In order to compare the durability of four different brands of golf balls (ALPHA, BEST, CENTURY, and DIVOT), the National Golf Association randomly selects five balls of each brand and places each ball into a machine that exerts the force produced by a 250-yard drive. The number of simulated drives needed to crack or chip each ball is recorded. The results are given in the following table.
Alpha Best Century Divot
281 270 218 364
220 334 244 302
274 307 225 325
242 290 273 337
251 331 249 355
Provide the EXCEL printout of ONEWAY ANOVA using either DATA ANALYSIS or PTSTAT2 to determine whether the mean number of stimulated drives differs by brand of golf ball.
Use the ANOVA printout to write-up the six step hypothesis testing procedure for determining whether the average number of simulated drives differs by brand of golf ball. What is the p-value for this problem and what does it mean? Assume alpha =.05.

As our computed test statistic falls in critical region, so we reject our null hypothesis. Also p-value=0.0009 is less than significance level, so we reject our null hypothesis and conclude that the average number of simulated drives differs by brand of golf ball at 5% significance level.

Step-by-step explanation:

Excel output print out

Anova: Single Factor

SUMMARY

Groups Count Sum Average Variance

Alpha 5 1268 253.6 609.3

Best 5 1532 306.4 740.3

Century 5 1209 241.8 469.7

Divot 5 1583 316.6 1235.3

ANOVA

Source of Variation SS df MS F P-value F crit

Between Groups 20960.4 3 6986.8 9.149217573 0.000925804 3.238871517

Within Groups 12218.4 16 763.65

Total 33178.8 19

The six steps of hypothesis are

1. Null hypothesis: The average number of simulated drives are same by brand of golf ball

Alternative hypothesis: The average number of simulated drives differs by brand of golf ball

2. Significance level: α=0.05

3. Test statistic: F-ratio for one way anova

4. Computations

F=9.149

5. Critical region: F>3.239

6. Conclusion:

As our computed test statistic falls in critical region, so we reject our null hypothesis. Also p-value=0.0009 is less than significance level, so we reject our null hypothesis and conclude that the average number of simulated drives differs by brand of golf ball at 5% significance level.