The FMA Company has designed a new type of 16 lb. bowling ball. The company know

Question

The FMA Company has designed a new type of 16 lb. bowling ball. The company knows that the average man who bowls in a scratch league with the company's old ball has a bowling average of 155. The variance of these averages is 100. The company asks a random sample of 100 men bowling in scratch leagues to bowl for five weeks with their new ball. The mean of bowling averages for these men with the new ball is 170. There is no reason to believe the variance is any different with the new ball. Test the null hypothesis that the new does not improve a bowler's average at the 5% level of significance?

Explanation / Answer

ans=

"true mean" of bowlers’ averages to a 0.05 level of significance are not improved above 155 with the new ball.
SINGLE SAMPLE TEST, ONE-TAILED, 7 - Step Procedure for t Distributions, "one-tailed test"

1. Parameter of interest: "" = population mean of bowling averages. (statistical “average” different than “bowling averages”)

2. Null hypothesis Ho: = 155

3. Alternative hypothesis Ha: > 155 (“one-tailed”)

4. Test statistic formula: t = (x-bar - )/(SQRT(variance))
x-bar = estimate of the Population Mean (statistical mean of the sample) [170]
n = number of individuals in the sample [100]
= Population Mean [155] (used for Test statistic)

5. Computation of Test statistic formula t = 1.5

6. Determination of the P-value: The test is based on n -1 = 99 df (degrees of freedom). Table "look-up" value shows area under the 99 df curve to the right of t = 1.5 is (approximately) 0.068

7. Conclusion: with level of significance value = 0.05 (5%) above P-value > , [0.068 > 0.05]. Null hypothesis Ho: = 155 should not be rejected. "true mean" of bowlers’ averages to a 0.05 level of significance are not improved above 155 with the new ball.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.