Question

A random sample of nequals9 values taken from a normally distributed population with a population variance of 16 resulted in the sample values shown below. Use the sample values to construct a 95​% confidence interval estimate for the population mean. 54 45 55 44 44 52 47 59 50

The 95​% confidence interval is -------.------- ​(Round to two decimal places as needed. Use ascending​ order.)

Answer 1

Solution:

Sample size = n = 9

54 45 55 44 44 52 47 59 50

Let be the sample mean

= (54+45+55+.......+59+50)/9 = 50

Given that

Population variance = ² = 16

So ,

Population SD = = 4

Note that, Population standard deviation() is known..So we use z distribution.

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025 and 1- /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

= 1.96   

The margin of error is given by

E =  /2 * ( / n )

= 1.96 * (4 / 9)

= 2.61

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

(50 - 2.61)   <   <  (50 + 2.61)

47.39 <   < 52.61

The 95​% confidence interval is  47.39 , 52.61