Question

We have the survey data on the body mass index (BMI) of 651 young women. The mean BMI in the sample was x( with bar over it)=26.3. We treated these data as an SRS from a Normally distributed population with standard deviation σ=7.3.

Give confidence intervals for the mean BMI and the margins of error for 90%, 95%, and 99% confidence

Conf. level                         Interval                      margins of error

90%                                  ___to___     ________ 95%                                 ___to____                        _________ 95%                               ____to____                       __________

Answer 1

Solution :

Given that,

= 26.3

= 7.3

n = 651

(a)

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( /n)

= 1.645 * (7.3 / 651)

= 0.47

At 90% confidence interval estimate of the population mean is,

- E < < + E

26.3 - 0.47 < < 26.3 + 0.47

25.83 < < 26.77

25.83 to 26.77

b)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (7.3 / 651)

= 0.56

At 95% confidence interval estimate of the population mean is,

- E < < + E

26.3 - 0.56 < < 26.3 + 0.56

25.74 < < 26.86

(25.74 to 26.86)

c)

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z_/2* ( /n)

= 2.576 * (7.3 / 651)

= 0.74

At 99% confidence interval estimate of the population mean is,

- E < < + E

26.3 - 0.74 < < 26.3 + 0.74

25.56 < < 27.04

(25.56 to 27.04)