A report summarized the results of a survey of 308 U.S. businesses. Of these com

Question

A report summarized the results of a survey of 308 U.S. businesses. Of these companies, 202 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. 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 = P-value = Do not reject H_0. We do not have convincing evidence that more than 60% of U.S. businesses monitor employees' web site visits. Reject H_0. We do not have convincing evidence that more than 60% of U.S. businesses monitor employees' web site visits. Reject H_0. We have convincing evidence that more than 60% of U.S. businesses monitor employees' web site visits. Do not reject H_0. We have convincing evidence that more than 60% of U.S. businesses monitor employees' web site visits. 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 =

Explanation / Answer

ans=

This problem requires you to perform a one-proportion z-test.

(a) is asking to test whether more than 60% ( > 0.60) of businesses monitor site visits.
Ho: p = 0.60
Ha: p > 0.60

z = [ (202/308) - 0.60 ] / [ (0.60 * 0.40) / (308) ]
= [ 0.0558 ]/0.027
= 2.0666
P( z > 2.0666 ) = p-value
p-value 0.9999 1.0000

1.0000 is greater than = 0.01, so there is sufficient evidence.

(b) is asking to test whether a "majority" of businesses monitor site visits. A majority would constitute any percentage greater than 50%.

Ho: p = 0.50
Ha: p > 0.50

z = [ (202/308) - 0.50 ] / [ (0.50 * 0.50) / (308) ]
= 0.15584/ 0.00081
= 0.15584/0.028 =5.5657
P( z > 5.5657) = p-value
p-value 8.0571E-12 0.0000

0.0000 is less than = 0.01, so there is not sufficient evidence

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