Ch. 10 Par, INC.

Topics: Statistical hypothesis testing, Arithmetic mean, Sample size, Statistics / Pages: 2 (384 words) / Published: Oct 31st, 2013
Introduction
The purpose of this report is to show if the new golf ball coating design does induced resist cuts and provide a more durable ball.
Hypothesis
µ1 is the mean driving distance of currently produced golf balls µ2 is the mean driving distance of newly designed golf balls
Null hypothesis (H0): µ1- µ2 ≤ 0
Alternate hypothesis (Ha): µ1- µ2 > 0
Analysis

The sample mean for current golf balls is 270.275, while the new golf balls have a sample mean of 267.50. So on average the current golf balls have an advantage of 2.775 yards. Using a 5% significance level, a p value of 0.094 is obtained. As this is larger than the significance level, the null hypothesis cannot be rejected. The conclusion is that this data does not provide statistical evidence that the new golf balls have a lower mean driving distance.
However, this does not mean that the new golf balls do not have a better driving distance only that this set of data does not support it. A type II error could have been made, where we have not rejected the null hypothesis, but it is in fact false. Only more data can help give a better idea on whether there is a difference in the driving distance of the current and new golf balls. So it is recommended that more data be acquired and a new test be performed with a bigger sampling size.
Descriptive Summaries

Confidence Intervals (95%)

Confidence interval at 95% for the population mean driving distance of current golf balls.

The 95% confidence interval for the population mean driving distance of current golf balls is between 267.563 yards and 272.988 yards
Confidence interval at 95% for the population mean driving distance of new golf balls

The 95% confidence interval for the population mean driving distance of new golf balls is between 264.43 yards and 270.566 yards
The 95% confidence interval for the difference between the means of the two populations

The confidence interval at 95% for the difference between the means of

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