According to the research team at Par Inc. they have been investigating a new stronger coated golf ball on driving distances, there comparing distances to the current model ball. 40 balls of each type were used in the distance tests.
The two samples are independent In order to see the differences in distance between both samples of balls I have developed the Hypothesis test as follows.
Ho : µ1 - µ2 < 0
Ha : µ1 - µ2 > 0
Ho (The null hypothesis represents the new balls distance being further or as far as the current ball)
Ha ( The alternative hypothesis represents the new ball having less distance than the current ball)
The averages for the current golf ball of 270.275 and the averages for the new golf ball of 267.50 show us the old ball has a 2.775 yard advantage. But using a significance level of .05, a t-score of 1.328 and a p-value of 0.094 shows us we cannot reject (Ho) because our p-value of 0.094 is greater than 0.05 (significance level). Based on the my calculations which are in the attached Excel Files labeled the results do not show enough data to suggest that the current ball has a higher distance level then the new ball.
-All outputs( T-Test Two Sample) and (Descriptive Statistics) for both models are in the Excel attachment labeled.
-The 95% confidence interval for the population mean driving distance of each individual sample is for the current golf balls is 2.7993, for the new golf balls is 3.1651.
-The 95% confidence interval for the population mean driving distance of the two golf balls combined is 6.9343.
By taking a larger sample size the standard deviation would decrease and the mean point would show us a more precise estimate. But due to fact in this specific situation the calculated (z-value) is far away from our rejection area. Allowing Par Inc. to not need a larger sample size so our conclusion is the new balls can and should be taken into production as soon as possible.