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 21.70 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:

these two questions go together

Answer 1

Answer Date: 04/11/2019 To test the hypothesis is that the mean time of Facebook visits has increased 19.50 minutes at 5% significance level. The null and alternative hypothesis is, H:1 = 19.50 H:\>19.50 The z-test statistics is, Z=- - I-u (en) 21.70-19.50 (8/1/100) = 2.75 The z-test statistics is 2.75).

The p-value for this test is, From the standard normal distribution table, the associated probability is 0.9970. Since find the area to the right, so subtract from 1 to get, P(z> 2.75) =1-(z< 2.75) = 1-0.9970 = 0.0030 The p-value for this test is 0.0030). Decision The conclusion is that the significant p-value in this context is less than 0.05 which is 0.0030, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the mean time of Facebook visits has increased 19.50 minutes. The result is statistically significant. The answer option is 4.