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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* (/n)

= 1.960 * (4/ 50 )

= 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* (/n)

= 2.576* (4 / 50 )

= 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 4 hours ago
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