According to Nielsen Media Research, the average number of hoursof TV viewing per household per week in the United States is 50.4hours.
1 (a) Suppose the population standard deviation is 11.8 hours and arandom sample of 42 U.S. household is taken, what is theprobability that the sample mean TV viewing time is between 47.5and 52 hours?
1 (b) Suppose the population mean and sample size is still 50.4hours and 42, respectively, but the population standard deviationis unknown. If 72% of all sample means are greater than 49 hours,what is the value of the unknown population standarddeviation?
1(c) What is the result of part (a) if the sample only consists of5 households? Explain.


The average age of online consumers ten years ago was 23.3 years.As older individuals gain confidence with the Internet, it isbelieved that the average age has increased. We would like to testthis belief.
2(a) Write the appropriate hypotheses to be tested.
2(b) The online shoppers in our sample consisted of 40 individuals,had an average age of 24.2 years, with a standard deviation of 5.3years. What is the test statistic and p‐value for the hypothesesbeing tested in part (a)? (Remark: Report the p‐value using thestatistical table, but NOT Excel function.) 2 (c) What is thepractical implication of the conclusion of the hypothesis test ati. 5% level of significance, and ii. 10% level of significance?