Question

In: Statistics and Probability

In a comparison of the effectiveness of distance learning with traditional classroom instruction, 12 students took...

In a comparison of the effectiveness of distance learning
with traditional classroom instruction, 12 students
took a business administration course online, while
14 students took it in a classroom. The final
scores were as follows.
Online 64 66 74 69 75 72 77 83 77
91 85 88
Classroom 80 77 74 64 71 80 68 85 83
59 55 75 81 81
Can you conclude that the mean score differs between
the two types of course

Solutions

Expert Solution

Given that

(a)

From above diagram, it is observed that the normality assumption holds here.

Test and CI for Two Variances: Online, Classroom

Method

Null hypothesis Variance(Online) / Variance(Classroom) = 1
Alternative hypothesis Variance(Online) / Variance(Classroom) not = 1
Significance level Alpha = 0.05


Statistics

Variable N StDev Variance
Online 12 8.572 73.477
Classroom 14 9.250 85.566

Ratio of standard deviations = 0.927
Ratio of variances = 0.859


95% Confidence Intervals

CI for
Distribution CI for StDev Variance
of Data Ratio Ratio
Normal (0.518, 1.707) (0.269, 2.913)
Continuous (0.490, 1.968) (0.240, 3.875)


Tests

Test
Method DF1 DF2 Statistic P-Value
F Test (normal) 11 13 0.86 0.810
Levene's Test (any continuous) 1 24 0.05 0.831

Since p-value>0.05 so we can assume that the population variances are same.

Two-Sample T-Test and CI: Online, Classroom

Two-sample T for Online vs Classroom

N Mean StDev SE Mean
Online 12 76.75 8.57 2.5
Classroom 14 73.79 9.25 2.5


Difference = mu (Online) - mu (Classroom)
Estimate for difference: 2.96
95% CI for difference: (-4.30, 10.23)
T-Test of difference = 0 (vs not =): T-Value = 0.84 P-Value = 0.408 DF = 24
Both use Pooled StDev = 8.9457
Since p-value=0.408>0.05 so we conclude that the methods of learning are equally effective.

b.

Pooled version:

Two-Sample T-Test and CI: Online, Classroom

Two-sample T for Online vs Classroom

N Mean StDev SE Mean
Online 12 76.75 8.57 2.5
Classroom 14 73.79 9.25 2.5


Difference = mu (Online) - mu (Classroom)
Estimate for difference: 2.96
95% CI for difference: (-4.30, 10.23)
T-Test of difference = 0 (vs not =): T-Value = 0.84 P-Value = 0.408 DF = 24
Both use Pooled StDev = 8.9457
Unpooled version:

Two-Sample T-Test and CI: Online, Classroom

Two-sample T for Online vs Classroom

N Mean StDev SE Mean
Online 12 76.75 8.57 2.5
Classroom 14 73.79 9.25 2.5


Difference = mu (Online) - mu (Classroom)
Estimate for difference: 2.96
95% CI for difference: (-4.27, 10.20)
T-Test of difference = 0 (vs not =): T-Value = 0.85 P-Value = 0.405 DF = 23
Since in both cases p-value>0.05 so we conclude that the methods of learning are equally effective.

(c) 95% CI for difference: (-4.30, 10.23)


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