Question

State the null and alternative hypotheses, calculate the test statistic, determine the critical value for the given alpha, determine whether to reject or retain the null, and then draw a conclusion about the original study based on the significance test results.

4. The following data were observed when examining the reasons for using teasing statements in the discourse of college students.

Trying to Impress Not Trying to Impress

Males 30 10

Females 20 30

a. (2 point) State the null and alternative hypotheses.

b. (1 point) State what test should be used for this problem and why.

c. (11 points) Perform this test using all the steps listed above using α=.05

d. (2 points) What conclusion can you make about the reasons men and women tease? Can you determine whether men tease to impress more than women do?

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P₁< P₂
Alternative hypothesis: P₁ > P₂

Note that these hypotheses constitute a one-tailed test.

b)

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)

p = 0.5556

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.10541
z = (p₁ - p₂) / SE

z = 3.32

z_critical = + 1.645

Reject H0, if z > 1.645

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

Reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that men tease to impress more than women do.