Question

A recent broadcast of a television show had a15 ?share, meaning that among 6000 6000 monitored households with TV sets in? use,15?% of them were tuned to this program. Use a 0.01 significance level to test the claim of an advertiser that among the households with TV sets in? use, less than 25?% were tuned into the program. Identify the null? hypothesis, alternative? hypothesis, test? statistic, P-value, conclusion about the null? hypothesis, and final conclusion that addresses the original claim. Use the? P-value method. Use the normal distribution as an approximation of the binomial distribution. Identify the null and alternative hypotheses. Choose the correct answer below.

The? P-value is . ?(Round to four decimal places as? needed.)

Identify the conclusion about the null hypothesis and the final conclusion that addresses the original claim.

? Reject /Fail to reject H0. There ? is not/ is sufficient evidence to support the claim that less than 25?% of the TV sets in use were tuned to the program

Answer 1

(a) H0: Null Hypothesis: P 0.25

HA: Alternative Hypothesis: P < 0.25 (Claim)

(b)

P = Population proportion = 0.25

Q = 1 - P = 0.75

n = Sample size = 6000

p = Sample proportion = 0.15

SE =

Test statistic is:

Z = (p - P)/SE

= (0.15 - 0.25)/0.0056 = - 17.8571

(c)
Table of Area Under Standard Normal Curve gives area = 0.5 nearly.

So,

P-value = 0.5 - 0.5 nearly = 0 nearly

(d)

Since p-value = 0 nearly is less than = 0.001, Reject H0.

Conclusion:
There is sufficient evidence to support the claim that less than 25 % of the TV sets in use were turned to the program.