i need the answer of this question [using minitab. ] Par Products is a major man

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i need the answer of this question [using minitab. ]

Par Products 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-tasting 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. The results of the tests, with distances measured to the nearest metre, are available on the online platform, in the file 'Golf'. 1. Formulate and present the rationale for a hypo thesis test that Par could use to compare the driving distances of the current and new golf balls. 2. Analyze the data to provide the hypothesis test conclusion. What is the p-value for your test? What is your recommendation for Par Products? 3. Provide descriptive statistical summaries of the data for each model. 4. What is the 95 per cent confidence interval for the population mean of each model, and what is the 95 per cent confidence interval for the difference between the means of the two populations? 5. Do you see a need for larger sample sizes and more testing with the golf balls? Discuss.

Explanation / Answer

Please note that we dont have paid siftwares such as minitab. However , we shall provide an open source solution using R, the complete R code is accompanied

Hypothesis formulation is

H0 : There is no difference in the mean values for the current and the new method

H1 : There is a signifcant difference in the mean values for the current and the new method

Lets use the independent samples t test to run the analysis

current <- c(264,261,267,272,258,283,258,266,259,270,263,264,284,263,260,283,255,272,266,268,270,287,289,280,272,275,265,260,278,275,281,274,273,263,275,267,279,274,276,262)
new<- c(277,269,263,266,262,251,262,289,286,264,274,266,262,271,260,281,250,263,278,264,272,259,264,280,274,281,276,269,268,262,283,250,253,260,270,263,261,255,263,279)

##t test
t.test(current,new,confidence= 0.95)

The result is

Welch Two Sample t-test

data: current and new
t = 1.3284, df = 76.852, p-value = 0.188
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.384937 6.934937
sample estimates:
mean of x mean of y
270.275 267.500

as the p value is 0.188, which is not less than 0.05 , hence we fail to reject the null hypothesis and conclude that H0 : There is no difference in the mean values for the current and the new method

The decriptive summary stat is

> summary(current)
Min. 1st Qu. Median Mean 3rd Qu. Max.
255.0 263.0 270.0 270.3 275.2 289.0
> summary(new)
Min. 1st Qu. Median Mean 3rd Qu. Max.
250.0 262.0 265.0 267.5 274.5 289.0

we know that the confidence interval is given as

mean +- z*sd/sqrt(n) , where z = 1.96 from the z table for 95% Confidence

so we calculate the values as

mean(current) + 1.96*sd(current)/length(current)
mean(current) - 1.96*sd(current)/length(current)

> mean(current) + 1.96*sd(current)/length(current)
[1] 270.7039
> mean(current) - 1.96*sd(current)/length(current)
[1] 269.8461

for new

mean(new) + 1.96*sd(new)/length(new)
mean(new) - 1.96*sd(new)/length(new)

> mean(new) + 1.96*sd(new)/length(new)
[1] 267.9849
> mean(new) - 1.96*sd(new)/length(new)
[1] 267.0151

for the difference of the means

diff<- current-new
mean(diff) + 1.96*sd(diff)/length(diff)
mean(diff) - 1.96*sd(diff)/length(diff)

> mean(diff) + 1.96*sd(diff)/length(diff)
[1] 3.448455
> mean(diff) - 1.96*sd(diff)/length(diff)
[1] 2.101545

Please note that we can answer only 4 subparts of a question at a time , as per the answering guidelines.

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