Solution :
Given that,
= 63
s = 18
n = 13
Degrees of freedom = df = n - 1 = 13 - 1 = 12
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,12 =2.179
Margin of error = E = t/2,df * (s /n)
= 2.179 * (18 / 13)
= 10.88
Margin of error = 10.88
The 95% confidence interval estimate of the population mean is,
- E < < + E
63 - 10.88 < < 63 + 10.88
52.12 < < 73.88
(52.12, 73.88 )
Ample random sample of population mean 13 is drawn from a population that may dared the...
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