Question

A schoolteacher is concerned that her students watch more TV than the average American child. She reads that according to the American Academy of Pediatrics (AAP), the average American child watches 4 hours of TV per day (μ = 4.0 hours). She records the number of hours of TV each of her six students watch per day. The times (in hours) are 3.8, 4.6, 2.9, 5.5, 4.4, and 2.8.

(a) Test the hypothesis that her students watch more TV than the average American child using a 0.05 level of significance and a one-independent sample t-test. State the value of the test statistic. (Round your answer to three decimal places.)

State the decision to retain or reject the null hypothesis.

(b) Compute effect size using estimated Cohen's d. (Round your answer to two decimal places.)

Answer 1

a)

null hypothesis Alternate Hypothesis 4 4 for 0.05 level with right tailed test and n-1-6 degree of freedom, critical value of t- Decision rule 2.5706 reject Ho if test statistic t 2.571 brand B S. NO 1 2 4.6 4 5.5 sample mean std deviation sample size Population mean - 4.000 1.045 4 std error of mean - S.Е. 0.4266 test statistic t 0.0000

as test statistic is not in critical region; we can not reject null hypothesis

we do nt have suffciient evidence to conclude that her students watch more TV than the average American child.

b)

effect size d=|$\overline{x}-\mu$|/s =(4-4)/1.045 =0.000