Question

part a)

A report summarized the results of a survey of 312 U.S. businesses. Of these companies, 206 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 =

P-value =

part b)

(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	=

z=

p-value=

Answer 1

Solution :

Given that,

= 60% = 0.60

1 - = 1 - 0.60 = 0.40

n = 312

x = 206

Level of significance = = 0.01

Point estimate = sample proportion = = x / n = 206/312 = 0.660

This a right (One) tailed test.

A)

Ho: p = 0.20

Ha: p 0.20

Test statistics

z = ( - ) / *(1-) / n

= ( 0.660 - 0.60) / (0.60*0.40) / 312

= 2.17

B)

P-value = P(Z>z)

= 1 - P(Z <z )

= 1- P(Z < 2.17)

= 1 - 0.9850

= 0.0150

The p-value is p = 0.0150, and since p = 0.0150 < 0.01, it is concluded that the null hypothesis is fails to rejected.