Complete question is;
In a survey of 565 U.S. businesses, 430 of these companies 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 75% 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.)
Answer:
There's not sufficient evidence to suggest that more than 75% of U.S. businesses monitor employees website visits.
Explanation:
Population proportion; p^ = 430/565 = 0.761
Sample proportion; p = 75% = 0.75
Significance level; α = 0.01
Sample size; n = 560
Let's define the hypotheses;
Null hypothesis; H0: p = 0.75
Alternative hypothesis: p > 0.75
Formula for the test statistic is;
z = (p^ - p)/√[p^(1 - p^)/n]
z = (0.75 - 0.761)/√(0.75(1 - 0.75)/560)
z = -0.011/√0.00033482143
z = -0.011/0.0183
z = -0.60
From online p-value from z-score calculator attached, using; z = -0.60, significance level = 0.01, one tailed hypothesis, we have;
P-value ≈ 0.2743
The p-value is greater than the significance level and so we will fail to reject the null hypothesis and conclude that there's not sufficient evidence to suggest that more than 75% of U.S. businesses monitor employees website visits.