Question

A sample of 81 observations is taken from a normal population with a standard deviation of 5. The sample mean is 40. Determine the 95% confidence interval for the population mean.

Answer 1

38.9 ≤ U ≤ 41.1

Explanation:

Mean, m = 40; standard deviation, α = 5; Confidence limit, U = 95% or 0.95

N = 81

The standard error, α(m) = α/√(N) = 5/√81 =5/9

Using table: 0.95 = 0.0379

Z(0.95) = 2 - 0.0379 = 1.9621 or 1.96

Hence, confidence interval = { m - 1.96(α/√N) ≤ U ≤ m +1.96(α/√N)}

But, 1.96(α/√N) = 1.96 X 5/9 = 1.96 X 0.56 = 1.1

(40 - 1.1 ≤ U ≤ 40 + 1.1)

∴ the confidence interval = 38.9 ≤ U ≤ 41.1

Answer 2

38.911≤p≤41.089

Explanation:

The formula for calculating confidence interval for a population mean us as shown below;

CI = xbar ± Z×S/√N where;

xbar is the sample mean = 40

Z is the z score at 95% confidence interval = 1.96

S is the standard deviation = 5

N is the sample size = 81

Substituting this parameters in the formula we have;

CI = 40±1.96×5/√81

CI = 40±(1.96×5/9)

CI = 40±(1.96×0.556)

CI = 40±1.089

CI = (40-1.089, 40+1.089)

CI = (38.911, 41.089)

The 95% confidence interval for the population mean is 38.911≤p≤41.089