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

f(x) is going down by 4

step-by-step explanation:

(its either that or its going left by 4)

Answer from: Quest

wth is with the little mtn top and what are you talking about like fix your words and accually make sence with them and i like was in a completly different question like three times like it seems like your ilitteret

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