Question

A simple random sample of size n=15 is drawn from a population that is normally distributed. The sample mean is found to be x overbar=62 and the sample standard deviation is found to be s=19. Construct a 95% confidence interval about the population mean.

The 95% confidence interval is (_,_).

(Round to two decimal places as needed.)

Answer #1

Solution :

Given that,

Point estimate = sample mean = = 62

sample standard deviation = s = 19

sample size = n = 15

Degrees of freedom = df = n - 1 = 15 - 1 = 14

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t_{
/2,df} = t_{0.025,14} = 2.145

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

= 2.145 * (19 / 15)

= 10.52

The 95% confidence interval estimate of the population mean is,

- E < < + E

62 - 10.52 < < 62 + 10.52

51.48 < < 75.52

The 95% confidence interval is (51.48, 75.52)

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