Question

A sporting goods manufacturing company wanted to compare the distance traveled by golf balls produced using each of four different designs. Ten balls were manufactured with each design and were brought to the local golf course for the club professional to test. The order in which the balls were hit with the same club from the first tee was randomized so that the pro did not know which type of ball was being hit. All 40 balls were hit in a short period of time, during which the environmental conditions were essentially the same. The results (distance traveled in yards) for the four designs were as follows:

a. At the 0.05 level of significance, is there evidence of a significant difference in the mean distances traveled by the golf balls with different designs?

b. If results in (a) indicate that it is appropriate, use the Tukey – Kramer procedure to determine which designs differ in mean distances.

c. What assumptions are necessary in (a)?

e. What golf ball design should the manufacturing manager choose? Explain.

Design

1	2	3	4
206.32 207.94 206.19 204.45 209.65 203.81 206.75 205.68 204.49 210.86	217.08 221.43 218.04 224.13 211.82 213.90 221.28 229.43 213.54 214.51	226.77 224.79 229.75 228.51 221.44 223.85 223.97 234.30 219.50 233.00	230.55 227.95 231.84 224.87 229.49 231.10 221.53 235.45 228.35 225.09

Answer 1

Using excel, running a Single factor ANOVA,

To test:

Vs $H_{a}:$ Not all means are equal.

2 Anova: Single Factor 2 206.32 217.08 226.77 230.55 3 207.94221.43224.79 227.95 4 206.19 218.04229.75 231.84 5 204.45 224.13228.51 224.87 6 209.65 211.82221.44 229.49 7 203.81 213.9 223.85 231.1 8 206.75221.28223.97 221.53 9 205.68 229.43234.3 235.45 10 204.49 213.54219.5 228.35 11 210.86 214.51 12 13 14 15 16 17 18 SUMMARY Groups Count 10 10 10 10 Sum Average Variance 2066.14206.6145.246427 2185.16218.516 30.66718 2265.8 2286.22 228.622 16.10697 2 226.588 23.18213 4 233 225.09 ANOVA Source of Variation Between Groups Within Groups MS P-value F crit 2990.990 676.824 3 996.997 53.030 2.73E-132.866 36 18.801 Total 3667.814 39

a. Since the p-value of the test 0.000 < 0.05, we do not have sufficient evidence to support the null hypothesis.We may reject H₀ at 5% level of significance.

We may conclude that at the 0.05 level of significance, there is a significant difference in the mean distances traveled by the golf balls with different designs.

b. To identify which among them differs, we may use the Tukey – Kramer procedure.

The formula for Tukey's is given by:

where, q can be obtained from Studentized range tables,

MSE= Within mean sum of squares (Obtained from ANOVA Output)

n = No. of treatments compared

Substituting the values,

q = 3.809, MSE = 18.801, n = 4,

Comparing each of the pairwise differences with the HSD, if the former exceeds the latter, we conclude that the mean difference is significant.

* * *    *   

We find that means differs significantly from the rest at 5% level.Also, differs significantly from and ..

c. The main assumptions for One way ANOVA:

1. The samples are independent

2. Homogeneity of variances

3. The residuals are normally distributed

d. Comparing the 4 means, since the largest mean distance traveled is by ball design 4, manufacturing manager must choose golf ball design 4.