Question

From a recent company survey, it is known that the proportion of employees older than 55 and considering retirement is 8% . For a random sample of size 110 , what is standard deviation for the sampling distribution of the sample proportions, rounded to three decimal places?

Answer 1

`Sd(p)= \sqrt{(p(1-p))/(n)} = \sqrt{(0.08*(1-0.08))/(110)}= 0.026`

Explanation:

For this case we are interested in the true population proportion
`p` and the distribution for this parameter is given by:

`p \sim N (p, \sqrt{(p(1-p))/(n)})`

We know that n = 110 and p =0.08

Solution to the problem

For this case the standard deviation is given by:

`Sd(p)= \sqrt{(p(1-p))/(n)} = \sqrt{(0.08*(1-0.08))/(110)}= 0.026`