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.)
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 = t0.025,14 = 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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