Question

Question 10 (0.75 points) A study showed that in a certain month, the mean time spent per visit to Facebook was 19.5 minutes. Assume 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 21.13 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. Write down your P-value. You will need it for the next question. Round your answer to four decimal places (for example: 0.2305). Write only a number as your answer. Your Answer: Answer

Answer 1

Answer Date: 11/11/2019 To test the hypothesis is that the mean time spent per visit to Facebook was increased 19.50 minutes at 5% significance level. The null and alternative hypothesis is, Ho: [= 19.50 H: 4 >19.50 The z-test statistics is, 1- 2- Ton 21.13–19.50 (8/100) = 2.038 The z-test statistics is 2.038 The p-value for this test is,

From the standard normal distribution table, the associated probability is 0.9792. Since find the area to the right, so subtract from 1 to get, P(Z > 2.038)=1-(2<2.038) = 1-0.9792 = 0.0208 The p-value for this test is 0.0208). Decision The conclusion is that the significant p-value in this context is less than 0.05 which is 0.0208, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the mean time spent per visit to Facebook was increased 19.50 minutes. The result is statistically significant.