A survey of 410 workers showed that 172 said it was unethical to monitor employee e-mail. When 135 senior-level bosses were surveyed, 37 said it was unethical to monitor employee e-mail. At the 1% significance level, do the data provide sufficient evidence to conclude that the proportion of workers that say monitoring employee e-mail is unethical is greater than the proportion of bosses.
A. What is the Parameter of interest?
B. What is the underlying Distribution?

Respuesta :

Answer:

Step-by-step explanation:

Analysis of the Data Given:

For sample 1 ,  the sample size [tex]N_1 = 410[/tex]

number of favorable cases [tex]X_1 = 172[/tex]

thus ; the sample proportion is [tex]\bar{p_1} = \frac{X_1}{N_1}[/tex]

= [tex]\frac{172}{410}[/tex]

= 0.4195

For sample 2, the sample size [tex]N_2 = 135[/tex]

number of favorable cases [tex]X_2 = 37[/tex]

Then the sample proportion  is [tex]\bar{p_2} = \frac{X_2}{N_2}[/tex]

= [tex]\frac{37}{135}[/tex]

= 0.2741

The value of the pooled proportion is computed as [tex]\bar p = \frac{X_1+X_2}{N_1+N_2}[/tex]

= [tex]\frac{172+37}{410+135}[/tex]

= 0.3835

We are also given that the significance level is [tex]\alpha =0.01[/tex]

Null Hypothesis : [tex]H_0: p_1 =p_2[/tex]:  the proportion of workers that say monitoring employee e-mail is unethical is not greater than the proportion of bosses.

Alternative hypothesis : [tex]H_1 : p_1 > p_2[/tex]:  the proportion of workers that say monitoring employee e-mail is unethical is greater than the proportion of bosses.

[tex]H_0: p_1 =p_2[/tex]:

[tex]H_1 : p_1 > p_2[/tex]:

The above corresponds to the right-tailed test , for which a z-test for the two population proportions needs to be conducted .

Rejection Region:

From the given information ; the significance level is [tex]\alpha =0.01[/tex]

Then; the critical value for a right tailed test is [tex]z_c = 2.33[/tex]

The rejection region for this right tailed test is R = {z : z > 2.33}

Test Statistics:

The z-statistic is computed as:

[tex]z ={\frac{\arrow p_1 - \arrow p_2}{\sqrt{ \bar p(1- \bar p)(1/n_1 + 1/n_2)} }[/tex]

[tex]z ={\frac{0.4195- 0.2741}{\sqrt{ 0.3835(1-0.3835)(1/410 + 1/135)} }[/tex]

z = 3.014

Decision about the null hypothesis:

Since it is observed that z = 3.014 > [tex]z_c = 2.33[/tex] ; it is concluded that the null hypothesis is rejected

Using the P-value approach: The P-value p = 0.0013 and since p = 0.0013<0.01 , it is concluded that the null hypothesis is rejected.

Conclusion:

It is concluded that the proportion of workers that say monitoring employee e-mail is unethical is greater than the proportion of  bosses.