Question

Watching TV: The 2012 general Social Survey asked a large number of people how much time they spent watching TV each day. The mean number of hours was 3.09 with a standard deviation of 2.72. Assume that in a sample of 34 teenagers, the sample standard deviation of daily TV time is 1.9 hours, and that the population of TV watching times is normally distributed. Can you conclude that the population standard deviation of TV watching times for teenagers differs from 2.72? Use the a = 0.10 level of significance. Part: 0 / 5 Part 1 of 5 State the appropriate null and alternate hypotheses. H0:0 (Choose one) v H1:0 (Choose one) This hypothesis test is a (Choose one) test.

Answer 1

solution + aires the information u mean Uz population mean sumber of hours = 300g hours Ó = population standard deviation = 2.72 hours n = 34 5 = sample standard deviation log bowers A = level of significance 0.10 Part 1 70 test the claim that the population standood deviation of tv watching times fod teenagers differs from 2072 to test je. Ho: 2.72 Hii 6 + [2.72 Here Hi is not equal to type so test is two-tailed test. This hypothesis is a two-tailed test. Step 2' OR paat - 2 The formula for test statisticis, 2 يو (1-0) 82 (347) *clogg? (2.722 22 = 16.1021/ Step-3> or part 3) coitical value - 2014

dofi = degoee of forcedor En-/ 33 dof 2² - x = 233,040 2 x 320D = # 43.74512 233,000 43.7452 cround +04 decimal place) V soitical value = 43.7452 past4) Decision Rule Reject to shed 22 value of test statisticis poreater than or equal to coitical value and fail to reject otherwise at a % level of Significance 16:102) { 43.7452 Fail to reject Ho af 10 % level of significance part 5) conclusion we say that there is insufficient levidence to suppoot the claim that the Population standard deviation is different times for teenagers is diffesent from 2172 at 10% level of significance of a watching