The increased availability of light materials with high strength has revolutionized the design and manufacture of golf clubs, particularly drivers. One measure of drivers that result in much longer tee shots is known as the coefficient of restitution of the club. An experiment was performed in which 15 drivers produced by a particular club maker were selected at random and their coefficients of restitution measured. It is of interest to determine if there is evidence to support a claim that the mean coefficient of restitution exceeds 0.82. Assume values to be normally distributed. The following observations were obtained for the 15 drivers:
0.8411 0.8191 0.8182 0.8125 0.8750
0.8580 0.8532 0.8483 0.8272 0.7983
0.8042 0.8730 0.8282 0.8359 0.8660
Conduct the test using a significance level of 0.05.

Question

contestada

Respuesta :

fichoh fichoh · Answer 1 · 2021-07-26T05:32:47+02:00

Answer:

WE reject the Null and conclude that the mean coefficient of restitution exceeds 0.82

Step-by-step explanation:

This is a one sample t test :

The hypothesis :

H0 : μ = 0.82

H0 : μ > 0.82

Given the sample data:

0.8411 0.8191 0.8182 0.8125 0.8750

0.8580 0.8532 0.8483 0.8272 0.7983

0.8042 0.8730 0.8282 0.8359 0.8660

Sample size, n = 15

Sample mean = ΣX / n = 0.837

Sample standard deviation, s = 0.0246 (from calculator)

The test statistic :

T = (xbar - μ) ÷ (s/√(n))

T = (0.837 - 0.82) ÷ (0.0246/√(15))

T = 2.676

The critical value at α = 0.05

df = n - 1 ; 15 - 1 = 14

Tcritical(0.05, 14) = 1.761

Reject H0 if Test statistic > Tcritical

Since, 2.676 > 1.761 ; WE reject the Null and conclude that the mean coefficient of restitution exceeds 0.82