19. A STAT 200 instructor believes that the average quiz score is a good predict

Question

19. A STAT 200 instructor believes that the average quiz score is a good predictor of final exam score. A random sample of 10 students produced the following data where x is the average quiz score and y is the final exam score 72 70 94 96 63 70 10040 96 85 83 70 75 75 85 87 50 50 (a) Find an equation of the least squares regression line. Show all work, writing the correct equation, without supporting work, will receive no credit. (b)Based on the equation from part (a), what is the predicted final exam score if the average quiz score is 80? Show all work and justifv your answer.

Explanation / Answer

a) I am using R software to solve this problem.

First we can load the data in R using the below commands:

x <- c(72,94,55,63,100,40,85,70,75,85)

y <- c(70,96,50,70,96,50,83,75,77,87)

We can fit a linear model in R using the lm() function as below:

fit <- lm(y ~ x)

We can see the summary of the model using the summary() function as below:

summary(fit)

Call:

lm(formula = y ~ x)

Residuals:

Min 1Q Median 3Q Max

-9.030 -2.012 1.317 3.138 4.041

Coefficients:

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

(Intercept) 11.39256 6.20373 1.836 0.104   

x 0.86614 0.08175 10.595 5.5e-06 ***

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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 4.46 on 8 degrees of freedom

Multiple R-squared: 0.9335, Adjusted R-squared: 0.9252

F-statistic: 112.3 on 1 and 8 DF, p-value: 5.504e-06

So we can see that the linear equation is :

y = 0.86614 * x + 11.39256

b) If the average quiz score is 80, we can predict the final exam score by substituting the value of x as 80 in the above equation:

y = 0.86614 * 80 + 11.39256

y = 80.68376

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