statistics problem

Example-1

Case A Par, Inc. Write-up

Statement of the Problem

Par, Inc. manufactures golf equipment and is releasing a new line of cut-resistant, longer-

lasting golf balls. Par, Inc. has received complaints that the ball does not go as far. The

customers have been returning feedback that although Par, Inc.’s new line does resist cuts and

lasts longer, the driving distance for the new line of balls is not as far as their older line of balls.

Purpose of the Study

The purpose of this quantitative study is to test and compare the driving distances of Par,

Inc.’s old line of golf balls with their new longer-lasting line of golf balls. Par, Inc. wants to

determine if there is any significant difference in the mean driving distances of the two lines of

golf balls.

Research Questions

Q1. Is there a significant difference between the average driving distances of Par,

Inc.’s old line and their new line of resistant golf balls?

Hypotheses

µ0 – mean driving distance of old line of golf balls

µ1 – mean driving distance of new line of golf balls

H0. There is no difference in the driving distance between the new balls and the old ball

(i.e., µ0 = µ1)

H1. There is a difference in the driving distances between the new balls and old balls

(i.e., µ0 ≠ µ1)

Example-2

Methodology

For the study a 2-sided t-test was used to test the hypotheses at a sig.05. The two

variables are the current and new line driving distances. For the test, a sample size of 40 balls

from each line were subjected to driving tests by using a mechanical hitting machine on each

ball.

Findings

From the output results above from the Minitab t-test we find that the mean driving

distances of Par, Inc.’s two lines of golf balls is not significantly different. The test was run with

a confidence interval of 95%, giving an alpha of 0.05. The t-test results gave a t-value of 1.33

and a p-value of 0.188. From the p-value we find that the difference in driving distance of the

two golf ball lines is not significant because the p-value of 0.188 is greater than the alpha of

0.05. Therefore, there is no evidence to suggest that the null hypothesis should be rejected.

Further, the alternate hypothesis is rejected.

Business implications/recommendations

From the results of the study it is recommended that Par, Inc. should not recall their new

line of gold balls. Instead, it is recommended that Par, Inc. rethink their advertising strategy for

the new line of golf balls. Par, Inc. need to clearly show to customers that the new line of balls

drives just as far as their current line of balls through data or graphs comparing the two on the

boxes.

Example-3

Two-sample T for Current vs. New

N Mean StDev SE Mean

Current 40 270.27 8.75 1.4

New 40 267.50 9.90 1.6

Difference = mu (Current) - mu (New)

Estimate for difference: 2.77

95% CI for difference: (-1.39, 6.94)

T-Test of difference = 0 (vs. not =): T-Value = 1.33 P-Value = 0.188 DF = 76