## Related questions with answers

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.

Solution

VerifiedGiven:

$x_1=377$

$n_1=422$

$x_2=431$

$n_2=518$

The sample proportion is the number of successes divided by the sample size:

$\hat{p}_1=\dfrac{x_1}{n_1}=\dfrac{377}{422}\approx 0.8934$

$\hat{p}_2=\dfrac{x_2}{n_2}=\dfrac{431}{518}\approx 0.8320$

Formula for the standard error of the difference in sample proportions:

$SE=\sqrt{\dfrac{\hat{p}_1\hat{q}_1}{n_1}+\dfrac{\hat{p}_2\hat{q}_2}{n_2}}=\sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$

Evaluate:

$\begin{align*} SE&=\sqrt{\dfrac{\hat{p}_1\hat{q}_1}{n_1}+\dfrac{\hat{p}_2\hat{q}_2}{n_2}} \\ &=\sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \\ &=\sqrt{\dfrac{0.8934(1-0.8934)}{422}+\dfrac{0.8320(1-0.8320)}{518}} \\ &\approx 0.0223 \end{align*}$

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