Question

In: Statistics and Probability

An experiment is conducted to determine if classes offered in an online format are as effective...

An experiment is conducted to determine if classes offered in an online format are as effective as classes offered in a traditional classroom setting. Students were randomly assigned to one of the two teaching methods. Data below.

a. Test the claim that the standard deviations for the two groups are equal. What is the p-value of the test?

b. Construct a 95% confidence interval on the difference in expected final exam scores between the two groups. Does the data support the claim that there is no difference?

On-line Classroom
77 79
66 64
70 88
79 80
76 66
58 81
54 71
72 84
56 77
82 76
90 89
68 62
59 74
67 68
71 98
74 77
72 65
62 83
77
78
76
57
67
69
82
78
80
61
77
65
71
76
58
82
78
74

Solutions

Expert Solution

a)

Ho : σ1 = σ2

Ha: σ1╪σ2

using excel PHStat tool for f test for variance,

Level of Significance 0.05
Larger-Variance Sample
Sample Size 18
Sample Variance 94.41830065
Smaller-Variance Sample
Sample Size 36
Sample Variance 77.10714286
Intermediate Calculations
F Test Statistic 1.2245
Population 1 Sample Degrees of Freedom 17
Population 2 Sample Degrees of Freedom 35
Two-Tail Test
Upper Critical Value 2.1828
p-Value 0.5938
Do not reject the null hypothesis

p-value = 0.5938 >α=0.05 , fail to reject Ho

so, there is no enough evidence that  to reject the claim that standard deviations for the two groups are equal

b)

Ho :   µ1 - µ2 =   0
Ha :   µ1-µ2 ╪   0

mean of sample 1,    x̅1=   71.08333333
standard deviation of sample 1,   s1 =    8.7811
size of sample 1,    n1=   36
      
mean of sample 2,    x̅2=   76.77777778
standard deviation of sample 2,   s2 =    9.7169
size of sample 2,    n2=   18

Degree of freedom, DF=   n1+n2-2 =    52  
t-critical value =    t α/2 =    2.0066   (excel formula =t.inv(α/2,df)


          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    9.0976  
          


std error , SE =    Sp*√(1/n1+1/n2) =    2.6263  
margin of error, E =    t*SE =    5.2700  


          
difference of means =    x̅1-x̅2 =    -5.6944  


confidence interval is           
Interval Lower Limit=   (x̅1-x̅2) - E =    -10.9644  
Interval Upper Limit=   (x̅1-x̅2) + E =    -0.4245

since, 0 do not lie in confidence interval, so, there is enough evidence to reject the claim that there is no difference



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