Answer to Question #344007 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

52

70

76

77

77

86

y

74

56

51

74

72

68

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 a? Round your answers to the nearest hundredths.

Answer 1

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & X & Y & XY & X^2 & Y^2 \\ \hline & 52 & 74 & 3848 & 2704 & 5476 \\ \hdashline & 70 & 56 & 3920 & 4900 & 3136 \\ \hdashline & 76 & 51 & 3876 & 5776 & 2601 \\ \hdashline & 77 & 74 & 5698 & 5929 & 5476 \\ \hdashline & 77 & 72 & 5544 & 5929 & 5184 \\ \hdashline & 86 & 68 & 5848 & 7396 & 4624 \\ \hdashline Sum= & 438 & 395 & 28734 & 32634 & 26497 \\ \hdashline \end{array}$$ $$\bar{X}=\dfrac{1}{n}\sum _{i}X_i=\dfrac{438}{6}$$ $$=73$$ $$\bar{Y}=\dfrac{1}{n}\sum _{i}Y_i=\dfrac{395}{6}$$ $$=65.833333$$ $$SS_{XX}=\sum_iX_i^2-\dfrac{1}{n}(\sum _{i}X_i)^2$$ $$=32634-\dfrac{438^2}{6}=660$$ $$SS_{YY}=\sum_iY_i^2-\dfrac{1}{n}(\sum _{i}Y_i)^2$$ $$=26497-\dfrac{395^2}{6}=492.833333$$

$$SS_{XY}=\sum_iX_iY_i-\dfrac{1}{n}(\sum _{i}X_i)(\sum _{i}Y_i)$$ $$=28734-\dfrac{438(395)}{6}=−101$$ $$b=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{-101}{660}$$ $$=-0.15$$ $$a=\bar{Y}-b\bar{X}$$ $$a=\dfrac{395}{6}-(\dfrac{-101}{660})(73)=77.00$$

$$Y=77.00−0.15X$$