Question

Is the average time to complete an obstacle course different when a patch is placed over the right eye than when a patch is placed over the left eye? Thirteen randomly selected volunteers first completed an obstacle course with a patch over one eye and then completed an equally difficult obstacle course with a patch over the other eye. The completion times are shown below. "Left" means the patch was placed over the left eye and "Right" means the patch was placed over the right eye.

Time to Complete the Course

Right	48	44	46	48	41	41	45	50
Left	45	39	45	45	41	38	44	49

Assume a Normal distribution. What can be concluded at the the αα = 0.01 level of significance level of significance?

For this study, we should use Select an answer: z-test for the difference between two population proportions, z-test for a population proportion, t-test for the difference between two independent population means, t-test for the difference between two dependent population means, or t-test for a population mean

1. The null and alternative hypotheses would be:   
2.   

H0: Select an answer p1 μd μ1   Select an answer > < ≠ =   Select an answer p2 μ2 0   (please enter a decimal)   

H1: Select an answer p1 μd μ1  Select an answer > = ≠ <   Select an answer 0 μ2 p2  (Please enter a decimal)

2. The test statistic ? t z   =  (please show your answer to 3 decimal places.)
3. The p-value =   (Please show your answer to 4 decimal places.)
4. The p-value is ? ≤ >   αα
5. Based on this, we should Select an answer: reject, fail to reject, or accept   the null hypothesis.
6. Thus, the final conclusion is that ...
   • The results are statistically insignificant at αα = 0.01, so there is statistically significant evidence to conclude that the population mean time to complete the obstacle course with a patch over the right eye is equal to the population mean time to complete the obstacle course with a patch over the left eye.
   • The results are statistically significant at αα = 0.01, so there is sufficient evidence to conclude that the population mean time to complete the obstacle course with a patch over the right eye is not the same as the population mean time to complete the obstacle course with a patch over the left eye.
   • The results are statistically insignificant at αα = 0.01, so there is insufficient evidence to conclude that the population mean time to complete the obstacle course with a patch over the right eye is not the same as the population mean time to complete the obstacle course with a patch over the left eye.
   • The results are statistically significant at αα = 0.01, so there is sufficient evidence to conclude that the eight volunteers that were completed the course in the same amount of time on average with the patch over the right eye compared to the left eye.

* NOTE: Ignore the “thirteen” I think it’s a typo. This is a matched pair (dependent) hypothesis test so you will run the test on the differences This is the 2 sample that once you take the differences it becomes a single sample test run on the differences.

*** Please print or type clearly. Thank you so much

Answer 1

, t-test for the difference between two dependent population means

Sample #1	Sample #2	difference , Di =sample1-sample2	(Di - Dbar)²
48	45	3.00	0.77
44	39	5.00	8.27
46	45	1.00	1.27
48	45	3.00	0.77
41	41	0.00	4.52
41	38	3.00	0.77
45	44	1.00	1.27
50	49	1.00	1.27

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	363	346	17.000	18.875

Ho :   µd=   0                  
Ha :   µd ╪   0                  
                          
Level of Significance ,    α =    0.01       claim:µd=0          
                          
sample size ,    n =    8                  
                          
mean of sample 1,    x̅1=   45.375                  
                          
mean of sample 2,    x̅2=   43.250                  
                          
mean of difference ,    D̅ =ΣDi / n =   2.125                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    1.6421                  
                          
std error , SE = Sd / √n =    1.6421   / √   8   =   0.5806      
                          
t-statistic = (D̅ - µd)/SE = (   2.125   -   0   ) /    0.5806   =   3.660
                          
Degree of freedom, DF=   n - 1 =    7                  
  
p-value =        0.0080 [excel function: =t.dist.2t(t-stat,df) ]   

          
Conclusion:     p-value <α , Reject null hypothesis  

                  
The results are statistically significant at αα = 0.01, so there is sufficient evidence to conclude that the population mean time to complete the obstacle course with a patch over the right eye is not the same as the population mean time to complete the obstacle course with a patch over the left eye...............

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THANKS

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