Upon leaving college you decide to do some volunteer medical

Upon leaving college, you decide to do some volunteer medical service in a poor rural community. Many of the children seem to be undernourished and appear to be anemic. People with less than 12 grams of hemoglobin per deciliter of blood are considered anemic. You saw 36 children and found that the mean amount of hemoglobin was 11.3 g/dl and the standard deviation was 1.5 g/dl. Construct two 95% confidence intervals, one for the mean and one for the standard deviation, for the hemoglobin level for the children in this community. Comment on your findings relative to the standard deviation of anemic given above.

Solution

Confidence Interval For Mean
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=11.3
Standard deviation( sd )=1.5
Sample Size(n)=36
Confidence Interval = [ 11.3 ± t a/2 ( 1.5/ Sqrt ( 36) ) ]
= [ 11.3 - 2.03 * (0.25) , 11.3 + 2.03 * (0.25) ]
= [ 10.793,11.808 ]

Confidence Interval For Standard Deviation
CI = (n-1) S^2 / ^2 right < ^2 < (n-1) S^2 / ^2 left
Where,
S = Standard Deviation
^2 right = (1 - Confidence Level)/2
^2 left = 1 - ^2 right
n = Sample Size

Since aplha =0.05
^2 right = (1 - Confidence Level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
^2 left = 1 - ^2 right = 1 - 0.025 = 0.975
the two critical values ^2 left, ^2 right at 35 df are 53.2033 , 20.569
S.D( S^2 )=1.5
Sample Size(n)=36
Confidence Interval = [ 35 * 2.25/53.2033 < ^2 < 35 * 2.25/20.569 ]
= [ 78.75/53.2033 < ^2 < 78.75/20.5694 ]
Confidence Interval = [ 1.4802 < ^2 < 3.8285 ]
Confidence Interval S.D = [ Sqrt(1.4802) < ^2 < Sqrt(3.8285) ] = [ 1.216 <    < 1.9566 ]