Question
In: Math
A sample of 50 night-school students' ages is obtained in order to estimate the mean age...
A sample of 50 night-school students' ages is obtained in order to estimate the mean age of night-school students. x = 24.3 years. The population variance is 16.
(b) Find the 95% confidence interval for μ. (Give your
answer correct to two decimal places.)
Lower Limit
Upper Limit
(c) Find the 99% confidence interval for μ. (Give your answer correct to two decimal places.)
Lower Limit
Upper Limit
Solutions
Expert Solution
Solution :
Given that,
a ) 
=24.3
2 = 16.
= 4
n = 50
B ) At 95% confidence level the z is ,
= 1 - 95% =
1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z0.025 =
1.960
Margin of error = E = Z/2* (
= 1.960 * (4/
= 1.11
At 95% confidence interval estimate of the population mean is,
- E < 
< + E
24.3 - 1.11 < < 24.3 + 1.11
23.19 < < 25.41
Lower Limit = 23.19
Upper Limit = 25.41
C ) At 99% confidence level the z is ,
= 1 - 99% =
1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
Z/2 = Z0.005 =
2.576
Margin of error = E = Z/2* (
= 2.576* (4 /
= 1.46
At 99% confidence interval estimate of the population mean is,
- E < < + E
24.3 - 1.46 < < 24.3 + 1.46
22.84 < < 25.76
Lower Limit = 22.84
Upper Limit = 25.76
milcah answered 1 month agoRelated Solutions
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