Question

In a study of 413 nonprofits nationwide, 89 indicated that turnover has been the biggest employment challenge at their organization.

(a) Construct a 95% confidence interval for the population proportion of nonprofits that indicate turnover as the biggest employment challenge at their organization. Express the confidence interval as ≤ π ≤ (rounded to three decimal places as needed).

(b) Interpret the interval constructed in part (a). Choose the correct answer below that best describes the interpretation of the confidence interval.

(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 plus or minus 0.02 with 95% confidence, how many nonprofits would you survey? A sample of nonprofits should be surveyed. (Round up to the nearest integer.)

Answer 1

To construct a 95% confidence interval for the population proportion of nonprofits that identify turnover as their biggest employment challenge, the sample proportion and Z-value are used in the confidence interval formula. The interpretation is that we can be 95% confident the actual proportion falls within this range. For a follow-up study with a margin of error of ± 0.02, sample size calculation is needed with rounding up to ensure accuracy.

So, the best answer is d, 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 plus or minus 0.02 with 95% confidence, how many nonprofits would you survey? A sample of nonprofits should be surveyed. (Round up to the nearest integer.)

Step-by-step explanation:

To calculate the 95% confidence interval for the population proportion of nonprofits that indicate turnover as the biggest employment challenge, use the formula for confidence interval of a population proportion:

CI = p ± Z*(√(p(1-p)/n))

Where:
p = sample proportion (89/413),

Z = Z-value from the Z-table that corresponds to 95% confidence level (approximately 1.96),

n = sample size (413).

After calculations, you will get a confidence interval expressed as ≤ π ≤.

The interpretation of the interval is that we are 95% confident that the actual proportion of all nonprofits that identify turnover as their biggest challenge is between the lower and upper bounds of the obtained interval.

For the follow-up study, you'd need to calculate the required sample size to obtain a margin of error of ± 0.02 with 95% confidence:

n = (Z/Δp)^2 * p*(1-p)

Where:
Δp = desired margin of error (0.02),
The assumed population proportion p could be the estimate from the initial study or 0.5 when no prior estimate is available, for maximum variability, which provides the largest sample size.

After calculating n, always round up to the nearest whole number to ensure the margin of error is not exceeded.

So, the best answer is a, Construct a 95% confidence interval for the population proportion of nonprofits that indicate turnover as the biggest employment challenge at their organization. Express the confidence interval as ≤ π ≤ (rounded to three decimal places as needed).