Answer :
Answer:
99% confidence interval for the difference in the proportion of employees and the proportion of bosses who says that monitoring e-mail is seriously unethical is [-0.016 , 0.236].
Step-by-step explanation:
We are given that a survey of 436 workers showed that 192 of them said that it was seriously unethical to monitor employee e-mail.
When 121 senior-level bosses were surveyed, 40 said that it was seriously unethical to monitor employee e-mail.
Firstly, the Pivotal quantity for 99% confidence interval for the difference in population proportion is given by;
P.Q. = [tex]\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] ~ N(0,1)
where, [tex]\hat p_1[/tex] = sample proportion of employees who say that monitoring e-mail is seriously unethical = [tex]\frac{192}{436}[/tex] = 0.44
[tex]\hat p_2[/tex] = sample proportion of senior-level bosses who say that monitoring e-mail is seriously unethical = [tex]\frac{40}{121}[/tex] = 0.33
[tex]n_1[/tex] = sample of employees = 436
[tex]n_2[/tex] = sample of senior-level bosses = 121
Here for constructing 99% confidence interval we have used Two-sample z test for proportions.
So, 99% confidence interval for the difference in proportion, ([tex]p_1-p_2[/tex]) is ;
P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level
of significance are -2.58 & 2.58}
P(-2.58 < [tex]\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] < 2.58) = 0.99
P( [tex]-2.58 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] < [tex]{(\hat p_1-\hat p_2)-(p_1-p_2)}[/tex] < [tex]2.58 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] ) = 0.99
P( [tex](\hat p_1-\hat p_2)-2.58 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] < ([tex]p_1-p_2[/tex]) < [tex](\hat p_1-\hat p_2)+2.58 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] ) = 0.99
99% confidence interval for ([tex]p_1-p_2[/tex]) =
[[tex](\hat p_1-\hat p_2)-2.58 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] , [tex](\hat p_1-\hat p_2)+2.58 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex]]
= [ [tex](0.44-0.33)-2.58 \times {\sqrt{\frac{0.44(1-0.44)}{436}+\frac{0.33(1-0.33)}{121} } }[/tex] , [tex](0.44-0.33)+2.58 \times {\sqrt{\frac{0.44(1-0.44)}{436}+\frac{0.33(1-0.33)}{121} } }[/tex] ]
= [-0.016 , 0.236]
Therefore, 99% confidence interval for the difference in the proportion of employees and the proportion of bosses who says that monitoring e-mail is seriously unethical is [-0.016 , 0.236].