3. The overall distance traveled by a golf ball is tested by hitting the ball wi
ID: 3333735 • Letter: 3
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3. The overall distance traveled by a golf ball is tested by hitting the ball with Iron Byron, a mechanical golfer with a swing that is said to emulate the distance hit by the 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 275 286 287 271 283 271 279 275 263 267 Brand2 258 | 244 260 | 265 273 281 271 270 263 268 n, = 10, T, = 265.3, S, = 10.0449 (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): Brandic(275,286,287,271,283,271,279,275,263,267) Brand 2=c(258244,260,265,273,281,271,270,263,268) qqnorm(Brand1) qgnorm(Brand2) Brand 1 Brand 2 shapiro.test(Brand1) Shapiro-Wilk normality test data: Brand1 W = 0.958, p-value = 0.7629 shapiro.test(Brand2) Shapiro-Wilk normality test data: Brand2 W = 0.9583, p-value = 0.7663 co TTTTTTT 15 0505 15 15 0505 15 Theoretical Quartiles Theoretical quartilesExplanation / Answer
a. For both brand1 and brand2, the p value for Shapiro Wilk test are greater than 0.05. Thus failure to reject null hypothesis and conclude that population is normal. Thus, the two samples come from normal populations.
b. The test statistic is as follows:
t=(x1bar-x2bar)/sqrt[s1^2/n1+s2^2/n2], where, xbar is sample mean, s is sample standard deviation, n is sample size.
=(275.70-265.3)/sqrt[8.03^2/10+10^2/10]
=2.56
The t critical at 17 df [df is calculated using technology] at alpha=0.05 is 2.1098.
c. The 95% lower confidence bound for mu1-mu2=(x1bar-x2bar)-talpha/2 sqrt[s1^1/n1+s2^2/n2]
=(275.70-265.3)-2.1098*sqrt[8.0284^2/10+10.0449^2/10]=3.33
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