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. Final exam scores reported 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? Do not use mini tab

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) let σ1 and σ2 be population std of classroom and online classes.

Ho: σ1 = σ2

H1: σ1 ╪ σ2

class room

Sample Size = 18
Sample Variance,s1² = 94.418

online
Smaller-Variance Sample  
Sample Size = 36
Sample Variance,s2² = 77.1071

F stat = s1²/s2² = 94.418/77.1071 = 1.2245

df1 =n1-1=17

df2=n2-1 = 35

p value =0.5938 [excel fucntion :   =F.DIST.RT(1.2245,17,35) ]

p value >α=0.05,Do not reject the null hypothesis

so, there is not enough evidence to reject the claim

b)

mean of sample 1,    x̅1=   71.0833
standard deviation of sample 1,   s1 =    8.7811
size of sample 1,    n1=   36


      
mean of sample 2,    x̅2=   76.778
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, confidence interval do not contain 0, so, data support that there is no difference between two means


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