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

Answer:

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

Step-by-step explanation:

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

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

The margin of error for this interval is:

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

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:

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

       $n=[\frac{z_{\alpha/2}\times \sqrt{\hat p(1-\hat p)} }{MOE}]^{2}$

          $=[\frac{2.58\times \sqrt{0.50(1-0.50)} }{0.03}]^{2}\\\\=1849$    

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