A chi-square distribution has 10 degrees of freedom. What is its mean?
10
✓
20
4.47
5
Correct Answer
10
The mean of a chi-square distribution equals its degrees of freedom, so mu = 10. The value 20 is the variance, 2(df), not the mean. The value 4.47 is the standard deviation, sqrt(20). The value 5 is half the degrees of freedom, which no chi-square formula gives.
Category: Chi-Square Distribution
Question 2
For a fixed significance level such as 0.05, what happens to the right-tail chi-square critical value as the degrees of freedom increase?
It decreases
It increases
✓
It stays the same
It approaches zero
Correct Answer
It increases
The center of a chi-square curve is at its mean, which equals df, so the whole curve shifts right as df grows and the cutoff moves right with it. At alpha = 0.05 the critical value is about 3.84 for df = 1, 5.99 for df = 2 and 7.81 for df = 3. It decreases is the opposite of what happens. It stays the same ignores that each df gives a different curve. It approaches zero would mean the cutoff moves toward the left edge, which it does not.
Category: Chi-Square Distribution
Question 3
A genetics model predicts that plants show a dominant trait and a recessive trait in a 3:1 ratio. In a goodness-of-fit test on 80 plants, what is the expected count for the dominant trait?
20
40
60
✓
75
Correct Answer
60
A 3:1 ratio means the dominant trait is expected in 3/4 of the plants, so the expected count is 0.75 x 80 = 60. The value 20 is the expected count for the recessive trait, 1/4 of 80. The value 40 assumes the two traits are equally likely. The value 75 treats the 75 percent share as if it were a count.
Category: Goodness of Fit
Question 4
A die is rolled 120 times to test whether its six faces are equally likely. How many degrees of freedom does the goodness-of-fit test have?
6
119
4
5
✓
Correct Answer
5
For a goodness-of-fit test, df equals the number of categories minus 1, so six faces give df = 6 - 1 = 5. The value 6 forgets to subtract 1. The value 119 wrongly subtracts 1 from the number of rolls instead of the number of categories. The value 4 subtracts 2 instead of 1.
Category: Goodness of Fit
Question 5
A shop surveys 200 customers to test whether preference is spread evenly among 4 ice cream flavors. Under the null hypothesis, what is the expected count for each flavor?
50
✓
200
4
40
Correct Answer
50
If preference is spread evenly, each flavor gets 1/4 of the 200 customers, so each expected count is 200 / 4 = 50. The value 200 is the total sample, not a single flavor's share. The value 4 is the number of flavors. The value 40 divides by 5 instead of by the 4 flavors.
Category: Goodness of Fit
Question 6
In one category of a goodness-of-fit test, the observed count is 30 and the expected count is 20. What does this category contribute to the chi-square statistic? Round to two decimal places if needed.
10
5
✓
100
3.33
Correct Answer
5
Each category adds (O - E)^2 / E to the statistic, which here is (30 - 20)^2 / 20 = 100 / 20 = 5. The value 10 is the raw difference O - E, before squaring or dividing. The value 100 is the squared difference without dividing by E. The value 3.33 divides by the observed count, 100 / 30, instead of the expected count.
Category: Goodness of Fit
Question 7
A spinner with three equal sections is spun 60 times and lands on the sections 18, 22 and 20 times. The expected count is 20 for each section. What is the chi-square goodness-of-fit statistic?
8
0.4
✓
0.2
Correct Answer
0.4
The terms are (18 - 20)^2 / 20 = 0.2, (22 - 20)^2 / 20 = 0.2 and (20 - 20)^2 / 20 = 0, which add to 0.4. The value 0 comes from adding the raw differences, which always cancel to zero. The value 8 adds the squared differences without dividing each by 20. The value 0.2 counts only one of the two sections that differ from 20.
Category: Goodness of Fit
Question 8
A chi-square goodness-of-fit test is run at alpha = 0.05. Where on the chi-square curve is its rejection region?
Left tail only
Both tails
Center of the curve
Right tail only
✓
Correct Answer
Right tail only
The goodness-of-fit statistic adds up (O - E)^2 / E over the categories, so it grows as the observed counts move away from the expected counts, and only a large value signals a poor fit. The test therefore rejects H0 for values far out in the right tail. Left tail only would reject when the fit is unusually close, which is not evidence against H0. Both tails would also count very small statistics as evidence against H0. Center of the curve holds the typical values expected when H0 is true.
Category: Goodness of Fit
Question 9
A chi-square test of whether two categorical variables are related uses a table with 3 rows and 4 columns. How many degrees of freedom does the test have?
6
✓
12
7
11
Correct Answer
6
For a two-way table, df = (rows - 1)(columns - 1) = (3 - 1)(4 - 1) = 2 x 3 = 6. The value 12 is the number of cells, 3 x 4. The value 7 adds 3 and 4 instead of using the formula. The value 11 subtracts 1 from the number of cells.
Category: Test of Independence
Question 10
In a two-way table with a grand total of 200, one cell has a row total of 40 and a column total of 50. Under the null hypothesis, what is the expected count for that cell?
20
10
✓
45
2000
Correct Answer
10
The expected count for a cell is (row total)(column total) / grand total = (40)(50) / 200 = 2000 / 200 = 10. The value 20 divides by 100 instead of by the grand total of 200. The value 45 simply averages the row and column totals. The value 2000 multiplies the totals but never divides by the grand total.
Category: Test of Independence
Question 11
One sample is classified by two categorical variables to see whether the variables are related. Which statement is the null hypothesis of the chi-square test?
The two variables are dependent
The two population means are equal
The two variables are independent
✓
The data fit a uniform distribution
Correct Answer
The two variables are independent
The null hypothesis states that there is no relationship, meaning the two variables are independent. The two variables are dependent is the alternative hypothesis. The two population means are equal is a hypothesis about means, which chi-square tests of counts do not address. The data fit a uniform distribution is a goodness-of-fit null hypothesis for one variable.
Category: Test of Independence
Question 12
Separate random samples are taken at two colleges, and each student names a preferred study spot (library, home or cafe). Which test checks whether the distribution of preferences is the same at both colleges?
Test of independence
Goodness-of-fit test
Test of a single variance
Test for homogeneity
✓
Correct Answer
Test for homogeneity
A test for homogeneity asks whether two different populations have the same distribution of one categorical variable, which matches samples drawn separately from two colleges. Test of independence uses one sample classified on two variables. Goodness-of-fit test compares one sample to a fixed, claimed distribution. Test of a single variance concerns the spread of one quantitative variable.
Category: Choosing a Test
Question 13
One random sample of 300 adults is classified by age group and by preferred news source. Which test checks whether news source preference is related to age group?
Test of independence
✓
Test for homogeneity
Goodness-of-fit test
Test of a single variance
Correct Answer
Test of independence
A test of independence uses a single sample in which each member is classified on two categorical variables, then asks whether the variables are related. Test for homogeneity applies when separate samples come from different populations measured on one variable. Goodness-of-fit test involves only one variable compared with a claimed distribution. Test of a single variance concerns the spread of a quantitative variable.
Category: Choosing a Test
Question 14
A quality engineer wants to know whether the spread of bottle fill amounts is larger than a claimed population standard deviation. Which test fits this question?
Goodness-of-fit test
Test of a single variance
✓
Test of independence
Test for homogeneity
Correct Answer
Test of a single variance
A claim about the spread of one quantitative variable, stated as a standard deviation or variance, is checked with the chi-square test of a single variance. Goodness-of-fit test compares category counts with a claimed distribution. Test of independence asks whether two categorical variables are related. Test for homogeneity compares the distribution of a categorical variable across populations.
Category: Choosing a Test
Question 15
A sample of 21 measurements from a normal population has s^2 = 30. To test H0: sigma^2 = 20, what is the chi-square test statistic? Round to two decimal places if needed.
31.5
13.33
30
✓
1.5
Correct Answer
30
The statistic is (n - 1)s^2 / sigma^2 = (21 - 1)(30) / 20 = 600 / 20 = 30. The value 31.5 uses n = 21 instead of n - 1 = 20. The value 13.33 flips the ratio, putting the hypothesized variance on top. The value 1.5 is only s^2 / sigma^2, leaving out the factor n - 1.
Category: Single Variance
Question 16
Which condition must hold for the chi-square test about one population variance to be valid?
Every expected count is at least 10
The sample size is exactly 30
The population standard deviation is known
The population is normally distributed
✓
Correct Answer
The population is normally distributed
The chi-square test of a single variance assumes the underlying population is normal, and it can give misleading results when that assumption fails. Every expected count is at least 10 confuses this test with the count-based chi-square tests, which have expected counts. The sample size is exactly 30 is not a requirement of any chi-square test. The population standard deviation is known is wrong because that value is only the hypothesized value being tested.
Category: Single Variance
Question 17
A manufacturer wants evidence that the variance of its fill amounts is below 4. Which alternative hypothesis matches this goal?
Ha: sigma^2 < 4
✓
Ha: sigma^2 > 4
Ha: sigma^2 = 4
Ha: s^2 < 4
Correct Answer
Ha: sigma^2 < 4
The claim needing evidence goes in the alternative hypothesis, and it concerns the population variance, so Ha: sigma^2 < 4, a left-tailed test. Ha: sigma^2 > 4 points in the wrong direction. Ha: sigma^2 = 4 is the form of the null hypothesis, not an alternative. Ha: s^2 < 4 is about the sample variance, which is already known and is not what a hypothesis tests.
Category: Single Variance
Question 18
Two populations are compared on one categorical variable with 5 categories, using a chi-square test of whether their distributions match. How many degrees of freedom does the test have?
5
4
✓
8
10
Correct Answer
4
This is a test for homogeneity, and df = (rows - 1)(columns - 1) = (2 - 1)(5 - 1) = 4. The value 5 counts the categories without subtracting 1. The value 8 forgets to subtract 1 for the two populations, computing 2 x 4. The value 10 is the number of cells, 2 x 5.
Category: Test of Homogeneity
Question 19
In a goodness-of-fit test, one category has an expected count that is too small for the test to be reliable. What is the usual fix?
Delete that category's observations
Use a left-tailed test instead
Combine it with a neighboring category
✓
Replace its observed count with 5
Correct Answer
Combine it with a neighboring category
Merging a small category with an adjacent one raises the expected count while keeping every observation in the analysis. Delete that category's observations throws away data and changes the question being tested. Use a left-tailed test instead does nothing about the small expected count. Replace its observed count with 5 alters the real data.
Category: Goodness of Fit
Question 20
A test of whether class year and preferred exercise type are related gives p = 0.003 with alpha = 0.05. Which conclusion is correct?
Fail to reject H0; the variables are independent
Reject H0; the variables are independent
Fail to reject H0; the variables are dependent
Reject H0; the variables appear dependent
✓
Correct Answer
Reject H0; the variables appear dependent
Because 0.003 is less than 0.05, the null hypothesis of independence is rejected and the data support a relationship between the variables. Fail to reject H0; the variables are independent ignores that p is below alpha. Reject H0; the variables are independent states the opposite of what rejecting H0 means. Fail to reject H0; the variables are dependent pairs the wrong decision with a conclusion that does not follow from it.
Category: Test of Independence