Answer to Question #344008 in Statistics and Probability for wiwili

Question

Let x be a random variable that represents the percentage of successful free throws a professional basketball player makes in a season. Let y be a random variable that represents the percentage of successful field goals a professional basketball player makes in a season.

The table below shows the percentage of successful free throws and field goals a professional basketball player makes in a season:

x

66

75

53

75

90

79

y

73

56

84

80

71

97

Suppose you want to predict the percentage of successful field goals a professional basketball player using the percentage of successful free throws, what is the value of b? Round your answers to the nearest hundredths.

Answer 1

In order to compute the regression coefficients, the following table needs to be used:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & X & Y & XY & X^2 & Y^2 \\ \hline & 66 & 73 & 4818 & 4356 & 5329 \\ \hdashline & 75 & 56 & 4200 & 5625 & 3136 \\ \hdashline & 53 & 84 & 4452 & 2809 & 7056 \\ \hdashline & 75 & 80 & 6000 & 5625 & 6400 \\ \hdashline & 90 & 71 & 6390 & 8100 & 5041 \\ \hdashline & 79 & 97 & 7663 & 6241 & 9409 \\ \hdashline Sum= & 438 & 461 & 33523 & 32756 & 36371 \\ \hdashline \end{array}$$ $$\bar{X}=\dfrac{1}{n}\sum _{i}X_i=\dfrac{438}{6}$$ $$=73$$ $$\bar{Y}=\dfrac{1}{n}\sum _{i}Y_i=\dfrac{461}{6}$$ $$=76.833333$$ $$SS_{XX}=\sum_iX_i^2-\dfrac{1}{n}(\sum _{i}X_i)^2$$ $$=32756-\dfrac{438^2}{6}=782$$ $$SS_{YY}=\sum_iY_i^2-\dfrac{1}{n}(\sum _{i}Y_i)^2$$ $$=36371-\dfrac{461^2}{6}=950.833333$$

$$SS_{XY}=\sum_iX_iY_i-\dfrac{1}{n}(\sum _{i}X_i)(\sum _{i}Y_i)$$ $$=33523-\dfrac{438(461)}{6}=−130$$ $$b=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{-130}{660}$$ $$=-0.17$$ $$a=\bar{Y}-b\bar{X}$$ $$a=\dfrac{461}{6}-(\dfrac{-130}{660})(73)=88.97$$ $$Y=88.97−0.17X$$