Question

9. (10pts) NBC reports that Americans watch a mean of 30 minutes of television per evening. We conduct a research study in an attempt to prove that the mean is less. In a sample of 200 Americans, the sample mean was 24 minutes per evening with a standard deviation of 15 minutes.

(a) (2pts) Develop the null and alternative hypotheses:

(b) (2pts) Using α = 0.02, what is the rejection rule?

(c) (2pts) Find the test statistic.

(d) (2pts) Should H0 be rejected?

(e) (2pts) Based on your answer in part

(d), what is your conclusion as it applies to THIS SITUATION?

Answer 1

x̅ = 24, s = 15, n = 200  

a) Null and Alternative hypothesis:  

Ho : µ = 30

H1 : µ < 30

b) df = n-1 = 199

Critical value, t-crit = T.INV(0.02, 199) = -2.067

Rejection rule: Reject Ho if t < -2.067  

c) Test statistic:  

t = (x̅- µ)/(s/√n) = (24 - 30)/(15/√200) = -5.6569

d) Decision:  

Reject the null hypothesis  

e) Conclusion:  

There is enough evidence to conclude that the mean is less than 30 at 0.02 significance level.