We have the survey data on the body mass index BMI of 662 yo

We have the survey data on the body mass index (BMI) of 662 young women. The mean BMI in the sample was x=25.2. We treated these data as an SRS from a Normally distributed population with standard deviation =7.8 .

Find the margins of error for 95 % 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)=25.2
Standard deviation( sd )=7.8
Sample Size(n)=73
Margin of Error = Z a/2 * 7.8/ Sqrt ( 73)
= 1.96 * (0.913)
= 1.789

b)
Mean(x)=25.2
Standard deviation( sd )=7.8
Sample Size(n)=357
Margin of Error = Z a/2 * 7.8/ Sqrt ( 357)
= 1.96 * (0.413)
= 0.809

c)
Sample Size(n)=1623
Margin of Error = Z a/2 * 7.8/ Sqrt ( 1623)
= 1.96 * (0.194)
= 0.379