normal population?

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Question 15

A chi-squared distribution is symmetric.

(True/False)

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Question 16

A left-tailed area in the chi-squared distribution equals .90.For 10 degrees of freedom the table value equals 15.9871.

(True/False)

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Question 17

The rule of five states that in order to conduct the chi-squared goodness-of-fit test,the ____________________ value for each cell must be five or more.

(Short Answer)

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Question 18

Consider a multinomial experiment with 200 trials,where the outcome of each trial is classified into one of 5 categories.The number of degrees of freedom associated with the chi-squared goodness-of-fit test equals:

(Multiple Choice)

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Question 19

The number of ATVs sold by three salespersons over a 3-month period is shown below: Use the 5% level of significance and test for the independence of salesperson and type of product sold.

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Question 20

Showing 1 - 20 of 179

A chi-squared goodness-of-fit test is always conducted as a(n):

The only way the chi-squared test statistic can be zero is if the observed frequencies are all exactly the same as the expected frequencies.

The owner of a consumer products 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 age.Is there sufficient evidence to conclude that the level of job satisfaction is related to age? Use α = .10.

If two events A and B are independent,the P(A and B)= P(A)+ P(B).

A chi-squared test of a contingency table with 10 degrees of freedom results in a test statistic of 17.894.Using the chi-squared table,the most accurate statement that can be made about the p-value for this test is that .05 < p-value < .10.

In the test of a contingency table,the observed cell frequencies must satisfy the rule of 5.

The sampling distribution of the test statistic for a goodness-of-fit test with k categories is a:

In the test of a contingency table,the expected cell frequencies must satisfy the rule of 5.

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

If we want to perform a two-tail test for differences between two populations of nominal data with exactly two categories,we can employ either the z-test of p₁ − p₂,or the chi-squared test of a contingency table.

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

A chi-squared distribution is symmetric.

A left-tailed area in the chi-squared distribution equals .90.For 10 degrees of freedom the table value equals 15.9871.

The rule of five states that in order to conduct the chi-squared goodness-of-fit test,the ____________________ value for each cell must be five or more.

Consider a multinomial experiment with 200 trials,where the outcome of each trial is classified into one of 5 categories.The number of degrees of freedom associated with the chi-squared goodness-of-fit test equals:

The number of ATVs sold by three salespersons over a 3-month period is shown below: Use the 5% level of significance and test for the independence of salesperson and type of product sold.

What Is Statistics

Graphical Descriptive Techniques 1

Graphical Descriptive Techniques 2

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

Model Building

Nonparametric Statistics

Time-Series Analytics and Forecasting

Statistical Process Control

Decision Analysis

Conclusion

Filters

Exam 15: Chi-Squared Tests

A chi-squared goodness-of-fit test is always conducted as a(n):

The only way the chi-squared test statistic can be zero is if the observed frequencies are all exactly the same as the expected frequencies.

The owner of a consumer products 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 age.Is there sufficient evidence to conclude that the level of job satisfaction is related to age? Use α = .10.

If two events A and B are independent,the P(A and B)= P(A)+ P(B).

A chi-squared test of a contingency table with 10 degrees of freedom results in a test statistic of 17.894.Using the chi-squared table,the most accurate statement that can be made about the p-value for this test is that .05 < p-value < .10.

In the test of a contingency table,the observed cell frequencies must satisfy the rule of 5.

The sampling distribution of the test statistic for a goodness-of-fit test with k categories is a:

In the test of a contingency table,the expected cell frequencies must satisfy the rule of 5.

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

If we want to perform a two-tail test for differences between two populations of nominal data with exactly two categories,we can employ either the z-test of p1 − p2,or the chi-squared test of a contingency table.

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

A chi-squared distribution is symmetric.

A left-tailed area in the chi-squared distribution equals .90.For 10 degrees of freedom the table value equals 15.9871.

The rule of five states that in order to conduct the chi-squared goodness-of-fit test,the ____________________ value for each cell must be five or more.

Consider a multinomial experiment with 200 trials,where the outcome of each trial is classified into one of 5 categories.The number of degrees of freedom associated with the chi-squared goodness-of-fit test equals:

The number of ATVs sold by three salespersons over a 3-month period is shown below: Use the 5% level of significance and test for the independence of salesperson and type of product sold.

What Is Statistics

Graphical Descriptive Techniques 1

Graphical Descriptive Techniques 2

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

Model Building

Nonparametric Statistics

Time-Series Analytics and Forecasting

Statistical Process Control

Decision Analysis

Conclusion

Filters

If we want to perform a two-tail test for differences between two populations of nominal data with exactly two categories,we can employ either the z-test of p₁ − p₂,or the chi-squared test of a contingency table.