Question

A recent broadcast of a television show had a 15 share, meaning that among 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. O A. Ho p=0.75 OB. Hyp=0.25 H,: p0.75 He: p>0.25 06. Họ p= 0,75 OD. Hyp0.75 H, p>0.75 Hyp<0.75 O E. Ho: p=0.25 OF Họ p=0.25 H:D<0.25 Hp 0.25

Answer 1

Given : n=6000 , $\hat{p}$ =15%=0.15 , =25%=0.25

Po

The null and alternative hypothesis is ,

$H_0:p=0.25$

H:P 0.25

The test is one-tailed test.

The test statistic is ,

$Z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}=\frac{0.15-0.25}{\sqrt{\frac{0.25(1-0.25)}{6000}}}=-17.89$

The p-value is ,

p-value=$P(Z>|Z|)=P(Z>17.89)=1-P(Z\leq 17.89)=0.0000$ ; The Excel function is , =1-NORMDIST(17.89,0,1,TRUE)

Decision : Here , p-value<0.01

Therefore , reject Ho.

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