In: Statistics and Probability
10. The test taking times (in minutes) for Algebra students from my morning, afternoon and evening classes are listed below. Use a 0.05 significance level to test the claim that the different classes have the same test taking time.
Morning Class 24 21 23 30 23
Afternoon Class 25 31 23 20 28
Evening Class 23 27 22 30 20
USE A 0.05 SIGNIFICANCE LEVEL TO TEST THE CLAIM:
H0: All the means are the same. (generic null hypothesis)
H1:At least one mean is different. (generic alternative hypothesis)
Test Statistic: F=_________ Q37(ROUND TO 3 DECIMAL PLACES)
P-value: p=___________Q38(ROUND TO 4 DECIMAL PLACESzero place holders)
Decision/Conclusion: _________________________________Q39(CIRCLE ONE BELOW) Reject H0, There is sufficient evidence to warrant the rejection of the claim that the means are the same.
OR
Fail to Reject H0, There is not sufficient evidence to warrant the rejection of the claim that the means are the same.
Morning | Afternoon | Evening | Total | |
1 | 24 | 25 | 23 | |
2 | 21 | 31 | 27 | |
3 | 23 | 23 | 22 | |
4 | 30 | 20 | 30 | |
5 | 23 | 28 | 20 | |
Total | 121 | 127 | 122 | 370 |
Mean | 24.2 | 25.4 | 24.4 | |
Squares | 2975 | 3299 | 3042 | 9316 |
This is ANOVA - Single Factor. We first need to find the and . They are 'Total variance' and 'Variance in between the treatments' respectively.
We know
=
Grand total = 370 Sum of individual squares = 9316
= 189.333
=
Note: We have to separately sum for all the treatment.
=
= 9130.8
= 4.1333
Df (Bet) = k -1 ………..where k = no. of treatments
= 2
189.333 - 4.133
= 185.2
Df (res) = n – k
=15 - 3
= 12
ANOVA | ||||
Source of Variation | SS | df | MS (SS /df) | F |
Between Groups | 4.1333 | 2 | 2.0667 | 0.1339 |
Within Groups | 185.2000 | 12 | 15.4333 | |
Total | 189.3333 | 14 |
H0: All the mean test time for all the class are the same.
H1:At least one mean test time is different.
Test Stat = F-ratio
= MSB / MSW
= 2.0667/ 15.4333
Test Statistic: F=_0.1339_
p-value = P( > F-stat)
= P( > 0.13) ......................using f-dist with df1 = 2, df2 = 12
P-value: p=_0.8760_Q38(ROUND TO 4 DECIMAL PLACESzero place holders)
Since p-value > 0.05
Decision/Conclusion:
Fail to Reject H0, There is not sufficient evidence to warrant the rejection of the claim that the means are the same.
p-value is the probability that the null hypothesis is true by chance.