Question

Do people who work for non-profit organizations differ from those who work at for-profit companies when it comes to personal job satisfaction? Separate random samples were collected by a polling agency to investigate the difference. Data collected from 422 employees at non-profit organizations revealed that 377 of them were "highly satisfied." From the for-profit companies, 431 out 518 employees reported the same level of satisfaction. Find the standard error of the difference in sample proportions.

Answer 1

Answer:

0.0223

Step-by-step explanation:

Given the following data x(1) = 377, n(1) = 422, x(2) = 431, n(2) = 518

The sample proportion is the number of success divided by the sample.

P(1) = x(1)/n(1) = 377/422 = 0.8934

P(2) = x(2)/n(2) = 431/518 = 0.8320

Formular for the standard error of the difference in sample proportions

S.E = √p(1)q(1)/n(1)+P(2)q(2)/n(2)

S.E = √p1(1-P1)/n1+P2(1-P2)/n2

By substitution we have that,

S.E = √0.8934(1-0.8934)/422+0.8320(1-0.8320)/518

S.E = 0.0223

Answer 2

The standard error will be "0.0223".

Given values:

• $x(1) = 377$
• $x(2) = 431$
• $n(1) = 422$
• $n(2) = 518$

The sample proportion will be:

→ $P(1) = \frac{x(1)}{n(1)}$

           $= \frac{377}{422}$

           $= 0.8934$

→ $P(2) = \frac{x(2)}{n(2)}$

           $= \frac{431}{518}$

           $= 0.8320$

hence,

The standard error will be:

→ $S.E = \sqrt{\frac{P(1)q(1)}{n(1)} +\frac{P(2)q(2)}{n(2)} }$

By substituting the values, we get

         $= \sqrt{\frac{P(1)(1-P(1))}{n(1)} +\frac{P(2)(1-P(2))}{n(2} }$

         $= \sqrt{\frac{0.8934(1-0.8934)}{422} +\frac{0.8320(1-0.8320)}{518} }$        

         $= 0.0223$

Thus the solution above is right.

Learn more about probability here:

https://brainly.com/question/11234923