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 352 consumers who have owned a ColorSmart-5000 television set for five years. Of these 352 consumers, 310 say their ColorSmart-5000 television sets did not need a repair, whereas 42 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.) Using p and z0.005 2.576 n

Answer 1

total number pf customers (n) = 352

x = customers who say that their tv sets do not need repair in five years = 310

the proportion be:-

$p=\frac{x}{n}=\frac{310}{352}=0.88068$

critical value of z for 99 % confidence level = 2.576

margin of error (E) = 0.03

the necessary calculation be:-

using    p = 0.88068	calculations
sample size (n) = 82	$n=[\frac{z_{critical}*p*(1-p)}{E}]^2$   $=[\frac{2.576*0.88068*(1-0.88068)}{0.03}]^2$   $=81.4164$   $\approx 82$

***in case of doubt, comment below. And if u liked the solution, please like.