Question

Suppose weight is normally distributed. Nine men with a genetic condition that causes obesity entered a weight reduction program. After four months, their mean weight loss was 11.2 pounds lost with standard deviation σ = 9.0 Construct an 80 and 99 percent confidence interval for the mean.

Answer 1

Solution:

Given,

= 9.0 ........Sample standard deviation

= 11.2 ....... Sample mean

n = 9 ....... Sample size

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

1)

80% confidence interval

c = 0.80

= 1- c = 1- 0.80 = 0.20

  /2 = 0.10

Using z table

  = 1.282

The margin of error is given by

E =  /2 * ( / n )

= 1.282 * (9.0 / 9)

= 3.846

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

( - E ) <   <  ( + E)

(11.2 - 3.846)   <   <  (11.2 + 3.846)

7.354 <   <  15.046

Required 80% Confidence interval is (7.354  ​​​, 15.046)

2)

99% confidence interval

c = 0.99

= 1- c = 1- 0.99 = 0.01

  /2 = 0.01 2 = 0.005 and 1- /2 = 0.995

Search the probability 0.995 in the Z table and see corresponding z value

= 2.576   

The margin of error is given by

E =  /2 * ( / n )

= 2.576 * (9.0 / 9)

= 7.728

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

( - E ) <   <  ( + E)

(11.2 - 7.728)   <   <  (11.2 + 7.728)

3.472 <   <  18.928

Required 99% Confidence interval is (3.472 , 18.928)