Question

ole Facebook: A study showed that two years ago, the mean time spent per visit to Facebook was 20.2 minutes. Assume the standard deviation is o = 40 minutes. Suppose that a simple random sample of 107 visits was selected this year and has a sample mean of x = 19.2 minutes. A social scientist is interested to know whether the mean time of Facebook visits has decreased. Use the a=0.10 level of significance and the P-value method with the TI-84 calculator. (a) State the appropriate null and alternate hypotheses. (b) Compute the P-value. (c) State a conclusion. Use the a=0.10 level of significance. Part: 0/4 Part 1 of 4 (a) State the appropriate null and alternate hypotheses. H: H: This hypothesis test is a (Choose one) v test.

Answer 1

20.2 minutes (2) Hull hy nothesis states that the mean time spent per usit of Facebook is Alternate hypothesis states that the mean time spent visit Facebook is less than 20.2 minutes. per of Ho: M= 20.2 H, : ' < 2012 The hypothesis test is a single sample z test. 19.2- 20.2 (b) Test statistic = x-u TAN - = -2.586 4.0/107 P-value 0.0049 C Source - STATKEY Cimage attached 9 (c) Conclusion : As p-value (0.0049) < < (0-10) Reject Ho. Hence, we can conclude that the mean time Facebook is less than 20.2 minuter. spent per visit of

Statkey Theoretical Distribution Normal Distribution- Reset Plot Normal Distribution Left Tail Two-Tail Right Tail Mean Standard Deviation 0.40 0 1 0.35 Edit Parameters 0.30 0.25 0.20 0.0049 0.15 0.10 0.05 0.00 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 -2.586 Activate Windows

The null and alternate hypothesis is stated. The test statistic is calculated using the formula mentioned. The p value is obtained from STATKEY (image attached for reference). We compare the test statistic with alpha and make the required conclusion.