In: Math
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.
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