solution
Given that,
= 59
s =11
n = 20
Degrees of freedom = df = n - 1 =20 - 1 =169
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,19 = 2.093 ( using student t table)
Margin of error = E = t/2,df * (s /n)
= 2.093 * ( 11/ 20) = 5.15
The 95% confidence interval estimate of the population mean is,
- E < < + E
59 - 5.15< <59 + 5.15
53.85 < < 64.15
(lower bound =53.85 ,upper bound= 64.15 )
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