Please help this is my 3rd time asking :(
1 answer:
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Given:
n = 195, sample size.
x = 162, successes in the sample
The proportion is
p = x/n = 162/195 = 0.8308
n* p = 195*0.8308 = 162
n*(1-p) = 195*(1 - 0.8308) = 33
If n*p >= 10, and n*(1-p) >= 10, then the sample proportions will have a normal distribution. This condition is satisfied.
The proportion mean is
μ = 0.8308
The proportion standard deviation is
σ/√n = 0.0269/√195 = 0.00192
At the 95% confidence level, the interval for the population proportion is
(μ - 1.96(σ/√n), μ + 1.96(σ/√n))
= (0.8308 - 1.96*0.00192, 0.8308 + 1.96*0.00192)
= (0.827, 0.8345)
Answer: The 95% confidence interval is (0.827, 0.835)
n = 195, sample size.
x = 162, successes in the sample
The proportion is
p = x/n = 162/195 = 0.8308
n* p = 195*0.8308 = 162
n*(1-p) = 195*(1 - 0.8308) = 33
If n*p >= 10, and n*(1-p) >= 10, then the sample proportions will have a normal distribution. This condition is satisfied.
The proportion mean is
μ = 0.8308
The proportion standard deviation is
σ/√n = 0.0269/√195 = 0.00192
At the 95% confidence level, the interval for the population proportion is
(μ - 1.96(σ/√n), μ + 1.96(σ/√n))
= (0.8308 - 1.96*0.00192, 0.8308 + 1.96*0.00192)
= (0.827, 0.8345)
Answer: The 95% confidence interval is (0.827, 0.835)
