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.

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. =    ~ N(0,1)

where, = sample proportion of employees who say that monitoring e-mail is seriously unethical = = 0.44

= sample proportion of senior-level bosses who say that monitoring e-mail is seriously unethical = = 0.33

= sample of employees = 436

= 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, () 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 < < 2.58) = 0.99

P( < < ) = 0.99

P( < () < ) = 0.99

99% confidence interval for () =

[ , ]

 = [ , ]

 = [-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].

answer: check all that apply, where the options?

step-by-step explanation: