As the degrees of freedom increase, the chi-square curve becomes more symmetric and looks more like a normal curve.
Correct Answer
True
True. Chi-square curves with small df are strongly skewed to the right, and the skew fades as df grows. With very large df, the curve is close to a normal curve with mean df and standard deviation sqrt(2df).
Category: Chi-Square Distribution
Question 2
The mean of a chi-square distribution equals twice its degrees of freedom.
Correct Answer
False
False. The mean of a chi-square distribution equals its degrees of freedom. Twice the degrees of freedom, 2(df), is the variance, and its square root is the standard deviation.
Category: Chi-Square Distribution
Question 3
A chi-square random variable with k degrees of freedom can be built as the sum of k independent squared standard normal variables.
Correct Answer
True
True. Squaring k independent standard normal variables and adding them gives a chi-square variable with k degrees of freedom. This construction also explains why chi-square values can never be negative.
Category: Chi-Square Distribution
Question 4
In a goodness-of-fit test, the degrees of freedom equal the number of observations minus 1.
Correct Answer
False
False. In a goodness-of-fit test, df equals the number of categories minus 1, not the number of observations minus 1. A die rolled 600 times still has 6 categories, so df = 5.
Category: Goodness of Fit
Question 5
The null hypothesis of a goodness-of-fit test states that the population fits the claimed distribution.
Correct Answer
True
True. The goodness-of-fit null hypothesis says the data come from the stated distribution, such as a uniform distribution or a set of known percentages. The alternative says the population does not fit that distribution.
Category: Goodness of Fit
Question 6
In a test of independence, a cell's expected count equals its row total times its column total divided by the number of cells in the table.
Correct Answer
False
False. The expected count is the row total times the column total divided by the grand total, the total number of observations. Dividing by the number of cells would give far too large a value.
Category: Test of Independence
Question 7
A test of independence uses one sample in which each member is classified on two categorical variables.
Correct Answer
True
True. A test of independence takes a single sample and records two categorical variables for each member, arranged in a contingency table. It then asks whether the two variables are related.
Category: Test of Independence
Question 8
A test for homogeneity compares the means of a quantitative variable across two populations.
Correct Answer
False
False. A test for homogeneity compares the distributions of one categorical variable across two or more populations. Comparing means of a quantitative variable calls for a t-test or ANOVA, not a chi-square test.
Category: Test of Homogeneity
Question 9
The degrees of freedom for a test for homogeneity are found the same way as for a test of independence, as (rows - 1)(columns - 1).
Correct Answer
True
True. Both tests organize counts in a two-way table and use df = (rows - 1)(columns - 1). The expected counts are also computed the same way, as row total times column total divided by the grand total.
Category: Test of Homogeneity
Question 10
A chi-square test of a single variance can only be right-tailed.
Correct Answer
False
False. A test of a single variance can be left-tailed, right-tailed or two-tailed, depending on whether the claim is that the variance is smaller than, larger than or different from the hypothesized value.
Category: Single Variance
Question 11
When the population is normal, the statistic (n - 1)s^2 / sigma^2 follows a chi-square distribution with n - 1 degrees of freedom.
Correct Answer
True
True. This is the test statistic for a single variance, where s^2 is the sample variance and sigma^2 is the hypothesized population variance. The result depends on the normality assumption.
Category: Single Variance
Question 12
A very small goodness-of-fit statistic is strong evidence against the null hypothesis.
Correct Answer
False
False. A small statistic means the observed counts are close to the expected counts, which is consistent with H0. Evidence against H0 comes from a large statistic in the right tail.
Category: Goodness of Fit
Question 13
Observed counts in a chi-square test are whole numbers, but expected counts can be decimals.
Correct Answer
True
True. Observed counts are tallies of actual outcomes, so they are whole numbers. Expected counts come from formulas such as n times a proportion or row total times column total over grand total, so values like 12.5 are common and should not be rounded off.
Category: Goodness of Fit
Question 14
A test of independence on a 2 x 2 table has 4 degrees of freedom.
Correct Answer
False
False. The degrees of freedom are (rows - 1)(columns - 1) = (2 - 1)(2 - 1) = 1. The value 4 is the number of cells, not the degrees of freedom.
Category: Test of Independence
Question 15
A significant chi-square test of independence shows that two variables are associated, but it does not by itself show that one causes the other.
Correct Answer
True
True. Rejecting independence shows only that the variables are related in the population. Establishing that one causes the other requires a well-designed experiment, not just an observed association.
Category: Test of Independence
Question 16
In a goodness-of-fit test, the expected counts come from the sample data rather than from the claimed distribution.
Correct Answer
False
False. Expected counts come from the claimed distribution, found by multiplying the sample size by the proportion H0 assigns to each category. The sample supplies the observed counts.
Category: Goodness of Fit
Question 17
Hypotheses for a single variance test may be written in terms of the population standard deviation instead of the variance.
Correct Answer
True
True. Because the standard deviation is the square root of the variance, a claim such as sigma > 3 is equivalent to sigma^2 > 9. Either form leads to the same chi-square test.
Category: Single Variance
Question 18
In a test for homogeneity, the null hypothesis states that the two populations have different distributions.
Correct Answer
False
False. The null hypothesis of a test for homogeneity states that the populations have the same distribution. The claim that the distributions differ is the alternative hypothesis.
Category: Test of Homogeneity
Question 19
The standard deviation of a chi-square distribution with 18 degrees of freedom is 6.
Correct Answer
True
True. The standard deviation of a chi-square distribution is sqrt(2df). With df = 18, that is sqrt(36) = 6.
Category: Chi-Square Distribution
Question 20
A single variance test based on a sample of 25 values uses 25 degrees of freedom.
Correct Answer
False
False. A test of a single variance uses df = n - 1, so a sample of 25 values gives 24 degrees of freedom.
Category: Single Variance