Question

In a recent study, it was reported that the proportion of employees who miss work on Fridays is 0.15, and that the standard deviation of the sampling distribution of sample proportion ˆp is 0.025. However, the report did not indicate what the sample size n was. What was it?

A. 26
B. 204
C. 108
D. More information needed to determine the value of n

Answer 1

The sample size n for employees missing work on Fridays, given a population proportion of 0.15 and a standard deviation of the sample proportion
`(\( \hat{p} \))` as 0.025, is approximately 26, making the answer A. 26.

The relationship between the standard deviation of the sampling distribution of the sample proportion
`(\( \hat{p} \))`, the population proportion p, and the sample size n is given by the formula:

`\[ \text{Standard Deviation} (\sigma_{\hat{p}}) = \sqrt{(p(1-p))/(n)} \]`

In this case, p = 0.15 and
`\( \sigma_{\hat{p}} = 0.025 \)`. We need to find n.

`\[ 0.025 = \sqrt{(0.15 \cdot (1-0.15))/(n)} \]`

Solving for n:

`\[ n = (0.15 \cdot (1-0.15))/((0.025)^2) \]\[ n = (0.15 \cdot 0.85)/(0.000625) \]\[ n \approx 25.5 \]`

Since n must be a whole number, the closest option is:

`\[ \text{A. 26} \]`

So, the sample size is approximately 26.