Question

An accountant is concerned about the percentage of customers that have businesses which are operating at a loss . Of the last 10 years, the percentage of customers who showed a loss on their tax return is 66%. If the accountant conducts research on this concern, for what sample size n will the sampling distribution of sample proportions have a standard deviation of 0.09

Answer 1

Answer:

A sample size of 28.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

We need to find n for which s = 0.09, with p = 0.66. So

$s = \sqrt{\frac{p(1-p)}{n}}$

$0.09 = \sqrt{\frac{0.66*0.34}{n}}$

$0.09\sqrt{n} = \sqrt{0.66*0.34}$

$\sqrt{n} = \frac{\sqrt{0.66*0.34}}{0.09}$

$(\sqrt{n})^{2} = (\frac{\sqrt{0.66*0.34}}{0.09})^{2}$

$n = 27.7$

Rounding to the nearest whole number

A sample size of 28.