A recent broadcast of a television show had a 10 share, meaning that among 6000

Question

A recent broadcast of a television show had a 10 share, meaning that among 6000 monitored households with TV sets in use, 10% of them were turned 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 15% 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.

ldentify the null and alternative hypotheses. Choose the correct answer below A. Ho'p:015 p #0.15 ° C. Ho: p 0.85 H1 p 0.85 OE. Ho:p=0.15 H1. p>0.15 B. Hop:015 H1 p0.15 D. Hop-085 H1: p#085 OF. H: p 0.85 H1 p0.8:5 The test statistic is z = (Round to two decimal places as needed.) 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 H0. There sufficient evidence to support the claim that less than 15% of the TV sets in use were tuned to the program

Explanation / Answer

1)

Ho: p = 0.15

Ha: p < 0.15

option B) is correct

2)

n = 6000 , p^ = 0.110

Z = (p^ - p)/sqrt(pq/n) = (0.1 - 0.15)/sqrt(0.15 *0.85/6000)

= -10.846522

p-value = P(Z < -10.846522) = 0.00000

since p-value < 0.01

we reject the null hypothesis

there is sufficient evidence to support the claim

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