Question

The manufacturer of the ColorSmart-5000 television set claims 95 percent of its sets last at least five years without needing a single repair. In order to test this claim, a consumer group randomly selects 403 consumers who have owned a ColorSmart-5000 television set for five years. Of these 403 consumers, 330 say their ColorSmart-5000 television sets did not need a repair, whereas 73 say their ColorSmart-5000 television sets did need at least one repair.

Determine the sample size needed in order to be 99 percent confident that p, the sample proportion of ColorSmart-5000 television sets that last at least five years without a single repair, is within a margin of error of .03 of p, the population proportion of sets that last at least five years without a single repair. (Round your p answer to 5 decimal places. Round your n answer to the next whole number.)

Answer 1

Solution :

Given that,

n = 403

x = 330

Point estimate = sample proportion = = x / n = 330 / 403 = 0.81886

1 - = 1 - 0.81886 = 0.18114

margin of error = E = 0.03

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z_0.005 = 2.576

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (2.576 / 0.03)² * 0.81886 * 0.18114

= 1093.63

sample size = n = 1094