Solution :
Given that,
a) population mean = unknown
sample mean =
= 2.06
b) Population standard deviation =
= 0.05
Sample size = n = 59
At 99% confidence level
= 1 - 99%
= 1 - 0.99 =0.01
/2
= 0.005
Z/2
= Z0.005 = 2.576
Margin of error = E = Z/2
* (
/n)
= 2.576 * ( 0.05 / 59
)
= 0.017
At 99% confidence interval estimate of the population mean is,
± E
2.06 ± 0.017
( $ 2.043, $ 2.077 )
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