In order to help identify baby growth patterns that are unus

In order to help identify baby growth patterns that are unusual, we need to construct a confidence interval estimate of the mean head circumference of ail babies that arc two months old. A random sample of 100 babies is obtained, and the mean head circumference is found to be 40.6 cm. Assume that the population standard deviation IS known to be 1.6 cm, find a 99% confidence interval estimate of the mean head circumference of ail two-month-old babies. What is the margin of error? IF The mergin error was to be plusminus 1 what would the sample size be?

Solution

a)
Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Mean(x)=40.6
Standard deviation( sd )=1.6
Sample Size(n)=100
Margin of Error = Z a/2 * 1.6/ Sqrt ( 100)
= 2.58 * (0.16)
= 0.413

CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=40.6
Standard deviation( sd )=1.6
Sample Size(n)=100
Confidence Interval = [ 40.6 ± Z a/2 ( 1.6/ Sqrt ( 100) ) ]
= [ 40.6 - 2.58 * (0.16) , 40.6 + 2.58 * (0.16) ]
= [ 40.187,41.013 ]

b)
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.01% LOS is = 2.58 ( From Standard Normal Table )
Standard Deviation ( S.D) = 1.6
ME =1
n = ( 2.58*1.6/1) ^2
= (4.128/1 ) ^2
= 17.04 ~ 18