Question

hw18#4

A random sample of 80 observations produced a mean of x¯=21.1 from a population with a normal distribution and a standard deviation σ=4.54.

(a) Find a 90% confidence interval for ?μ
_____ ≤ ? ≤ ______

(b) Find a 99% confidence interval for ?μ
____ ≤ ? ≤ ______

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 21.1

Population standard deviation =    = 4.54

Sample size = n = 80

(a) At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z/2 * ( /n)

= 1.645 * ( 4.54 /  80 )

= 0.83

At 90% confidence interval estimate of the population mean is,

- E < < + E

21.1 - 0.83 <   < 21.2 + 0.83

20.27 <   < 21.93

( 20.27 , 21.93 )

The 90% confidence interval for is :  20.27 <   < 21.93

(b) At 99% confidence level

= 1 - 99%  

= 1 - 0.99 = 0.01

/2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z/2 * ( /n)

= 2.576 * ( 4.54 /  80 )

= 1.31

At 99% confidence interval estimate of the population mean is,

- E < < + E

21.1 - 1.31 <   < 21.2 + 1.31

19.79 <   < 22.41

( 19.79 , 22.41 )

The 99% confidence interval for is : 19.79 <   < 22.41