Question

1.   Par, Inc., 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-lasting 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 negative effect of the new coating on driving distances. Par would like the new cut-resistant ball to offer driving distances no worse than 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 yard, are in the file “Golf.csv”. Let the current-model golf balls be population 1 and the new cut-resistant balls be population 2. Complete the following.
a)   Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls.

b)   Provide descriptive statistical summaries of the data for each model; in particular, the sample mean, the sample standard deviation, and the sample size for each model.
Please copy your R code and the result and paste them here.

c)   Compute the standard error for your test.
Please copy your R code and the result and paste them here.

d)   Compute the degree of freedom for your test.
Please copy your R code and the result and paste them here.

e)   Compute the test statistic for your test.
Please copy your R code and the result and paste them here.

f)   Compute the p value for your test.
Please copy your R code and the result and paste them here.

g)   Suppose the significance level is set at 5%. What is your conclusion? Provide a practical interpretation of your conclusion in this case.

h)   Use the function t.test() in R to run the test directly to confirm your results above are correct. .

i)   What is the 95% confidence interval for the population mean driving distance of the current model?
Please copy your R code and the result and paste them here.

j)   What is the 95% confidence interval for the population mean driving distance of the new model?
Please copy your R code and the result and paste them here.

k)   What is the 95% confidence interval for the difference between the means of the two populations?
Please copy your R code and the result and paste them here.

Current	New
264	277
261	269
267	263
272	266
258	262
283	251
258	262
266	289
259	286
270	264
263	274
264	266
284	262
263	271
260	260
283	281
255	250
272	263
266	278
268	264
270	272
287	259
289	264
280	280
272	274
275	281
265	276
260	269
278	268
275	262
281	283
274	250
273	253
263	260
275	270
267	263
279	261
274	255
276	263
262	279

Answer 1

a) We have to test

b) The R code below for descriptive statistical summaries of the data for each model.

X1 <- 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)

X2 <- 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)

length(X1)

length(X1)

mean(X1)

mean(X2)

sd(X1)

sd(X2)

The sample means are , The sample standard deviations are

.

c) The standard error is

sqrt(var(X)/length(X)+var(Y)/length(Y))

d) The degree of freedom is

e) The test statistic is

f)The P-value of the test is

> 2*pt(-1.3284 ,78)
[1] 0.1879197

g) Since , we accept the null hypothesis. There is no change.

h) The R code for T-test below:

X1 <- 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)

X2 <- 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(X1, X2 , alternative = c("two.sided"), mu = 0, var.equal = FALSE, conf.level = 0.95)

The oputput:

Welch Two Sample t-test

data: X1 and X2

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

The results are correct.

j) The 95% CI for difference in means is .