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

Answer:

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]

Step-by-step 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]