solution
Given that,
= 120.2
s =12.7
n = 40
Degrees of freedom = df = n - 1 =40 - 1 = 39
At 99% confidence level the t is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
t /2 df = t0.005, 39= 2.708 ( using student t table)
Margin of error = E = t/2,df * (s /n)
= 2.708* ( 12.7/ 40) = 5.44
The 99% confidence interval estimate of the population mean is,
- E < < + E
120.2 - 5.44 < <120.2 + 5.44
114.76 < < 125.64
(lower bound= 114.76 , upper bound=125.64)
The upper bound is ___ A simple random sample of size n = 40 is drawn...
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