Question

We have the survey data on the body mass index (BMI) of 659 young women. The mean BMI in the sample was

x = 26.2. We treated these data as an SRS from a Normally distributed population with standard deviation σ = 8.1.

Give confidence intervals for the mean BMI and the margins of error for 90%, 95%, and 99% confidence. (Round your answers to two decimal places.)

confidence level interval margin of error

90% _____ to _____   _____

95%    _____ to _____   _____

99%        _____ to _____   _____

Answer 1

Given ,

Sample size = n = 659

Sample mean = = 26.2

standard deviation σ = 8.1

a)

Confidence level = 90% = 0.9

Significance level = = 1 - 0.9 = 0.1 , /2 = 0.05

Margin of error ( E ) :

Where , is critical value for given confidence level.

= = 1.6449 { Using Excel function , =NORMSINV(1-0.05) = 1.6449 }

So, margin of error is,

E 0.52

90% Confidence interval is ,

Lower limit = - E = 26.2 - 0.52 = 25.68

Upper limit = + E = 26.2 + 0.52 = 26.72

b)

Confidence level = 95% = 0.95

Significance level = = 1 - 0.95 = 0.05 , /2 = 0.025

Margin of error ( E ) :

Where , is critical value for given confidence level.

= = 1.96 { Using Excel function , =NORMSINV(1-0.025) = 1.96}

So, margin of error is,

E 0.62

90% Confidence interval is ,

Lower limit = - E = 26.2 - 0.62 =25.58

Upper limit = + E = 26.2 + 0.62= 26.82

c)

Confidence level = 99% = 0.99

Significance level = = 1 - 0.99 = 0.01 , /2 = 0.005

Margin of error ( E ) :

Where , is critical value for given confidence level.

= = 2.5758    { Using Excel function , =NORMSINV(1-0.005) =2.5758 }

So, margin of error is,

E 0.81

90% Confidence interval is ,

Lower limit = - E = 26.2 - 0.81 =25.39

Upper limit = + E = 26.2 + 0.81= 27.01

Answers :