In: Statistics and Probability
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 |
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