We have the survey data on the body mass index BMI of 663 yo

We have the survey data on the body mass index (BMI) of 663 young women. The mean BMI in the sample was x¯=25.5. We treated these data as an SRS from a Normally distributed population with standard deviation =7.6 .

Find the margins of error for 99 % confidence based on SRSs of N young women.

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)=25.5
Standard deviation( sd )=7.6
Sample Size(n)=71
Margin of Error = Z a/2 * 7.6/ Sqrt ( 71)
= 2.58 * (0.902)
= 2.327
b)
Sample Size(n)=408
Margin of Error = Z a/2 * 7.6/ Sqrt ( 408)
= 2.58 * (0.376)
= 0.971
c)

Margin of Error = Z a/2 * 7.6/ Sqrt ( 1560)
= 2.58 * (0.192)
= 0.496