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In a study of 404 nonprofits nationwide, 87 indicated that turnover has been the biggest employment challenge at their organization. Complete parts (a) through (C).

a. Construct a 95% confidence interval for the population proportion of nonprofits that indicate turnover as the biggest employment challenge at their organization.

b. Interpret the interval constructed in part (a).

c. If you wanted to conduct a follow-up study to estimate the population proportion of nonprofits that indicate turnover as the biggest employment challenge at their
organization to within +/- 0.05 with 95% confidence, how many nonprofits would you survey? Assume that the population proportion (pi) is the calculated sample proportion "p" in the case below!

A sample of ___ nonprofits should be surveyed. 
 

Normal distribution approximation applies because the sample size is large enough, such the data is normally distributed even is the population is not normally distributed. 
The sample statistic to estimate the population proportion is , pi = 87/404

pi _ estimate = 87/404 = 0.215

Proportion confidence interval formula is, p_estimate +/- Zc * sqrt(p_estimate * (1- p_estimate)/ sample size). 

Zc can be easily obtained from Excel, =NORM.S.INV(1-0.05/2) = 1.96

Margin of error  = 1.96 * sqrt(0.215 * (1-0.215)/404) = 0.0401

Confidence interval = 0.215 +/- 0.0401 

The confidence interval is (0.1749, 0.2551). 

we are 95% confident that the true proportion associated with the population parameter of interest is within 0.1749 and 0.2551. 

What if a margin of error of 0.05 and a 95% confidence interval is targeted. 

Sample size for population proportion confidence interval is, 

n  = p_estimate * (1-p_estimate) * (Zc/Error)^2

n = 0.215 * (1-0.215) *(1.96/0.05)^2 =  260

A sample of 260 nonprofits should be surveyed. 


 

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