The university conducted a study to observe the efficiency of students to see whether students do better when they study all at once or in intervals.

Isa Trevino 2020-11-26 Answered
The university conducted a study to observe the efficiency of students to see whether students do better when they study all at once or in intervals.
One group of 12 participants participated in the survey after studying for one hour continuously.
The other group of 12 participated in the survey after studying for three twenty-minute sessions.
As per the survey, results group 1 had a mean score of 75 and a variance of 120.
Second group had a mean score of 86 and a variance of 100. Assuming the normal populations, independent samples, and equal population variances conditions hold, are the mean test scores of these two groups significantly different at the 0.05 level?
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Expert Answer

Sally Cresswell
Answered 2020-11-27 Author has 91 answers

The objective is to check whether the mean test scores of the two groups significantly differ at the 0.05 level.
Let μ1 denote the mean of the participants participated in the survey after studying for one hour continuously.
Let μ2 denote the mean of the participants participated in the survey after studying for three twenty-minute sessions.
The null hypothesis is that there is no difference between the mean test of scores between the two groups. That is,
H0 : μ1=mu2
The alternative hypothesis is that there is a difference between the mean test of the score between the two groups. That is,
H1 : μ1  μ2
The level of significance is,
α=0.05
Group 1:
Sample mean, x¯1=75
Sample variance, S21=120
Sample Size, n1=12
Group 2:
Sample mean, x¯2=86
Sample variance, S22=100
Sample Size, n2=12
t=(x¯1  x¯2)  (μ1  μ2)S2p(1n 1 + 1n 2)
The formula for pooled variance is,
S2p=(n1  1)S21 + (n2  1)S22n1 + n2  2
=(12  1)120 + (12  1)10012 + 12  2
=11(120) + 11(100)24  2
=1320 + 110022
=242022
=110
The test statistic is computed below.
t=(x¯1  x¯2)  (μ1  μ2)S2p(1n 1 + 1n 2)
=75  86  0110(112 + 112)
=11110(0.08333 + 0.08333)
=11110(166667)
=1118.33333
=114.281744
= 2.569047
The degrees of freedom is,
df=n1 + n2  2=12 + 12  2=24  2=22
Computation of p value:
from the alternative hypothesis,it is clear that the test is a two- tailed test.
t represents the test statistic and df is the degrees of freedom p value
=t dist(t, df tails)
=t dist(2.569047, 22, 2)
0.017503
Decision rule:
Reject the null hypothesis if p value is less than the level of significance.
That is, p value < 0.05
Conclusion:
Here the p value is 0.017503 is less than 0.05. So, reject the null hypothesis and conclude that there is a significant difference between the means of the two groups.

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