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. A random sample of n = 6 professional basketball players gave the following information.

x	87	88	70	84	78	76
y	53	57	50	51	46	50

Given that S_{e =} 3.054, a= 16.547, b= 0.425, and , find the predicted percentage of successful field goals for a player with x= 73% successful free throws.

Answer 1

In R find the regression model

Rcode is

x <- c(87,88,70,84,78,76)
y <- c(53,57,50,51,46,50)
rmod1 <- lm(y~x)
summary(rmod1)

output:

Call:

lm(formula = y ~ x)

Residuals:

1 2 3 4 5 6

-0.4121 3.2424 2.4606 -1.3758 -4.3030 0.3879

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 23.3576 15.6745 1.49 0.21

x 0.3455 0.1941 1.78 0.15

Residual standard error: 3.054 on 4 degrees of freedom

Multiple R-squared: 0.4419, Adjusted R-squared: 0.3024

F-statistic: 3.168 on 1 and 4 DF, p-value: 0.1497

y=23.3576+ 0.3455 *x

slope=0.3455

y intercept=23.3576

For x=73

we get

y=23.3576+ 0.3455 *73

y= 48.5791

y=49

rounding to nearest integer

the predicted percentage of successful field =49