Question

3. value: 10.00 points 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. (a) Find a 99 percent confidence interval for the proportion of all ColorSmart-5000 television sets that have lasted at least five years without needing a single repair. (Round your answers to 3 decimal places.) The 99 percent confidence interval is [ (b) Does this confidence interval provide strong evidence that the percentage of all ColorSmart-5000 television sets that last at least five years without a single repair is less than the 95 percent claimed by the manufacturer? (Click to select) , the interval is (Click to select) 4.95.

Answer 1

a)

sample proportion, = 0.8807
sample size, n = 352
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.8807 * (1 - 0.8807)/352) = 0.0173

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

CI(Proportion) = (0-2/34, P), P+=2 / 84,7"))

Margin of Error, ME = zc * SE
ME = 2.58 * 0.0173
ME = 0.0446

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.8807 - 2.58 * 0.0173 , 0.8807 + 2.58 * 0.0173)
CI = (0.836 , 0.925)

b)

yes, the interval is not contains 0.95

2)

a)
The following information is provided,
Significance Level, α = 0.05, Margin or Error, E = 1, σ = 10.05

The critical value for significance level, α = 0.05 is 1.96.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (zc *σ/E)^2
n = (1.96 * 10.05/1)^2
n = 388.01

Therefore, the sample size needed to satisfy the condition n >= 388.01 and it must be an integer number, we conclude that the minimum required sample size is n = 389
Ans : Sample size, n = 389

b)

sample mean, xbar = 252
sample standard deviation, σ = 10.05
sample size, n = 389

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 10.05/sqrt(389)
ME = 1

CI = (x-2x 18+zx )

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (252 - 1.96 * 10.05/sqrt(389) , 252 + 1.96 * 10.05/sqrt(389))
CI = (251 , 253)