Mathematics, 27.05.2020 14:57 marrizza7

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.

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Answer from: simihehe

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

where, $\hat p_1$ = sample proportion of employees who say that monitoring e-mail is seriously unethical = $\frac{192}{436}$ = 0.44

$\hat p_2$ = sample proportion of senior-level bosses who say that monitoring e-mail is seriously unethical = $\frac{40}{121}$ = 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 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $2.58 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ) = 0.99

P( $(\hat p_1-\hat p_2)-2.58 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < ( p_1-p_2 ) < $(\hat p_1-\hat p_2)+2.58 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ) = 0.99

99% confidence interval for ( p_1-p_2 ) =

[ $(\hat p_1-\hat p_2)-2.58 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ , $(\hat p_1-\hat p_2)+2.58 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ]

= [ $(0.44-0.33)-2.58 \times {\sqrt{\frac{0.44(1-0.44)}{436}+\frac{0.33(1-0.33)}{121} } }$ , $(0.44-0.33)+2.58 \times {\sqrt{\frac{0.44(1-0.44)}{436}+\frac{0.33(1-0.33)}{121} } }$ ]

= [-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].

Answer from: Quest

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all

Answer from: Quest

step-by-step explanation: