Question

We wish to determine if a two sections of the same introductory course have significantly different “success rates” (defined as the proportion of students who receive a course grade of A, B, or C). The first section meets in the early morning, the second section meets in the late afternoon. Each section has 70 students. Among the early morning section, 59 receive an A, B, or C. Among the late afternoon section, 49 receive an A, B, or C.. Assume these can be treated as independent simple random samples from their respective populations. Use this sample data to test the claim H0:(p1−p2)=0 against HA:(p1−p2)≠0, using a significance level of 5%.

The value of the test statistic is z=

(round to at least four decimal places).

The P-value for this sample is

A social media platform wants to determine if there is a significant difference between the average weekly usage (number of minutes spent on the site per week) of female users and male users, and plans to conduct a Hypothesis Test at the 5% significance level. Let μ1 be the average daily usage among all male users, and μ2 be the average daily usage among all female users.

An appropriate alternative hypothesis is:

In a simple random sample of 35 male users, the mean daily usage is 115.9 mintues, with a standard deviation of 8.07 minutes. An independent simple random sample of 46 female users has a mean daily usage of 113 minutes and a standard deviation of 7.24 minutes.

The value of the test statistic is t=

. The P-Value is

Answer 1

1) = 59/70 = 0.843

    = 49/70 = 0.7

The pooled proportion(P) = ( * n1 + * n2)/(n1 + n2)

                                         = (0.843 * 70 + 0.7 * 70)/(70 + 70)

                                         = 0.7715

SE = sqrt(P(1 - P)(1/n1 + 1/n2))

      = sqrt(0.7715 * (1 - 0.7715) * (1/70 + 1/70))

      = 0.071

The test statistic z = ()/SE

                            = (0.843 - 0.7)/0.071

                            = 2.01

P-value = 2 * P(Z > 2.01)

             = 2 * (1 - P(Z < 2.01))

             = 2 * (1 - 0.9778)

             = 0.0444

2) The test statistic

       

     

P-value = 2 * P(T > 1.674)

             = 2 * (1 - P(T < 1.674))

              = 2 * (1 - 0.9506)

              = 0.0988