Answer:
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.
