Question

 

1.       A study showed that in a certain month, the mean time spent per visit to Facebook was 19.5 minutes. Assumes the standard deviation of the population is 8 minutes. Suppose that a simple random sample of 100 visits in that month has a sample mean of 21.52 minutes. A social scientist is interested in knowing whether the mean time of Facebook visits has increased. Perform the hypothesis test and compute the P-value.

Based on the P value, what is the conclusion we test at 0.05 level significance?

2.       A random sample of 64 second graders in a certain school district are given a standardized mathematics skills test. The sample mean score is 51.58. Assume the standard deviation for the population of test scores is 15. The nationwide average score on this test us 50. The school superintendent wants to know whether the second graders in her school district have greater math skills than the nationwide average. Perform the hypothesis test and compute the P value.

Based on your P value, what is the conclusion if we test at 0.05 level of significance?

3.       Suppose that the mean price of a home in Denver, Colorado in 2008 was 225.3 thousand dollars. A random sample of 49 homes sold in 2010 had a mean price of 200.1 thousand dollars. A real estate firm wants to test to see if the mean price of 2010 differs from the mean price in 2008. Assume that the population standard deviation is 140. Perform the hypothesis test and compute the P value.

Based on your P value, what is the conclusion if we test at the 0.05 level of significance?

Answer 1

1) H₀: = 19.5

    H₁: > 19.5

The test statistic z = ()/()

                              = (21.52 - 19.5)/(8/)

                              = 2.53

P-value = P(Z > 2.53)

             = 1 - P(Z < 2.53)

             = 1 - 0.9943

             = 0.0057

As the P-value is less than the significance level(0.0057 < 0.05), we should reject the null hypothesis.

So there is sufficient evidence to support the claim that the mean time of Facebook visits has increased.

2) H₀: = 50

    H₁: > 50

The test statistic z = ()/()

                              = (51.58 - 50)/(15/)

                              = 0.84

P-value = P(Z > 0.84)

             = 1 - P(Z < 0.84)

             = 1 - 0.7995

             = 0.2005

As the P-value is greater than the significance level(0.2005 > 0.05), we should not reject the null hypothesis.

So there is not sufficient evidence to support the claim of the school superintendent that the second graders in her school district have greater math skills than the nation wide average.

c) H₀: = 225.3

    H₁: 225.3

The test statistic z = ()/()

                              = (200.1 - 225.3)/(140/)

                              = -1.26

P-value = 2 * P(Z < -1.26)

             = 2 * 0.1038

             = 0.2076

As the P-value is greater than the significance level(0.2076 > 0.05), we should not reject the null hypothesis.

So there is not sufficient evidence to support the claim that the mean price of 2010 differs from the mean price in 2008.