Question

In: Statistics and Probability

8. I am interested in whether or not students who fail this statistics course one semester...

8. I am interested in whether or not students who fail this statistics course one semester do better on their tests the second time they take the class. To examine this, I record the test scores for a sample of 10 students who took the class last semester and are repeating the class this semester. The test scores for these 10 students are reported below for last semester and this semester. Conduct the appropriate hypothesis test to determine whether or not student scores are significantly greater for students taking the class the second semester compared to their scores the first semester. State a type of a test, the null and research hypotheses, the critical value, obtained t statistic, your conclusion, and interpret your results (Alpha = .05). (6 pt)       

Student

Last Semester

This Semester

A

65

78

B

70

72

C

54

66

D

66

57

E

42

50

F

69

82

G

70

70

H

64

62

I

39

55

J

53

60

Solutions

Expert Solution

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 <   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   59.20                  
standard deviation of sample 1,   s1 =    11.55                  
size of sample 1,    n1=   10                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   65.20                  
standard deviation of sample 2,   s2 =    10.30                  
size of sample 2,    n2=   10                  
                          
difference in sample means =    x̅1-x̅2 =    59.2000   -   65.2   =   -6.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    10.9473                  
std error , SE =    Sp*√(1/n1+1/n2) =    4.8958                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -6.0000   -   0   ) /    4.90   =   -1.226
                          
Degree of freedom, DF=   n1+n2-2 =    18                  

p-value =        0.1181 [ excel function: =T.DIST(t stat,df) ]              
Conclusion:     p-value>α , Do not reject null hypothesis                      
                          
There is not enough evidence to say taht student scores are significantly greater for students taking the class the second semester compared to their scores the first semester.

.................

THANKS

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