Question

A random sample of 25 with a mean of 80 and a standard deviation of 30 is taken from a population of 1000 that is normally distributed. Find a) the 90% b) the 95% and c) the 99% confidence intervals for the unknown population mean.

Answer 1

Let  α₁ = 1 - 0.90 = 0.1. α₂ = 1-0.95=0.05, α₃ = 1 - 0.99=0.01 are significance levels, degree of freedom

d = 25 - 1 = 24.

For α₁ = 0.1 and d = 24 critical t-value t₁ = 1.711.

For α₂ = 0.05 and d = 24 critical t-value t₂ = 2.064,

For α₃ = 0.1 and d = 24 critical t-value t₃ = 2.797.

We will find corresponding confidence intervals, using expression

$$(\mu-\frac{t\cdot s}{\sqrt{n}}, \mu+\frac{t\cdot s}{\sqrt{n}})$$

$$where\ \mu\ is \ sample\ mean ,\ t\ is\ a\ critical\ value,$$

$$n\ is\ the\ size\ and\ s\ is\ standard\ deviation\ of\ the\ sample.$$

a) Confidence interval is

$$(80 - \frac{1.711\cdot 30}{\sqrt{25}}, 80 + \frac{1.711\cdot 30}{\sqrt{25}})=$$

$$(80-10.266,\ 80+10.266)=$$

$$(69.734,\ 90.266)$$

b) Confidence interval is

$$(80 - \frac{2.064\cdot 30}{\sqrt{25}}, 80 + \frac{2.064\cdot 30}{\sqrt{25}})=$$

$$(80-12.384,\ 80+12.384)=$$

$$(67.616,\ 92.384)$$

c) Confidence interval is

$$(80 - \frac{2.797\cdot 30}{\sqrt{25}}, 80 + \frac{2.797\cdot 30}{\sqrt{25}})=$$

$$(80-16.782,\ 80+16.782)=$$

$$(63.218,\ 96.782)$$

Answer: a) (69.734, 90.266), b) (67.616, 92.384), c) (63.218, 96.782)