In: Math
For the following data values below, construct a 98% confidence interval if the sample mean is known to be 9.808 and the population standard deviation is 5.013. (Round to the nearest thousandth) (Type your answer in using parentheses!Use a comma when inputing your answers! Do not type any unnecessary spaces! List your answers in ascending order!) for example: (0.45,0.78)
6.6, 2.2, 18.5, 7.0, 13.7, 5.4, 5.3, 5.9, 4.7, 14.5
2.0, 14.8, 8.1, 18.6, 4.5, 17.7, 15.9, 15.1, 8.6, 5.2
15.3, 5.6, 10.0, 8.2, 8.3, 9.9, 13.7, 8.5, 8.2, 7.9
17.2, 6.1, 13.7, 5.7, 6.0, 17.3, 4.2, 14.7, 15.2, 3.3
3.2, 9.1, 8.0, 18.9, 14.2, 5.1, 5.7, 16.4, 10.1, 6.4
Solution :
Given that,
Point estimate = sample mean =
= 9.808
Population standard deviation =
= 5.013
Sample size = n =50
At 98% confidence level the z is ,l
= 1 - 98%
= 1 - 0.98 =0.02
/2
= 0.01
Z/2
= Z0.01= 2.326
Margin of error = E = Z/2
* (
/
n)
= 2.326 * (5.013 /
50)
= 1.65
At 98% confidence interval estimate of the population mean is,
- E <
<
+ E
9.808 - 1.65 <
<9.808 + 1.65
8.158 <
< 11.458
( 8.158 , 11.458 )