We use cookies to give you the best experience possible. By continuing we’ll assume you’re on board with our cookie policy

Check Writers' Offers

What's Your Topic?

Hire a Professional Writer Now

The input space is limited by 250 symbols

What's Your Deadline?

Choose 3 Hours or More.
Back
2/4 steps

How Many Pages?

Back
3/4 steps

Sign Up and Get Writers' Offers

"You must agree to out terms of services and privacy policy"
Back
Get Offer

Par Inc. Case Study Essay

Paper type: Case study Pages: 2 (471 words)

The management at Par Inc. believes that with the introduction of a cut-resistant, longer-lasting golf ball could increase their market share. A new golf ball coating designed to resist cuts and provide a more durable ball have been developed and tested. A sample of 40 balls of both the new and current models were tested with a mechanical hitting machine so that any difference between the mean distances for the two models could be attributed to a difference in the two models.

Therefore, the hypothesis test that Par could use to compare the driving distances of the current and new golf balls can be formulated as follows:

It is set to be a two-tailed test which refers to the difference between the mean distances. Here, refers to the mean driving distance of the new golf ball whilst refers to the mean driving distance of the current golf ball. When the difference between the mean distances is equal to 0, therefore, the null hypothesis rejected.

This shows that the new golf ball is better.

With the formula, the value of the test statistics was computed. This is a necessary step to find the p-value. Next step would be computing the degree of freedom using the formula as such: . The findings presented a test statistics of and the degree of freedom of 76. Therefore, the p-value is greater than the level of significance which was chosen to be 0.05 and the null hypothesis is not rejected. It is recommended for Par Inc. to produce new golf balls with the coating as to increase their market share.

Current
New
Mean

270.275
267.5
Variance

76.61474291
97.94871868
Standard Deviation

8.7529848
9.8969045
Standard Error

1.383968415
1.564838

Using the formula, the interval estimation was calculated. Therefore, with confidence at 95% the differences between the mean distances are in between -1.385740214 and 6.935740214 yard.

The sample size is said to have an inverse relationship with standard error: as the sample size increases, the standard error decreases. It is believed that increasing the sample size gives a diminished return because the increased accuracy will be negligible. Therefore, it is suggested for Par Inc. to have a larger sample sizes for having more information delivers much accurate result.

b. The mean distance for current model, .
The mean distance for new model, .

Microsoft Excel was used to compute the standard deviation for both models. The standard deviation was formulated as follows: For the current model, =STDEV(A2:A41) giving the answer .
For the new model, =STDEV(B2:B41) giving the answer .

The test statistics was then formulated as follows:
Next, the degree of freedom was computed using the formula: With , the result shows the degree of freedom of 76.
To conclude, because the p-value , therefore there is not sufficient statistical evidence to infer that the null hypothesis is true.

Cite this page

Par Inc. Case Study. (2016, Apr 25). Retrieved from https://studymoose.com/par-inc-case-study-essay

How to Avoid Plagiarism
  • Use multiple resourses when assembling your essay
  • Use Plagiarism Checker to double check your essay
  • Get help from professional writers when not sure you can do it yourself
  • Do not copy and paste free to download essays
Get plagiarism free essay

Not Finding What You Need?

Search for essay samples now

image

Your Answer is very helpful for Us
Thank you a lot!