Question

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

Answer 1

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