In: Statistics and Probability
A report summarized the results of a survey of 309 U.S. businesses. Of these companies, 203 indicated that they monitor employees' web site visits. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of businesses in the United States. (a) Is there sufficient evidence to conclude that more than 60% of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of 0.01. (Round your test statistic to two decimal places and your P-value to four decimal places.)
z =
State your conclusion
. Reject H0. We have convincing evidence that more than 60% of U.S. businesses monitor employees' web site visits
. Reject H0. We do not have convincing evidence that more than 60% of U.S. businesses monitor employees' web site visits.
Do not reject H0. We do not have convincing evidence that more than 60% of U.S. businesses monitor employees' web site visits.
Do not reject H0. We have convincing evidence that more than 60% of U.S. businesses monitor employees' web site visits.
(b) Is there sufficient evidence to conclude that a majority of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of 0.01. (Round your test statistic to two decimal places and your P-value to four decimal places.)
z = P-value =
State your conclusion.
Reject H0. We have convincing evidence that a majority of U.S. businesses monitor employees' web site visits.
Do not reject H0. We have convincing evidence that a majority of U.S. businesses monitor employees' web site visits.
Do not reject H0. We do not have convincing evidence that a majority of U.S. businesses monitor employees' web site visits.
Reject H0. We do not have convincing evidence that a majority of U.S. businesses monitor employees' web site visits.
1)
null Hypothesis: Ho: p | = | 0.600 |
alternate Hypothesis: Ha: p | > | 0.600 |
sample success x = | 203 | |
sample size n = | 309 | |
std error se =√(p*(1-p)/n) = | 0.0279 | |
sample proportion p̂ = x/n= | 0.6570 | |
test stat z =(p̂-p)/√(p(1-p)/n)= | 2.04 | |
p value = | 0.0207 |
since p value >0.01
Do not reject H0. We do not have convincing evidence that more than 60% of U.S. businesses monitor employees' web site visits.
b)
test stat z =(p̂-p)/√(p(1-p)/n)= | 5.5200 |
p value = | 0.0000 |
Reject H0. We have convincing evidence that a majority of U.S. businesses monitor employees' web site visits.