Question

A father is concerned that his teenage son is watching too much television each day, since his son watches an average of 2 hours per day. His son says that his TV habits are no different than those of his friends. Since this father has taken a stats class, he knows that he can actually test to see whether or not his son is watching more TV than his peers. The father collects a random sample of television watching times from boys at his son's high school and gets the following data:

1.9

2.3

2.2

1.9

1.6

2.6

1.4

2.0

2.0

2.2

Is the father right? Perform a hypothesis test (by hand) using α=0.05 to check the father's claim that other boys average less than 2 hours of television per day. Include the following:

Part 1 (2 pts): Write your hypotheses in both symbols and words.

Part 2 (4 pts): Test an appropriate hypothesis "by hand" using an alpha level of 0.05. Assume conditions are met (so don't check them). Be sure to include the correct probability notation and use the editor for all work.

Part 3 (4 pts): Support your test with the corresponding confidence interval, again "by hand". Use the editor to show all steps.

Part 4 (3 pts): Write your final conclusion using the test results and interval results. Include a sentence that states whether the test and interval results agree.

Answer 1

Let $\mu$ denotes the average time spend watching television per day.

To test null hypothesis Ho 2.0 against alternative hypothesis H1 p< 2.0 Let denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, 2.0100, s2 0.1188, s = 0.3446, n = 10 The test statistic can be written as Vn (2.0) t S which under Ho follows a t distribution with n -1 df. We reject Ho at 0.05 level of significance if P-value 0.05 or if tobs<-to.050,n-1 Now, The value of the test statistic 0.091756 associated degrees of freedom The critical value = -1.833113 9 Р- value P(tg<tobs)= P(tg < 0.091756) 0.535549 Since p-value> 0.05 and tobsteritical = -1.833113, so we fail to reject Ho at 0.05 level of significance Hence, we can conclude that population mean is not significantly less than 2.0

Conclusion : There is not sufficient evidence to support the father's claim that other boys average less than 2 hours of television per day.

Here, 2.010, s0.1188, s 0.3446, n = 10 The margin of error is: 0.345 0.345 S ME 2.262157 0.246542 0.247 V10 to.025,9 V10 t 1-0.950 2 a 95 percent confidence interval population mean u is t 1-0.950 ,9 2 0.345 = 2.010 to.025,9 V10 0.345 = 2.010 t 2.262157 V10 = 2.010 t 0.246542 = (1.763458, 2.256542) (1.76, 2.26)

Since upper bound of the confidence interval > 2, so at 95% level of confidence or at 5% level of significance, we can conclude that there is not sufficient evidence to support the father's claim that other boys average less than 2 hours of television per day.