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

A sample size of 28.

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 = (\sigma)/(√(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{(p(1-p))/(n)}`

In this question:

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

`s = \sqrt{(p(1-p))/(n)}`

`0.09 = \sqrt{(0.66*0.34)/(n)}`

`0.09√(n) = √(0.66*0.34)`

`√(n) = (√(0.66*0.34))/(0.09)`

`(√(n))^(2) = ((√(0.66*0.34))/(0.09))^(2)`

`n = 27.7`

Rounding to the nearest whole number

A sample size of 28.