Question

In a study of 443 nonprofits nationwide, 130 indicated that the greatest diversity staffing
challenge they face is retaining younger staff (those under 30).
(a) Construct a 95% confidence interval for the population proportion of nonprofits that
indicate retaining younger staff is the greatest diversity staffing challenge for their
organization.
(b) If you wanted to conduct a follow-up study to estimate the population proportion of
nonprofits that indicate retaining younger staff is the greatest diversity staffing
challenge for their organization to within ±0.01 with 98% confidence, how many
nonprofits would you survey? (i.e. Find the sample size)

Answer 1

Solution :

Given that,

n = 443

x = 130

Point estimate = sample proportion = = x / n = 130/443 = 0.293

1 - = 1 -0.293 = 0.707

a) At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z_{0.025 = 1.960}

Z_/2 = 1.960

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.960 * ((0.293*(0.707) /443 )

= 0.042

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.293 - 0.042 < p < 0.293 + 0.042

0.251 < p < 0.336  

( 0.251 ,0.336 )

b)

= 0.293

1 - = 1-0.293 = 0.707

margin of error = E = 0.01

At 98% confidence level

= 1-0.98% =1-0.98 =0.02

/2 =0.02/ 2= 0.01

Z/2 = Z_{0.01 = 2.326}

Z/2 = 2.326

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (2.326/0.01)² *0.293*0.707

  = 11207

sample size = n = 11207