Case Problem Par, Inc.

Par, Inc., is a major manufacturer of golf equipment. Management believes that Par’s market share could be increased with the introduction of a cut-resistant, longer-lasting golf ball. Therefore, the research group at Par has been investigating a new golf ball coating designed to resist cuts and provide a more durable ball. The tests with the coating have been promising.

One of the researchers voiced concern about the effect of the new coating on driving distances. Par would like the new cut-resistant ball to offer driving distances comparable to those of the current-model golf ball. To compare the driving distances for the two balls, 40 balls of both the new and current models were subjected to distance tests. The testing was performed 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. Theresults of the tests, with distances measured to the nearest yard, follow. These data are available in the file Golf.

Do:

• Run the t-Test: Two-Sample Assuming Unequal Variances for the Data File Golf (Chapter 10) using the video How to Add Excel’s Data Analysis ToolPak (Links to an external site.) for assistance.

In a managerial report, use the methods of hypothesis testing to

• Provide descriptive statistical summaries of the data for each model, written in complete sentences.
• Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls. Clearly state the null and alternative hypothesis.
• Analyze the data to provide the hypothesis testing conclusion. What is the p-value for your test? What is your recommendation for Par, Inc.?
• Discuss whether you see a need for larger sample sizes and more testing with the golf balls.

Discuss

Based on your hypothesis testing conclusion, what are your recommendations for Par, Inc? Support your recommendations with findings from your managerial report.

Post by classmate 1

 

Hello everyone,

Here are my findings and managerial report after using the methods of hypothesis testing:

After reviewing the data for both current and new golf ball driving distance, a t-Test was performed to determine if the current golf balls travel a longer distance than the new ones in a sample size of 40. Assuming that the data is normally distributed, the t-Test displayed a sample mean for the current golf ball of 270.28 and a sample mean for the new golf ball of 267.5. The difference between the means is not significant. The variance for the current golf ball is 76.61, and the variance for the new golf ball is 97.95. The p-value is 0.094. The standard deviation for the current golf ball is 8.75, and the standard deviation for the new golf ball is 9.9. The “t Critical one-tail” is 1.66, and the “t Critical two-tail” is 1.99.

The hypothesis test was needed to determine if there is enough evidence to support that the current golf balls travel a greater distance than the new ones proposed for Par, Inc (null hypothesis) or if, in fact, they are the same (alternative hypothesis).

The p-value is a random variable derived from the distribution and is a measure of evidence against the null hypothesis (Hung et al., 1997). In this case, the p-value is 0.094, which is less than 0.05, indicates strong evidence against the null hypothesis as there is less than a 5% probability that the null would be correct; hence, the p-value is greater than the significance level, and one should fail to reject the null hypothesis (McLeod, 2019). There appears no evidence to support that the current golf balls travel a more significant distance than the new ones. Given the existing data, there is no evidence to support that the alternative hypothesis is correct, as one would need more data.

Based on my hypothesis testing conclusion, my recommendations for Par, Inc would be to utilize a more extensive-sized data sampling for current and new golf balls. An increased sample size would decrease the standard error of associated sampling distributions (Anderson et al., 2021).

References

Anderson, D. R., Sweeney, D. J., Williams, T. A., Camm, J. D., Cochran, J. J., Fry, M. J., & Ohlmann. J. W. (2021). Essentials of modern business statistics with Microsoft® Excel® (8th ed.). Cengage Learning

H. M. James Hung, O’Neill, R. T., Bauer, P., & Kohne, K. (1997). The Behavior of the P-Value When the Alternative Hypothesis is True. Biometrics, 53(1), 11–22. https://doi.org/10.2307/2533093 (Links to an external site.)

McLeod, S. (2019). What a p-value tells you about statistical significance. P. https://www.simplypsychology.org/p-value.html

Comment by instructor

 

I like reading your analysis of the test.

To write a report on hypothesis test, the very first step is to clearly state the null and alternative hypothesis. Can you add that part in?

Post by classmate 2

 

Wk2 Discussion

• Provide descriptive statistical summaries of the data for each model, written in complete sentences.
  • From the data that was given we compared two different types of golf balls to see if there is a difference in the current ones being used and the new durable cut resistant ones. The data compares 40 different golf ball drives with each test being for each of the golf balls. When comparing the statistical data given, The current golf ball had an average distance of 270.28 versus the new golf balls having an average distance of 267.48 which is a drop of 1.04% for averages. For highest drive distance per choice was the same at 289, while the low had a difference of 255 for the old ones versus 250 for the newer gold balls, this also had a decrease of 1.96%. The median between the two golf balls was similar to the average at 270 for the current and 265 for the newer ones. On average the change between the current golf balls to the new, more durable golf balls was each of the drives lost an average of 2.8. The p-value is 0.094, T-Critical one-tail is 1.66, and the T-Critical two-tail is 1.99.
• Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls. Clearly state the null and alternative hypothesis.
  • Null hypothesis: Current golf balls that Par uses travel a greater distance than the more durable, cut resistant golf balls.
  • Alternate hypothesis: New golf balls travel the same distance as current ones.
• Analyze the data to provide the hypothesis testing conclusion. What is the p-value for your test? 
  • With the p-value (0.094) being more than the (0.05) 5% probability making the null hypothesis not proven and would lean more towards the alternate hypothesis. This means that the data is not showing a lower than 5% accuracy making the null hypothesis true.
• What is your recommendation for Par, Inc.? Discuss whether you see a need for larger sample sizes and more testing with the golf balls.
  • With a test of this size showing almost identical results with little differences, a large scale would give a more accurate account. Based on there really only having such a small change in distance, many variable could have been at play to skew these results. If all the hits were the same order, golfers fatigue, and wind speed being some of these variables. If the scale was out of 100 than the results would supply more accurate result since it would give them an out of 100% answer they would need to confirm this theory. They would have to test the on almost identical condition days with the golfer hitting the ball in one order the first time and opposite order the second time. This would eliminate a lot of the “what if” variables that affect the overall numbers on this (mythbusters logic).
• Reference:
• Anderson, D. R., Sweeney, D. J., Williams, T. A., Camm, J. D., Cochran, J. J., Fry, M. J., & Ohlmann. J. W. (2021). Essentials of modern business statistics with Microsoft® Excel® (8th ed.). Cengage Learning

Comment by instructor

 

like how this post is broken into subsections. It’s very easy to read.

Your alternative hypothesis is incorrect. Can you fix it?

Post by classmate 3

 

rovide descriptive statistical summaries of the data for each model, written in complete sentences.

The sample mean for the current golf ball is 270.28. The sample mean for the new golf ball is 267.5. On average the current golf ball had a 2.775-yard advantage. You get this number by subtracting the distance between the two means.

The value of the test statistic is 1.3284.

Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls. Clearly state the null and alternative hypothesis.

Par is looking into designing a new golf ball that will be cut resistant and offer driving distances comparable to those of the current model golf balls. To compare the driving distance of the two balls, 40 balls of both new and current models are subject to distance tests. A one tailed hypothesis test is to be conducted to see whether that mean driving distance for the current golf balls is shorter than the new gold ball.

Letting:

M1= Population mean driving distance for the current golf ball.

M2= Population mean driving distance for the new golf ball.

Hypothesis test:

H0: M1-M2 (<=) 0

H2= M1-M2 (>=) 0

Analyze the data to provide the hypothesis testing conclusion. What is the p-value for your test? What is your recommendation for Par, Inc.?

The P-value of the test is 0.0940. If the P-value is less than or equal to the level of significance 0.05, the null hypothesis can be rejected. Therefore, in this test the null hypothesis cannot be rejected.

Discuss whether you see a need for larger sample sizes and more testing with the golf balls.

With more data, par should have a good idea of the difference between the means of the two golf balls. Continued study for this case should be very easy using a mechanical hitting machine that can hit several hundred golf balls without much trouble and would help to come a good data-based conclusion.

Based on your hypothesis testing conclusion, what are your recommendations for Par, Inc? Support your recommendations with findings from your managerial report.

The data does not provide statistical evidence that the new golf ball has a lower mean driving distance than the current golf ball. Failing to reject H0 means the research finding is inconclusive. The data did not show the new golf ball with a significant lower mean driving distance, which means the researcher should not be ready to conclude the mean distance for the new golf ball is equal to or better than the current golf ball. A potential for a type 2 error exists with this conclusion.

Post by classmate 4

 

Provide descriptive statistical summaries of the data for each model, written in complete sentences.

The descriptive statistical summaries from the data for each model using the T Test: Two-Sample Assuming Unequal Variances lead me to the current values; Current Mean comes out to 270.28, Current Variance is 76.61, Current Observation is 40, DF is 76, T Stat is 1.33, P(T<=t) one 0.09, T Critical one 1.66, P(T<=t) two 0.19 and lastly T Critical two being at 1.99. New Mean is 267.5, New Variance is 97.95 and the New Observation is also at a 40.

Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls. Clearly state the null and alternative hypothesis.

“It is not always obvious how the null and alternative hypotheses should be formulated. Care must be taken to structure the hypotheses appropriately so that the hypothesis testing conclusion provides the information the researcher or decision maker wants.” (Ch. 9.1)

Ho: u < 270  Ha: u > 270

Ho: u < 76.61 Ha: u > 76.61

From the current to the new mean the value between is 2.78 in favor of the current mean and the current variance versus the new variance is -21.34 in favor of the new variance.

Analyze the data to provide the hypothesis testing conclusion. What is the p-value for your test? What is your recommendation for Par, Inc.?

The p- value is listed at 0.09 and is currently 0.05% in value which in turn means the null hypothesis cannot be rejected.

Discuss whether you see a need for larger sample sizes and more testing with the golf balls.

With a larger sample size of testing more golf balls, there will be a larger margain of error as well.

References:

Anderson, D. R., Sweeney, D. J., Williams, T. A., Camm, J. D., Cochran, J. J., Fry, M. J., & Ohlmann. J. W. (2021). Essentials of modern business statistics with Microsoft® Excel® (8th ed.). Cengage Learning

Post by classmate 5

 

Provide descriptive statistical summaries of the data for each model, written in complete sentences.

Par, Inc. wants to know if their current golf balls travel a greater distance than their new golf balls. A sample size of 40 golf balls from both the current and new golf balls. After running a t-test these are the figures I got. The mean for the current golf balls is 270.275 and for the new golf balls it is 267.5. The variance of the current golf balls is 76.61 and the variance for the new ones is 97.94. The t critical one tail is 1.66, while the t critical two tail is at 1.99.

Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls. Clearly state the null and alternative hypothesis.

A hypothesis test will have to be done to determine which golf balls travel a greater distance.

The null hypothesis is that the current golf balls travel a greater distance than the new golf balls.

The alternative hypothesis is that the new golf balls travel a greater distance than the current ones

Analyze the data to provide the hypothesis testing conclusion. What is the p-value for your test? What is your recommendation for Par, Inc.?

The p value for the one tailed test is 0.094. The p value is more than the level of significance of 0.05 which means that the null hypothesis is not rejected.

Discuss whether you see a need for larger sample sizes and more testing with the golf balls.

I would suggest that the sample size does increase. This will provide more information about both golf balls and cut down on possible errors.