a.) Fit a multiple linear regression model relating the number of games won to t
ID: 2923111 • Letter: A
Question
a.) Fit a multiple linear regression model relating the number of games won to the team's passing yardage ( x2), the percentage of rushing plays ( x7 ), and the opponents' yards rushing ( x8 ).Report the information from summary.
(b) Provide an interpretation of your estimated regression coecient for the variable x2 which is passing yardage. Include units.
(c) Using only the information provided from the summary function for the output from a linear model and a t-table, construct a 95% confidence interval for the regression coefficient in part (b). Show enough work so that it is clear that you did this using the information from summary. You may verify your interval by using the confint function.
(d) Construct the design matrix X (do not forget the column of 1's), and the vector of observations y, and use matrix multiplication to find . Provide enough output/code in R to verify that your estimates were obtained this way.
(e) Using matrix multiplication and the estimate for MSres which can be obtained from summary, give the covariance matrix of . Verify that the standard errors for the 's provided by summary arise from this matrix.
(f) What is strange about the response variable in this problem?
y x1 x2 x3 x4 x5 x6 x7 x8 x9 10 2113 1985 38.9 64.7 4 868 59.7 2205 1917 11 2003 2855 38.8 61.3 3 615 55 2096 1575 11 2957 1737 40.1 60 14 914 65.6 1847 2175 13 2285 2905 41.6 45.3 -4 957 61.4 1903 2476 10 2971 1666 39.2 53.8 15 836 66.1 1457 1866 11 2309 2927 39.7 74.1 8 786 61 1848 2339 10 2528 2341 38.1 65.4 12 754 66.1 1564 2092 11 2147 2737 37 78.3 -1 761 58 1821 1909 4 1689 1414 42.1 47.6 -3 714 57 2577 2001 2 2566 1838 42.3 54.2 -1 797 58.9 2476 2254 7 2363 1480 37.3 48 19 984 67.5 1984 2217 10 2109 2191 39.5 51.9 6 700 57.2 1917 1758 9 2295 2229 37.4 53.6 -5 1037 58.8 1761 2032 9 1932 2204 35.1 71.4 3 986 58.6 1709 2025 6 2213 2140 38.8 58.3 6 819 59.2 1901 1686 5 1722 1730 36.6 52.6 -19 791 54.4 2288 1835 5 1498 2072 35.3 59.3 -5 776 49.6 2072 1914 5 1873 2929 41.1 55.3 10 789 54.3 2861 2496 6 2118 2268 38.2 69.6 6 582 58.7 2411 2670 4 1775 1983 39.3 78.3 7 901 51.7 2289 2202 3 1904 1792 39.7 38.1 -9 734 61.9 2203 1988 3 1929 1606 39.7 68.8 -21 627 52.7 2592 2324 4 2080 1492 35.5 68.8 -8 722 57.8 2053 2550 10 2301 2835 35.3 74.1 2 683 59.7 1979 2110 6 2040 2416 38.7 50 0 576 54.9 2048 2628 8 2447 1638 39.9 57.1 -8 848 65.3 1786 1776 2 1416 2649 37.4 56.3 -22 684 43.8 2876 2524 0 1503 1503 39.3 47 -9 875 53.5 2560 2241Explanation / Answer
The R snippet fis as follows
# read the data into R dataframe
data.df<- read.csv("C:\Users\586645\Downloads\Chegg\sample.csv",header=TRUE)
str(data.df)
## regression model
fit <- lm(y~x2 +x7 +x8,data=data.df)
summary(fit)
par(mfrow=c(2,2))
plot(fit)
The results are
> summary(fit)
Call:
lm(formula = y ~ x2 + x7 + x8, data = data.df)
Residuals:
Min 1Q Median 3Q Max
-3.0370 -0.7129 -0.2043 1.1101 3.7049
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -1.808372 7.900859 -0.229 0.820899
x2 0.003598 0.000695 5.177 2.66e-05 ***
x7 0.193960 0.088233 2.198 0.037815 *
x8 -0.004816 0.001277 -3.771 0.000938 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.706 on 24 degrees of freedom
Multiple R-squared: 0.7863, Adjusted R-squared: 0.7596
F-statistic: 29.44 on 3 and 24 DF, p-value: 3.273e-08
The coefficient of x2 is 0.003598 , this means that for every unit change in x2 , the value of y changes by 0.003598 units.
The 95% confidence interval is given as
esitmate +- 1.96*SE
0.003598 +- 1.96*0.000695 , the CI is
0.0022, 0.00496
> confint(fit,"x2",level=0.95)
2.5 % 97.5 %
x2 0.002163664 0.005032477
The confidence interval matches with what we calculated. Please note that we can answer only 4 subparts of a question at a time , as per the answering guildelines
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.