In the chi-square formula, the letter O stands for the observed count that was actually recorded in the sample.
Missing Word
observed
The observed count is the number of sample results that actually fell in a category or cell. The chi-square statistic measures how far these observed counts are from the counts predicted by H0.
Category: Test Statistic
Question 2
The letter E in the chi-square formula stands for the expected count, the number predicted for a category if H0 is true.
Missing Word
expected
The expected count is what a category or cell should contain if the null hypothesis is true. It is found from the claimed distribution or, in a two-way table, from row total times column total over grand total.
Category: Test Statistic
Question 3
The chi-square test that asks whether two separate populations share the same distribution of one categorical variable is the test for homogeneity.
Missing Word
homogeneity
Homogeneity means sameness. The test for homogeneity takes samples from different populations and checks whether one categorical variable is distributed the same way in each.
Category: Comparing Tests
Question 4
The chi-square test that uses one sample to decide whether two categorical variables are related is the test of independence.
Missing Word
independence
The test of independence classifies each member of a single sample on two variables. Its null hypothesis is that the variables are unrelated, meaning knowing one tells nothing about the other.
Category: Comparing Tests
Question 5
A two-way table that shows how many sample members fall into each combination of two categorical variables is also called a contingency table.
Missing Word
contingency
A contingency table lists the counts for every combination of the categories of two variables. Chi-square tests of independence and homogeneity are computed from these tables.
Category: Comparing Tests
Question 6
The statistic (n - 1)s^2 / sigma^2 is used to test a claim about a population variance.
Missing Word
variance
The variance measures spread as the average squared distance from the mean. The statistic compares the sample variance s^2 with the hypothesized population value sigma^2 and follows a chi-square distribution with n - 1 degrees of freedom.
Category: Single Variance
Question 7
Testing a claim about the spread of one population with a chi-square statistic requires that the population have a normal distribution.
Missing Word
normal
The chi-square test about a single population's spread assumes the population is normal. Unlike many tests about means, it is sensitive to departures from this assumption.
Category: Single Variance
Question 8
The chi-square curve is not symmetric; its long tail stretches toward large values, so it is skewed to the right.
Missing Word
right
A distribution skewed to the right has a long tail on the side of larger values. Chi-square curves start at zero and trail off to the right, with the skew lessening as the degrees of freedom increase.
Category: Chi-Square Distribution
Question 9
A goodness-of-fit test can check whether the number of successes in a fixed number of independent trials follows a binomial distribution.
Missing Word
binomial
A binomial distribution describes the count of successes in a fixed number of independent trials with the same success probability. A goodness-of-fit test compares observed counts of 0, 1, 2 and more successes with the counts that distribution predicts.
Category: Goodness of Fit
Question 10
A goodness-of-fit test that checks whether all categories are equally likely compares the data with a uniform distribution.
Missing Word
uniform
A uniform distribution gives every category the same probability. Under this null hypothesis, each expected count is the sample size divided by the number of categories.
Category: Goodness of Fit
Question 11
When the p-value of a goodness-of-fit test falls below the significance level, the researcher should reject H0.
Missing Word
reject
To reject H0 means the data give enough evidence against it. A p-value below alpha shows that counts this far from the claimed distribution would be unlikely if H0 were true.
Category: Goodness of Fit
Question 12
For a two-way table, the degrees of freedom equal (number of rows minus 1) times (number of columns minus 1).
Missing Word
columns
The columns of a two-way table hold the categories of one variable and the rows hold the categories of the other. Both tests built on such tables use df = (rows - 1)(columns - 1).
Category: Test Statistic
Question 13
Each term of the chi-square statistic takes the difference O - E, which is then squared and divided by E.
Missing Word
squared
The difference is squared so that positive and negative gaps cannot cancel each other out. Dividing by E scales each gap relative to the size of the count expected in that category.
Category: Test Statistic
Question 14
Because each of its terms is a square divided by a positive count, a chi-square statistic can never be negative.
Missing Word
negative
A squared number is zero or positive, and dividing it by a positive expected count keeps it that way. The chi-square statistic is therefore zero or positive, and it equals zero only when every observed count matches its expected count.
Category: Test Statistic
Question 15
In a goodness-of-fit test, the alternative hypothesis states that the data do not fit the claimed distribution.
Missing Word
alternative
The alternative hypothesis is the claim the test looks for evidence to support. In a goodness-of-fit test it is the statement that the population does not follow the distribution named in H0.
Category: Goodness of Fit