Question

A cellular telephone manufacturer that entered the post-regulation market too quickly has an initial problem with excessive customer complaints and consequent returns of the cell phones for repair or replacement. The manufacturer wants to determine the magnitude of the problem in order to estimate its warranty liability. How many cellular telephones should the company randomly sample from its warehouse and check in order to estimate the fraction defective, ?, to within 0.01 with 90% confidence?

Answer 1

Here the fraction defective= p

margin of error = 0.01

As we know

Margin of error = critical test statsitic * standard error of mean

critical test statistic for 90% confidence interval = 1.645

margin of error = sqrt [p * (1 -p)/n]

Here as sample proportion is not given, we assume it 0.5

so,

0.01 = 1.645 * sqrt [0.5 * 0.5/n]

n= (1.645/0.01)² * 0.5 * 0.5

n = 6763.86 or 6764

so here sample size required = 6764