Question

In: Statistics and Probability

10. The test taking times (in minutes) for Algebra students from my morning, afternoon and evening...

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.

Solutions

Expert Solution

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.


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