3.The overall distance traveled by a golf ball is tested by hitting the ball wit

Question

3.The overall distance traveled by a golf ball is tested by hitting the ball with Iron Byron, a mechanical golferwith a swing that is said to emulate the distance hit bythe legendary champion, Byron Nelson. Ten randomly selected balls of two different brands are tested and the overall distance measured. The data are given below: Brand1 275 286 287 271 283271279275 263 267 n1-10,x S1 8.0284 273 s2- 10.0449 X1-275.7 Brand2258 244 260 265 281 271 270 263 268 n2- 10 x2 265.3,s (a) Normal probability plots of the two data sets are given below. Results of the Shapiro-Wilk test of Normality are also given below. Are we justified in assuming that the two data sets come from Normal populations? Explain briefly and use any of the results below to support your answer. R commands (in bold): Brand1-c(275,286,287,271,283,271,279,275,263,267) Brand2-c(258,244,260,265,273,281,271,270,263,268) qqnorm(Brand1)

Explanation / Answer

R- Command

> Brand1=c(275,286,287,271,283,271,279,275,263,267)
> Brand1
[1] 275 286 287 271 283 271 279 275 263 267
> Brand2=c(258,244,260,265,273,281,271,270,263,268)
> Brand2
[1] 258 244 260 265 273 281 271 270 263 268
> t.test(Brand1,Brand2,var.equal=F,alternative="greater")

        Welch Two Sample t-test

data: Brand1 and Brand2
t = 2.5575, df = 17.166, p-value = 0.01014
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
3.330011      Inf
sample estimates:
mean of x mean of y
    275.7     265.3

> RR=qt(0.05,17,lower.tail=F)         ## Rejection Region
> RR
[1] 1.739607

Lower of the 95% confidence interval is 3.330011

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