A boutique fitness studio would like to test whether their program is affective

Question

A boutique fitness studio would like to test whether their program is affective at weight loss. They randomly select 7 female participants to adhere to their diet and exercise regimen for 12 weeks. A before and after weight in pounds was recorded for each participant. Consider all conditions to be met. Use a significance level of 0.10 and assume the population of differences is normal.

Participant 1 2 3 4 5 6 7

Before 160.1 154.4 145.2 128.1 137.4 124.8 188.9

After 158.3 143.5 139.4 124.6 129.7 122.2 190.1

d. (4 points) Calculate the test statistic. State the degrees of freedom and p-value

e. (3 points) Calculate and interpret a 90% confidence interval for the average difference of weight in lbs

Explanation / Answer

d)

The differences are

-1.8
-10.9
-5.8
-3.5
-7.7
-2.6
1.2

Formulating the null and alternative hypotheses,              
              
Ho:   ud   >=   0  
Ha:   ud   <   0  

At level of significance =    0.01          

As we can see, this is a    left   tailed test.      
              
Calculating the standard deviation of the differences (third column):              
              
s =    3.751716779          
              
Thus, the standard error of the difference is sD = s/sqrt(n):              
              
sD =    1.418015655          
              
Calculating the mean of the differences (third column):              
              
XD =    -4.442857143          
              
As t = [XD - uD]/sD, where uD = the hypothesized difference =    0   , then      
              
t =    -3.133150982 [ANSWER, TEST STATISTIC]

************************************
          
              
As df = n - 1 =    6   [ANSWER, DEGREES OF FREEDOM]

***********************************      
              
Also, using p values, as this is left tailed,              
              
p =        0.01012167   [ANSWER, P VALUE]

***********************************  
              
e)

For the   0.9   confidence level,      
              
alpha/2 = (1 - confidence level)/2 =    0.05          
t(alpha/2) =    1.943180281          
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    -7.198317201          
upper bound = [X1 - X2] + t(alpha/2) * sD =    -1.687397084          
              
Thus, the confidence interval is              
              
(   -7.198317201   ,   -1.687397084   ) [ANSWER]

Hence, we are 90% confident that the true weight difference (after-before) is between -7.198317201   and -1.687397084 lbs. [CONCLUSION]

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