Par, Inc. Case Problem Managerial Report
This report is case study of Par, Inc.
Par, Inc., is a manufacturer of golf equipment. The company wants to check whether this new type of golf ball is comparable to the old ones. Therefore they did a test with 40 samples of both new balls and old ones. The testing was performed with a mechanical hitting machine so that any difference between the mean distances for the two models could be measure to evaluate the difference in the two models, presented with the sample size of 40 and confidence level of 95%.
The table below depicts a descriptive statistical summary for the case study. | Current Model | New Model | Sample Mean | 268.89 | 272.98 | Standard Error | 1.30 | 1.55 | Standard Deviation | 8.20 | 9.81 | Standard variances | 67.28 | 96.28 | Sample Size | 40 | 40 | Confidence Level (95%) | 2.62 | 3.14 |
The hypothesis for Par, Inc. to compare the driving distance of the current and new golf balls is
H0 : µ1 - µ2 = 0 (they are the same)
Ha : µ1 - µ2 ≠ 0 (the are not the same).
P-Value for this two tailed test is 0.06, which is greater than level of significance α (0.05).
Hence, H0 will not be rejected which shows that Par, Inc. should take a new ball in production as the P value indicates that there is no significant difference between estimated population mean of current as well new sample model.
The 95% confidence interval for the population mean of the current model is 266.26 to 276.11 and of the new model is 269.84 to 276.11. It means that the estimated population mean for Par, Inc. should lie within this range for consistent result. However, the 95% confidence interval for the difference between the means of the two populations is -8.13 to -0.05.
The calculated test statistic value is far from the rejection area. The larger the sample size the smaller the standard deviations which means point estimator of mean will become more precise. Hence there is no need to take larger sample size.
Conclusion:-
In a nut shell, the
Par, Inc., is a manufacturer of golf equipment. The company wants to check whether this new type of golf ball is comparable to the old ones. Therefore they did a test with 40 samples of both new balls and old ones. The testing was performed with a mechanical hitting machine so that any difference between the mean distances for the two models could be measure to evaluate the difference in the two models, presented with the sample size of 40 and confidence level of 95%.
The table below depicts a descriptive statistical summary for the case study. | Current Model | New Model | Sample Mean | 268.89 | 272.98 | Standard Error | 1.30 | 1.55 | Standard Deviation | 8.20 | 9.81 | Standard variances | 67.28 | 96.28 | Sample Size | 40 | 40 | Confidence Level (95%) | 2.62 | 3.14 |
The hypothesis for Par, Inc. to compare the driving distance of the current and new golf balls is
H0 : µ1 - µ2 = 0 (they are the same)
Ha : µ1 - µ2 ≠ 0 (the are not the same).
P-Value for this two tailed test is 0.06, which is greater than level of significance α (0.05).
Hence, H0 will not be rejected which shows that Par, Inc. should take a new ball in production as the P value indicates that there is no significant difference between estimated population mean of current as well new sample model.
The 95% confidence interval for the population mean of the current model is 266.26 to 276.11 and of the new model is 269.84 to 276.11. It means that the estimated population mean for Par, Inc. should lie within this range for consistent result. However, the 95% confidence interval for the difference between the means of the two populations is -8.13 to -0.05.
The calculated test statistic value is far from the rejection area. The larger the sample size the smaller the standard deviations which means point estimator of mean will become more precise. Hence there is no need to take larger sample size.
Conclusion:-
In a nut shell, the