Math  /  Data & Statistics

QuestionA 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 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 20\% 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. A. H0:p=0.80H_{0}: p=0.80 B. H0:p=0.80H_{0}: p=0.80 H1:p<0.80H_{1}: p<0.80 H1:p>0.80H_{1}: p>0.80 C. H0:p=0.20H_{0}: p=0.20 D. H0:p=0.20H_{0}: p=0.20 H1:p0.20H_{1}: p \neq 0.20 H1:p<0.20H_{1}: p<0.20 E. H0:p=0.80H_{0}: p=0.80 F. H0:p=0.20H_{0}: p=0.20 H1:p0.80H_{1}: p \neq 0.80 H1:p>0.20H_{1}: p>0.20
The test statistic is z=\mathrm{z}= \square . (Round to two decimal places as needed.) Clear all Check answer Help me solve this Get more help -

Studdy Solution
The test statistic is z19.38z \approx -19.38.
The P-value is approximately 00.
We reject the null hypothesis and conclude there is sufficient evidence to support the claim that less than 20% of households were tuned to the program.

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