Question

A manufacturing company wants to use a sample to estimate the proportion of defective tennis balls that are produced by a machine and the company wants the margin of error to be within 3% ( .03 ) of the population proportion of all defective tennis balls and to be 99% confident. There is no previous survey on this. What is the least number of tennis balls needed for the sample?

Answer 1

The least number of tennis balls needed for the sample is 1849.

Explanation:

The (1 - α) % confidence interval for population proportion 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)}`

Assume that the proportion of all defective tennis balls is p = 0.50.

The information provided is:

MOE = 0.03

Confidence level = 99%

α = 1%

Compute the critical value of z for α = 1% as follows:

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

*Use a z-table.

Compute the sample size required 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)) )/(0.03)]^(2)\\\\=1849`

Thus, the least number of tennis balls needed for the sample is 1849.