Question

The Body Mass Index (BMI) is a value calculated based on the weight and the height of an individual. In a small European city, a survey was conducted one year ago to review the BMI of the citizens. In the sample with 200 citizens, the mean BMI was 23.3 kg/m2 and standard deviation was 1.5 kg/m2 . It is reasonable to assume the BMI distribution is a normal distribution.

(a) Find the point estimate of the population mean BMI one year ago.

(b) Calculate the sampling error at 90% confidence.

(c) Construct a 90% confidence interval estimate of the population mean BMI one year ago. This city launched a healthy exercise program to reduce citizen’s BMI after last year’s survey. Suppose the program effectively reduces the BMI of each citizen by 2.5%.

(d) Construct a 98% confidence interval estimate of the population mean BMI after the healthy exercise program. (Hint: find the updated sample mean and sample standard deviation of the BMI of the sample with 200 citizens selected last year)

Answer 1

a) The point estimate of the population mean one year ago is the mean of sample population, which = 23.3 kg/m2.

b) The Z score corresponding to 90% confindence, z = 1.65 (from the Z Table).

    Hence, the sampling error = +/-z X sigma/square root (n) (where sigma is the point estimate of the population standard deviation and n is the sample size) = +/-1.65 X 1.5/square root(200) = +/-0.175 (please note that we have assumed value (point estimate) of population standard deviation (sigma) to be the same as that of sample standard deviation)

c) Therefore, the 90% confidence interval estimate of population BMI one year ago = 23.3 +/- 0.175 = 23.125 to 23.475

d) Z score corresponding to 98% confidence interval = 2.33.

Post the program, the new estimate of population mean, n1 = 23.3 X (100-2.5)/100 = 22.72.

The new estimate of population standard deviation = 1.5 X (100-2.5)/100 = 1.46

Thus, the sampling error = +/- 2.33 X 1.46/square root (200) = +/-0.237

Hence, the estimate of population mean post the program with 98% confidence interval = 22.72 +/- 0.237 = 22.483 to 22.957.