Solution :
Given that,
= 50
s = 6.5
n = 25
Degrees of freedom = df = n - 1 = 25 - 1 = 24
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,24 = 2.064
standard error = s / n = 6.5 / 25 = 6.5 / 5 = 1.3
The 95% confidence interval for is ,
+/- t /2,df * S.E.
50 +/- 2.064(1.3)
Option D) is correct .
If a sample of size n = 25 is chosen from a normal population, which one...
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