a. the study was a goodness-of-fit test with n = 12?
b. the study was a test of independence between two categorical variables, the row variable with 3
values and the column variable with 4 values?Answer:
a. n = 12 df = 12 – 1 = 11 0.025 < P < 0.05.
(Using the TI-83/84: χ 2 cdf(20,1000,11) = 0.045.)
b. r = 3, c = 4 df. = (3 – 1)(4 – 1) = 6 0.0025 < P < 0.005.
(Using the TI-83/84: χ 2 cdf(20,1000,6) = 0.0028.)2–4. The following data were collected while conducting a chi-square test for independence:
What null and alternative hypotheses are being tested?
Answer:
H 0 : Gender and Preference are independent (or: H 0 : Gender and Preference are not associated).
H (^) A : Gender and Preference are not independent (H (^) A : Gender and Preference are associated).
What is the expected frequency of the cell marked with the X?
Answer: Identifying the marginals on the table we have
Since there are 34 values in the column with the X, we expect to find of each row total in thecells of the first column. Hence, the expected value for the cell containing X is .
How many degrees of freedom are involved in the test?
Answer: df = (2 – 1)(3 – 1) = 2.
- The null hypothesis for a chi-square goodness-of-fit test is given as:
H 0 : p 1 = 0.2, p 2 = 0.3, p 3 = 0.4, p 4 = 0.1. Which of the following is an appropriate alternative
hypothesis?
a. H (^) A : p 1 ≠ 0.2, p 2 ≠ 0.3, p 3 ≠ 0.4, p 4 ≠ 0.1.