Question

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. Find 99% confidence interval to test the claim that for those saying that monitoring e-mail is seriously unethical, the proportion of employees and the proportion of bosses are different.

Answer 1

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].

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. =
`\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` ~ N(0,1)

where,
`\hat p_1` = sample proportion of employees who say that monitoring e-mail is seriously unethical =
`(192)/(436)` = 0.44

`\hat p_2` = sample proportion of senior-level bosses who say that monitoring e-mail is seriously unethical =
`(40)/(121)` = 0.33

`n_1` = sample of employees = 436

`n_2` = 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, (
`p_1-p_2`) 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 <
`\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` < 2.58) = 0.99

P(
`-2.58 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` <
`{(\hat p_1-\hat p_2)-(p_1-p_2)}` <
`2.58 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` ) = 0.99

P(
`(\hat p_1-\hat p_2)-2.58 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` < (
`p_1-p_2`) <
`(\hat p_1-\hat p_2)+2.58 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` ) = 0.99

99% confidence interval for (
`p_1-p_2`) =

[
`(\hat p_1-\hat p_2)-2.58 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` ,
`(\hat p_1-\hat p_2)+2.58 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }`]

= [
`(0.44-0.33)-2.58 * {\sqrt{(0.44(1-0.44))/(436)+(0.33(1-0.33))/(121) } }` ,
`(0.44-0.33)+2.58 * {\sqrt{(0.44(1-0.44))/(436)+(0.33(1-0.33))/(121) } }` ]

= [-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].