Question

A study is to be conducted of the percentage of homeowners who own at least two television sets. At least how large a sample is required if we wish to be 99% confident that the error in estimating this quantity is less than 0.019

Answer 1

The minimum sample size required is 4610.

Explanation:

The (1 - α)% confidence interval for population proportion p is:

`CI=\hat p\pm z_(\alpha/2)* \sqrt{(\hat p(1-\hat p))/(n)}`

The margin of error for this interval is:

`MOE= z_(\alpha/2)* \sqrt{(\hat p(1-\hat p))/(n)}`

It is provided that a 99% confidence interval is computed to estimate the percentage of homeowners who own at least two television sets.

Assume that 50% of the homeowners own at least two television sets.

So, the sample proportion is,
`\hat p=0.50`.

The critical value of z for 99% confidence interval is:

`z_(\alpha/2)=z_(0.01/2)=z_(0.005)=2.58`

*Use a z-table for the critical value.

The margin of error is given as, MOE = 0.019.

Compute the value of n as follows:

`MOE= z_(\alpha/2)* \sqrt{(\hat p(1-\hat p))/(n)}`

`n=[(z_(\alpha/2)* √(\hat p(1-\hat p)) )/(MOE)]^(2)`

`=[(2.58* √(0.50(1-0.50)))/(0.019)]^(2)`

`=(67.8947)^(2)\\=4609.6903\\\approx4610`

Thus, the minimum sample size required is 4610.