CASE PROBLEM Men ParProducts isamajormanufacturer of golf equpment. Management b

Question

CASE PROBLEM Men ParProducts isamajormanufacturer of golf equpment. Management believes that Pars market share couldbe increased with the introduction loneeriasting eoir ball. Therefore, the resead goup designed to resist aksandproide a more durabiebal, 203 Thetests the coating have been promising one of the researchers voiced conorm about the effect or the new coating on diving dstances. Par 2es would like the new out esistant to omor to those of the current model bat to compare the distances for the 2ee two bats, 40 bats of both the new end current to estance tests. The test chine so that any dee e ee bet een the mean 1. formutata and the ationale for a hypo tost that Pw could use to compare datances for the two models couis oe buted to distances ofe current and new bats daterenee m the two model The .esuts of the 2. Analyze the to the meanest data to provide the Kyaothesis sest conclusion What is the pvalue for your test? me ine on on ne platform, the What is your recommendaion for Par Products? Provide descriptive statistical summanes What 95 percent confidence intera tor mean of each modes, and ahat is the per cent cont aeroe inter for the between the means of the two popu ations

Explanation / Answer

1)

H0: D = 0

H1: D ¹ 0

2)

D=EDi/n

   =-111/40

   =-2.775

S=E(Di-D)^2/n-1

     =7366.975/39

     =13.74

tSTAT=D-m/SD/n

        =-2.775-0/13.74/40

        =-1.277

The P-Value is .209153.

The result is not significant at p < .05.

3)

4)

Confidence interval:-

D+- ta/2*SD/n

-2.775+-2.02*2.17

LCL=-2.775-4.3834=-7.1584

UCL=-2.775+4.3834=1.6084

(-7.1584,1.6084)

5)

Based on the above results, we can see that the results are not significant. Hence, a larger sample might help although a normal distribution requires sample of 30 or more.

Current Mean 270.275 Standard Error 1.383968421 Median 270 Mode 272 Standard Deviation 8.752984839 Sample Variance 76.61474359 Kurtosis -0.762587222 Skewness 0.306169371 Range 34 Minimum 255 Maximum 289 Sum 10811 Count 40 Largest(1) 289 Smallest(1) 255 Confidence Level(95.0%) 2.799340358 New Mean 267.5 Standard Error 1.564837994 Median 265 Mode 263 Standard Deviation 9.896904463 Sample Variance 97.94871795 Kurtosis -0.514586551 Skewness 0.239810448 Range 39 Minimum 250 Maximum 289 Sum 10700 Count 40 Largest(1) 289 Smallest(1) 250 Confidence Level(95.0%) 3.165183603

4)

Confidence interval:-

D+- ta/2*SD/n

-2.775+-2.02*2.17

LCL=-2.775-4.3834=-7.1584

UCL=-2.775+4.3834=1.6084

(-7.1584,1.6084)

5)

Based on the above results, we can see that the results are not significant. Hence, a larger sample might help although a normal distribution requires sample of 30 or more.

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