Question

The hemoglobin level in blood of men is normally distributed with the standard deviation of 2.0 g/dl. 12 men were selected in a sample. a. The researcher calculated the confidence interval between 13 and 15 g/dl. What is the level of confidence for this interval? b. Calculate the 95% confidence interval from this sample

Answer 1

Solution:

Given ,

= 2.0

n = 12

a)

Given a confidence interval (13 , 15)

Lower Limit = 13

Upper Limit = 15

Confidence interval is nothing but    E

So ,

= (Upper Limit + Lower Limit)/2 = (15 + 13)/2 = 14

E = (Upper Limit - Lower Limit)/2 = (15 - 13)/2 = 1

But we know ,

E =  /2 * ( / n )

1 =   /2 * (2.0 / 12 )

/2 = 1.73

Use z table , see the column of z . See 1.7 and then see corresponding value in the row at .03

The value is 0.9584

So , 1 - (/2) = 0.9584

(/2) = 0.0416

= 0.0832

Now , confidence level = 1 - = 1 - 0.0832 = 0.9168 = 0.92 = 92%

Confidence level is 92%

b)

Now , let c = 95% = 0.95

= 1 - c = 1 - 0.95 = 0.05

/2 = 0.025

/2 = 1.96

E =  /2 * ( / n )

= 1.96 * (2.0 / 12 )

= 1.13

Confidence interval is given by

- E < < +  E

14 - 1.13 < < < 14 + 1.13

12.87 < < < 15.13

Required 95% confidence interval is (12.87 , 15.13)