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 2.7, 4.8, 4.4, 2.4, 4.1, and 5.6.

(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.)
t =  

(a) State the decision to retain or reject the null hypothesis.

Retain the null hypothesis.

Reject the null hypothesis.    

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

Answer 1

= (2.7 + 4.8 + 4.4 + 2.4 + 4.1 + 5.6)/6 = 4

s = sqrt(((2.7 - 4)^2 + (4.8 - 4)^2 + (4.4 - 4)^2 + (2.4 - 4)^2 + (4.1 - 4)^2 + (5.6 - 4)^2)/5) = 1.2345

a) H₀: = 4

    H₁: > 4

The test statistic t = ()/(s/)

                             = (4 - 4)/(1.2345/)

                             = 0

At alpha = 0.05, the critical value is t^* = 2.015

Since the test statistic value is not greater than the critical value (0 < 2.015), so we should not reject the null hypothesis.

b) d = ()/s

       = (4 - 4)/1.2345 = 0