Question

In: Math

10. In a study conducted to determine whether the role that sleep disorders play in academic...

10. In a study conducted to determine whether the role that sleep disorders play in academic performance, researcher conducted a survey of 1800 college students to determine if they had a sleep disorder. Of the 500 students with a sleep disorder, the mean GPA was 2.51 with a standard deviation of 0.85. Of the 1300 students without a sleep disorder, the mean GPA is 2.85 with a standard deviation of 0.78. Test the claim that sleep disorder adversely affects one’s GPA at the 0.05 level of significance?

11. In one experiment, the participant must press a key on seeming a blue screen and reaction time (in seconds) to press the key is measured. The same person is then asked to press a key on seeing a red screen, again with reaction time measured. The results for six randomly sampled study participants are as follows:

Participant

1

2

3

4

5

6

Blue

0.582

0.481

0.841

0.267

0.685

0.450

Red

0.408

0.407

0.542

0.402

0.456

0.522

Construct a 99% confidence interval about the population mean difference. Assume the differences are approximately normally distributed.

Solutions

Expert Solution

10)

Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 <   0          
                  
Level of Significance ,    α =    0.05          
                  
Sample #1   ---->   1          
mean of sample 1,    x̅1=   2.51          
standard deviation of sample 1,   s1 =    0.85          
size of sample 1,    n1=   500          
                  
Sample #2   ---->   2          
mean of sample 2,    x̅2=   2.850          
standard deviation of sample 2,   s2 =    0.78          
size of sample 2,    n2=   1300          
                  
difference in sample means = x̅1-x̅2 =    2.510   -   2.8500   =   -0.3400
                  
std error , SE =    √(s1²/n1+s2²/n2) =    0.0437          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -0.3400   /   0.0437   ) =   -7.7736

p-value =        0.0000   [ excel function: =T.DIST(t stat,df) ]
Conclusion:     p-value<α , Reject null hypothesis      
          
There is enough evidence that to support the claim that  sleep disorder adversely affects one’s GPA at the 0.05 level of significance


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