12. A survey of 61 comma 648 people included several questions about office relationships. Of the respondents, 26.7% reported that bosses scream at employees. Use a 0.01 significance level to test the claim that more than 1 divided by 4 of people say that bosses scream at employees. How is the conclusion affected after learning that the survey is an online survey in which Internet users chose whether to respond? 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 = 0.25
H1: p > 0.25
B.H0: p = 0.75
H1: p > 0.75
C.H0: p = 0.75
H1: p ≠ 0.75
D.H0: p = 0.75
H1: p < 0.75
E.H0: p = 0.25
H1: p ≠ 0.25
F.H0: p = 0.25
H1: p < 0.25
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.
(Reject/Fail to reject) H0. There (is not/is) sufficient evidence to support the claim that more than 1 divided by 4 of people say that bosses scream at employees. If the sample is a voluntary response sample, the conclusion (might not be/is) valid.
The null and alternative hypothesis are
H0: p = 0.25
Ha: p > 0.25
Test statistics
z = - p / sqrt( p( 1 - p) / n)
= 0.267 - 0.25 / sqrt( 0.25 * 0.75 / 61648)
= 9.75
This is test statistics value.
p-value = P( Z > z)
= P( Z > 9.75)
= 0 (From Z table)
Since p-value < 0.01 level, we have sufficient evidence to reject H0.
We conclude that,
Reject H0, There is sufficient evidence to support the claim that more than 1/4 of people says that
bosses scream at employees. If sample is voluntary response sample, the conclusion might not be valid.
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