Question

The police and faculty conducted several experiments with the possible murder weapons (even the murder hornet). In one test, they simulated the use of each weapon and recorded the amount of time (in minutes) required to complete the simulated murder task (i.e. the “time to kill”). The times for each weapon are normally distributed.

	Sample size (n)	Sample mean (x̄)	Sample Variance ()
Stun gun	10	13.6667	3.5152
Poisoned M&M	10	12.25	3.8409
Murder hornet	10	14.0833	0.5184

(a) Perform an F test for one-way ANOVA with α = 0.05 to determine whether there is a significant difference among the mean times. For simplicity, assume that all assumptions are satisfied.

(b) If ANOVA indicates that there is a difference among the means, perform the Tukey-Kramer test with α = 0.05 on each pair of means.  If the “time to kill” of at least two of the weapons significantly differ, you can eliminate the weapon with the longest “time to kill.”

Answer 1

The Hypothesis:

H0: There is no difference between the means

Ha: Not all the means are equal.

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The ANOVA table is as below.

Source	SS	DF	Mean Square	F	Fcv	p
Between	18.47	2	9.24	3.52	3.354	0.0439
Within/Error	70.87	27	2.625
Total	89.34	29

The p value is calculated for F = 3.52 for df1 = 2 and df2 = 27

The F critical is calculated at = 0.05 for df1 = 2 and df2 = 27

The Decision Rule:

If F test is > F critical, Then Reject H0.

Also if p-value is < , Then reject H0.

The Decision:

Since F test (3.52) is > F critical (3.354), We Reject H0.

Also since p-value (0.0439) is < (0.05), We Reject H0.

The Conclusion: There is sufficient evidence at the 95% level of significance to conclude that the mean of at least one time is different from the others.

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Calculations For the ANOVA Table:

Overall Mean : Since n is equal , the overall mean = mean of means

Overall Mean = (13.6667 + 12.25 + 14.0833) / 3 = 13.33

SS treatment = SUM [n* ( Individual Mean - overall mean)²] = 10 * (13.6667 - 13.33)² + 10 * (12.25 - 13.33)² + 10 * (14.0833 - 13.33)² = 18.47

df1 = k - 1 = 3 - 1 = 2

MSTR = SS treatment / df1 = 18.47 / 2 = 9.24

SS error = SUM [(n - 1) * Variance] = 9 * (3.5152 + 3.8409 + 0.5184) = 70.87

df2 = N - k = 30 - 3 = 27

Therefore MS error = SS error / df2 = 70.87 / 27 = 2.625

F = MSTR / MSE = 9.24 / 2.625 = 3.52

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Tukeys Post Hoc Table

Where Mi and Mj are the 2 means being compared and their positive (absolute) differences is taken.

n = number of replicates in each sample. Here n = 10, MSerror = 2.625

The Rule is that if Tukeys observed is > Tukeys critical, then there is a significant difference between groups.

Tukeys critical is found from the critical value tables for = 0.0, for k (# of columns) = 3 on the horizontal and df error = 27on the vertical.

The Critical Value = 3.506

Stun Gun = 13.6667, Poisoned M & M = 12.25, Murder Hornet = 14.0833

For Stun Gun and Poisoned M & M:

HSD = ABS(13.6667 - 12.25) / 0.51 = 2.78 which is < 3.506, hence the difference between the means is not significant.

For Stun Gun and Murder Hornet:

HSD = ABS(13.6667 - 14.0833) / 0.51 = 0.816 which is < 3.506, hence the difference between the means is not significant.

For Poisoned M & M and Murder Hornet:

HSD = ABS(1.25 - 14.0833) / 0.51 = 3.6 which is > 3.506, hence the difference between the means is significant.