Question

Advertisements for the Sylph Physical Fitness Center claim that completion of their course will result in a loss of weight (measured in pounds). A random sample of 8 recent students revealed the following body weights before and after completion of SPF course.

Student 1 2 3 4 5 6 7 8
Before 155 228 141 162 211 185 164 172
After 154 207 147 157 196 180 150 165
The above data summarizes to the following (Note that "Difference = Before - After").
Mean Std Dev
Before 177.25 29.325
After 169.50 22.431
Difference 7.75 8.598
Construct a 90% confidence interval for the mean weight loss for the population represented by this sample assuming that the differences are coming from a normally distributed population.

Answer 1

90% confidence interval for the mean weight loss for the population represented by this sample assuming that the differences are coming from a normally distributed population is [1.989 ,13.5105]

Explanation:

The data given is

Mean Std Dev

Before 177.25 29.325

After 169.50 22.431

Difference 7.75 8.598

Hence d`= 7.75 and sd= 8.598

The 90% confidence interval for the difference in means for the paired observation is given by

d` ± t∝/2(n-1) *sd/√n

Here t∝/2(n-1)=1.895 where n-1= 8-1= 7 d.f

and ∝/2= 0.1/2=0.05

Putting the values

d` ± t∝/2(n-1) *sd/√n

7.75 ±1.895 * 8.598 /√8

7.75 ± 5.7605

1.989 ,13.5105

90% confidence interval for the mean weight loss for the population represented by this sample assuming that the differences are coming from a normally distributed population is [1.989 ,13.5105]