Question

To test a newly formulated protein bar designed to improve running performance, 10 distance runners performed two trials. All the participants ate the new protein formula bar on the first trial. In the second trial, the same participants ate the regular (old) protein bar. Thirty minutes after consuming the bar the participant ran for 2 hours on a treadmill.

At the end of the two trials, performed on consecutive days for each participant, the distance each participant ran (in km) was recorded. The data is below.

Runner	New formula	Old formula
1	10.6	10.5
2	11.2	10.9
3	10.3	10.2
4	11.9	11.6
5	9.8	9.7
6	11.1	10.8
7	12.1	11.9
8	10.9	10.8
9	9.7	9.8
10	10.3	10.3

a. Comment on the validity of the study design and any ways you can think of to improve the design?

b. Calculate a 99% confidence interval for the population mean difference in run distance after eating the newly formulated protein bar compared to the old protein bar.

c. Carry out the appropriate hypothesis test using a two-sided alternative - at a 0.01 significance level.

H₀: ________________________________

H₁: ________________________________

Name of test:   __________________________

Test Statistic:

Define the rejection region:

Calculate the p-value:

Conclusion in context of the study:

Answer 1

Sample #1	Sample #2	difference , Di =sample1-sample2	(Di - Dbar)²
10.6	10.5	0.10	0.00
11.2	10.9	0.30	0.03
10.3	10.2	0.10	0.00
11.9	11.6	0.30	0.03
9.8	9.7	0.10	0.00
11.1	10.8	0.30	0.03
12.1	11.9	0.20	0.00
10.9	10.8	0.10	0.00
9.7	9.8	-0.10	0.06
10.3	10.3	0.00	0.02

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	107.9	106.5	1.400	0.164

sample size ,    n =    10          
Degree of freedom, DF=   n - 1 =    9   and α =    0.01  
t-critical value =    t α/2,df =    3.2498   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    0.1350          
                  
std error , SE = Sd / √n =    0.1350   / √   10   =   0.0427
margin of error, E = t*SE =    3.2498   *   0.0427   =   0.1387
                  
mean of difference ,    D̅ =   0.140          
confidence interval is                   
Interval Lower Limit= D̅ - E =   0.140   -   0.1387   =   0.001
Interval Upper Limit= D̅ + E =   0.140   +   0.1387   =   0.279
                  
so, confidence interval is (   0.0013   < µd <   0.2787   )  

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Ho :   µd=   0                  
Ha :   µd ╪   0                  
                          
Level of Significance ,    α =    0.01       claim:µd=0          
                          
sample size ,    n =    10                  
                          
mean of sample 1,    x̅1=   10.790                  
                          
mean of sample 2,    x̅2=   10.650                  
                          
mean of difference ,    D̅ =ΣDi / n =   0.140                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    0.1350                  
                          
std error , SE = Sd / √n =    0.1350   / √   10   =   0.0427      
                          
t-statistic = (D̅ - µd)/SE = (   0.14   -   0   ) /    0.0427   =   3.280
                          
Degree of freedom, DF=   n - 1 =    9                  
t-critical value , t* =    ±   3.250   [excel function: =t.inv.2t(α,df) ]               
                          
p-value =        0.009535   [excel function: =t.dist.2t(t-stat,df) ]               
Conclusion:     p-value <α , Reject null hypothesis      

newly formulated protein bar designed to improve running performance

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THANKS

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