Problem 2-3 ERA The following is the split pitching record (ERA) for Wei-Yin Che
ID: 3337984 • Letter: P
Question
Problem 2-3 ERA
The following is the split pitching record (ERA) for Wei-Yin Chen during the last 2 seasons in Bal- timore (2014 and 2015) by opponents. Let (x)'s be his ERAs in the year 2014 and (y's be his ERAs in the year 2015. Let(d)'s be the difference between these two years (d *),-X)The data are pro- vided in the followingtable· Opponent 202015 )d(d-d, -X- vs. BOS vs. CLE vs. KC -y) 0.57 0.1107 0.95 0.5080 1.10 1.7883 2.55 5.3487 1.35 1.2382 0.12 0.1276 2.55 7.7689 0.75 0.2629 1.78 4.0694 2.51 5.1653 0.52 0.5735 36.78134.17-2.61' 26.9614 2.92 2.35 3.95 3.00 2.19 3.29 4.632.08 .91 3.56 .97 4.09 0.593.14 3.00 2.25 2.19 3.97 4.762.25 3.674.19 0.3204 0.1795 0.5721 0.0645 0.3677 0.0113 0.2118 1.3309 0.0337 1.32031.6547 1.0534 0.7106 2.45350.2058 0.6161 0.3923 0.9675 0.0926 7.5825 0.0011 0.2943 0.1181 0.7334 0.9963 1.3309 0.7459 1.2129 2.0061 0.7334 0.3537 0.10651.174 -1.6035 17.5227 6.2319 4 vs. MIN vs. NY vs. OÄK vs. S s. TEX vs/HOU vs. WAS -l a3.3413.1140,241 Consider the following hypothesis test for questions (a) and (b) Average PA+owspry During the regular games in 2015, Chen didn't pitch CWS like he did in 2014. Based on Chen's ERA v.s. CWS in 2014, which is 4.15, what ERA would you expect him to have (v.s. CWS) if he did pitch CWS in the 2015 based on the estimated regression equation? (2 pts) e) (f Based on the point estimate in question (e), please provide an interval estimate (prediction interval estimate). Furthermore, please also give a confidence interval estimate for Chen's ERA vs. CWS if he did pitch them in 2015. (6 pts)Explanation / Answer
e) regression line is
y=3.41233 -0.09151 x
now given that x=4.15, so we get y= 3.41233 -0.09151 *4.15=3.0325335
f) confidence interval with predicted values
fit lwr upr
1 3.145130 2.553538 3.736721
2 3.050877 2.428668 3.673085
3 3.211930 2.452105 3.971755
4 2.988652 2.187851 3.789452
5 2.963030 2.069160 3.856899
6 3.049047 2.422938 3.675155
7 3.358342 2.012324 4.704360
8 3.137809 2.556552 3.719066
9 3.211930 2.452105 3.971755
10 2.976756 2.133750 3.819762
11 3.076499 2.497211 3.655787
prediction interval with predicted values
fit lwr upr
1 3.145130 1.1932196 5.097040
2 3.050877 1.0894702 5.012283
3 3.211930 1.2026251 5.221235
4 2.988652 0.9634964 5.013807
5 2.963030 0.8993011 5.026758
6 3.049047 1.0863995 5.011694
7 3.358342 1.0623158 5.654368
8 3.137809 1.1890062 5.086612
9 3.211930 1.2026251 5.221235
10 2.976756 0.9345434 5.018968
11 3.076499 1.1282825 5.024715
R-codes for this problem is
x=c(2.92,3.95,2.19,4.63,4.91,3.97,.59,3,2.19,4.76,3.67)
y=c(2.35,3,3.29,2.08,3.56,4.09,3.14,2.25,3.97,2.25,4.19)
fit=lm(y~x)
fit
predict(fit,interval="confidence")
predict(fit,interval="predict")
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