Question

A simple random sample of size n=23 is drawn from a population that is normally distributed. The sample mean is found to be x bar=68 and the sample standard deviation is found to be s=15. Construct a 90?% confidence interval about the population mean.

The 90?% confidence interval is ?(nothing?,nothing?).

Answer #1

Solution :

Given that,

= 68

s = 15

n = 23

Degrees of freedom = df = n - 1 = 23 - 1 = 22

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t_{ /2,df} = t_{0.05,22} =
1.717

Margin of error = E = t_{/2,df} * (s /n)

= 1.717 * (15 / 23)

= 5.37

The 90% confidence interval estimate of the population mean is,

- E < < + E

68 - 5.37 < < 68 + 5.37

62.63 < < 73.37

(62.63 , 73.37)

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