Construct the confidence interval for the population mean mu. cequals0.98, x overbar equals 4.2, sigmaequals0.6, and nequals43 A 98% confidence interval for mu is (Round to two decimal places as needed)
Solution :
Given that,
Point estimate = sample mean =
= 4.2
Population standard deviation =
= 0.6
Sample size = n =43
At 98% confidence level the z is ,
= 1 - 98% = 1 - 0.98 = 0.02
/ 2 = 0.02 / 2 = 0.01
Z/2 = Z0.01 = 2.326
Margin of error = E = Z/2
* (
/n)
= 2.326 * ( 0.6/ 43
)
= 0.21
At 98% confidence interval estimate of the population mean
is,
- E <
<
+ E
4.2 - 0.21<
< 4.2 + 0.21
3.99 <
< 4.41
( 3.99 , 4.41 )
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