Question 1

Which of the following statements is true for chi-squared tests?

Accepted Answer

A)  Testing for equal proportions is identical to testing for goodness-of-fit. 
B)  The number of degrees of freedom in a test of a contingency table with r rows and c columns is (r -1)(c-1). 
C)  The number of degrees of freedom in a goodness-of-fit test with k categories is k - 1. 
D)  All of these choices are true. 
A)  Testing for equal proportions is identical to testing for goodness-of-fit. 
B)  The number of degrees of freedom in a test of a contingency table with r rows and c columns is (r -1)(c-1). 
C)  The number of degrees of freedom in a goodness-of-fit test with k categories is k - 1. 
D)  All of these choices are true.

Question 2

The degrees of freedom for the test statistic in a test of a contingency table is (r - 1)(c -1) where r is the number of rows in the table, and c is the number of columns.

Accepted Answer

A) True 
 B)False

Question 3

When we test for differences between two populations of nominal data with two categories, we can use only one technique, namely, the chi-squared test of a contingency table.

Accepted Answer

A) True 
 B)False

Question 4

In 2011, Brand A MP3 Players had 45% of the market, Brand B had 35%, and Brand C had 20%.This year the makers of brand C launched a heavy advertising campaign.A random sample of electronic stores shows that of 10,000 MP3 Players sold, 4,350 were Brand A, 3,450 were Brand B, and 2,200 were Brand C.Has the market changed? Test at $\alpha$ = .01.

Accepted Answer

H₀: p₁ = .45, p₂ = .35, p₃ = .20 H₁: At least

Question 5

A major insurance firm interviewed a random sample of 1,200 college students to find out the type of life insurance preferred, if any.The results follow:
   Is there evidence that life insurance preference of male students is different than that of female students? Test using the 5% level of significance.

Accepted Answer

H₀: Gender and life insurance preference

Question 6

To test for normality, the ____________________ hypothesis specifies probabilities of certain intervals within the normal distribution.

Accepted Answer

The answer of To test for normality, the ____________________ hypothesis...

Question 7

The rule of five requires that the:

Accepted Answer

A)  observed frequency for each cell must be at least 5. 
B)  degrees of freedom for the test must be at least 5. 
C)  expected frequency for each cell must be at least 5. 
D)  difference between the observed and expected frequency for each cell must be at least 5. 
A)  observed frequency for each cell must be at least 5. 
B)  degrees of freedom for the test must be at least 5. 
C)  expected frequency for each cell must be at least 5. 
D)  difference between the observed and expected frequency for each cell must be at least 5.

Question 8

In 2011, the student body of a state university in Alabama consists of 30% freshmen, 25% sophomores, 27% juniors, and 18% seniors.A sample of 400 students taken from the 2012 student body showed that there are 138 freshmen, 88 sophomores, 94 juniors, and 80 seniors.Test with 5% significance level to determine whether the student body proportions have changed.

Accepted Answer

H₀: p₁ = .30, p₂ =.25, p₃ = 0.27, p₄ = .18 H₁:

Question 9

Of the values for a chi-squared test statistic listed below, which one is most likely to lead to rejecting the null hypothesis in a goodness-of-fit test?

Accepted Answer

A)  0 
B)  .05 
C)  1.96 
D)  45 
A)  0 
B)  .05 
C)  1.96 
D)  45

Question 10

The alternative hypothesis of a goodness-of-fit test states that ____________________ of the proportions is not equal to its value specified in H0.

Accepted Answer

Question 11

A sports preference poll showed the following data for adults and children:
   Use the 5% level of significance and test to determine whether sports preferences depend on age group.

Accepted Answer

H₀: Age group and sports preferences are

Question 12

In a chi-squared test of a contingency table, the value of the test statistic was  = 15.652, and the critical value at $\alpha$ = .025 was 11.1433.Thus, we must reject the null hypothesis at $\alpha$ = .025.

Accepted Answer

A) True 
 B)False

Question 13

You cannot use a chi-squared goodness-of-fit test when there are only two possible outcomes for each trial in your experiment.

Accepted Answer

A) True 
 B)False

Question 14

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 = .10, pS1U1B12S1U1B0 = .25, pS1U1B13S1U1B0 = .30, pS1U1B14S1U1B0 = .20, pS1U1B15S1U1B0 = .15.
-{Student Absenteeism Narrative} Test the hypothesis at the 5% level of significance with the following frequencies:

Accepted Answer

Rejection region: @#IMG-DLM& > @#IMG-DLM& .05,4S1U1B0 = 9.48

Question 15

To calculate the expected values in a test of a contingency table, you assume that the null hypothesis is true.

Accepted Answer

A) True 
 B)False

Question 16

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} How many degrees of freedom are associated with the proper test in this experiment?

Accepted Answer

df = (2 -

Question 17

The number of degrees of freedom associated with a chi-squared test of a contingency table is ____________________.

Accepted Answer

(r-1)(c-1)

Question 18

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 200 observations.In addition, the researcher used 5 standardized intervals to test for normality.Using a 10% level of significance, the critical value for this test is 4.60517.

Accepted Answer

A) True 
 B)False

Question 19

The chi-squared test for normality must follow the rule of ____________________ regarding expected values.

Accepted Answer

The answer of The chi-squared test for normality must follow...

Question 20

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

Accepted Answer

A) True 
 B)False

Which of the following statements is true for chi-squared tests?

The degrees of freedom for the test statistic in a test of a contingency table is (r - 1)(c -1) where r is the number of rows in the table, and c is the number of columns.

When we test for differences between two populations of nominal data with two categories, we can use only one technique, namely, the chi-squared test of a contingency table.

To test for normality, the ____________________ hypothesis specifies probabilities of certain intervals within the normal distribution.

The rule of five requires that the:

Of the values for a chi-squared test statistic listed below, which one is most likely to lead to rejecting the null hypothesis in a goodness-of-fit test?

The alternative hypothesis of a goodness-of-fit test states that ____________________ of the proportions is not equal to its value specified in H₀.

A sports preference poll showed the following data for adults and children: Use the 5% level of significance and test to determine whether sports preferences depend on age group.

You cannot use a chi-squared goodness-of-fit test when there are only two possible outcomes for each trial in your experiment.

To calculate the expected values in a test of a contingency table, you assume that the null hypothesis is true.

The number of degrees of freedom associated with a chi-squared test of a contingency table is ____________________.

The chi-squared test for normality must follow the rule of ____________________ regarding expected values.

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

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

Which of the following statements is true for chi-squared tests?

The degrees of freedom for the test statistic in a test of a contingency table is (r - 1)(c -1) where r is the number of rows in the table, and c is the number of columns.

When we test for differences between two populations of nominal data with two categories, we can use only one technique, namely, the chi-squared test of a contingency table.

To test for normality, the ____________________ hypothesis specifies probabilities of certain intervals within the normal distribution.

The rule of five requires that the:

Of the values for a chi-squared test statistic listed below, which one is most likely to lead to rejecting the null hypothesis in a goodness-of-fit test?

The alternative hypothesis of a goodness-of-fit test states that ____________________ of the proportions is not equal to its value specified in H0.

A sports preference poll showed the following data for adults and children: Use the 5% level of significance and test to determine whether sports preferences depend on age group.

In a chi-squared test of a contingency table, the value of the test statistic was = 15.652, and the critical value at α\alphaα = .025 was 11.1433.Thus, we must reject the null hypothesis at α\alphaα = .025.

You cannot use a chi-squared goodness-of-fit test when there are only two possible outcomes for each trial in your experiment.

To calculate the expected values in a test of a contingency table, you assume that the null hypothesis is true.

The number of degrees of freedom associated with a chi-squared test of a contingency table is ____________________.

The chi-squared test for normality must follow the rule of ____________________ regarding expected values.

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

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

The alternative hypothesis of a goodness-of-fit test states that ____________________ of the proportions is not equal to its value specified in H₀.