A recent broadcast of a television show had a 10 share, meaning that among the 6

Question

A recent broadcast of a television show had a 10 share, meaning that among the 6000 mentored households with TV sets in use, 10% if them were to this program. Use a 0.01 significance level to test the claris of an that among the households with TV sets in use, less than 15% were tuned into the program. Identify the null hypothesis, alternative hypothesis, 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 on appropriate of the binomial distribution. Identify the null and alternative hypotheses. Choose the correct answer below. A. H_0: p = 0.85 H_1: p 0.85 The test static 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. H_0: There sufficient evidence to support the claim that less than 15% of the TV sets in use were tuned to the program.

Explanation / Answer

The claim is that less than 15% were using tv sets

so the hypothesis formulation is

h0 : p = 0.15

ha : p <0.15 , hence C

b)

The test stat z is calculated as

now pooled sample proportion is given as

p = (p1 * n1 + p2 * n2) / (n1 + n2)

p = (0.1*6000 + 0.15*6000)/(6000+6000) = 0.125

now standard error is

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

SE = sqrt( 0.125 * ( 1 - 0.125 ) * [ (1/6000) + (1/6000 ] ) = 0.006

now z stat is

z = (p1 - p2) / SE

(0.1-0.15)/0.006 = -8.33

now we check the p value from the z table for 1 sided test as

the p value is 0

now as the p value is less than 0.01 , hence we reject the null hypothesis and cconclude that p <0.15

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