The average number of push-ups a united states marine does daily is 300, with a standard deviationof 50. a random sample of 36 m
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Answer:
Step-by-step explanation:
Hello!
Your study variable is X: "number of ColorSmart-5000 that didn't need repairs after 5 years of use, in a sample of 390"
X~Bi (n;ρ)
ρ: population proportion of ColorSmart-5000 that didn't need repairs after 5 years of use. ρ = 0.95
n= 390
x= 303
sample proportion ^ρ: x/n = 303/390 = 0.776 ≅ 0.78
Applying the Central Limit Theorem you approximate the distribution of the sample proportion to normal to obtain the statistic to use.
You are asked to estimate the population proportion of televisions that didn't require repairs with a confidence interval, the formula is:
^ρ±* √[(^ρ(1-^ρ))/n]
=
= 2.58
0.78±2.58* √[(0.78(1-0.78))/390]
0.0541
[0.726;0.834]
With a confidence level of 99% you'd expect that the interval [0.726;0.834] contains the true value of the proportion of ColorSmart-5000 that didn't need repairs after 5 years of use.
I hope it helps!