Question

Do students who read more books for pleasure tend to earn higher grades in English? The boxplots show data from a simple random sample of 79 students at a large high school. Students were classified as light readers if they read fewer than 3 books for pleasure per year. Otherwise, they were classified as heavy readers. Each student’s average English grade for the previous two

marking periods was converted to a GPA scale, where A=4.0, A−=3.7, B+=3.3

Reading and grades (3.2) The scatterplot shows the number of books read and the English grade for all 79 students in the study. The least-squares regression line y^=3.42+0.024x has been added to the graph.

a. Interpret the slope and y intercept.

b. The student who reported reading 17 books for pleasure had an English GPA of 2.85. Calculate and interpret this student’s residual.

c. For this linear model, r2=0.083 Interpret this value.

Short answer

Part (a) The y-intercept explain the average grade is $3.42$when $0$books read.

The slope explains the $0.02$rise in average English grade per book. read.

Part (b) When utilizing the lease square regression equation to predict the average grade, we overestimated it by $0.9831$

Part (c) The model employing the number of books read explains $8.3$ percent of the difference in the average English grade.

Step-by-step solution

Part (a) Step 1: Given information

The box plot:

Scatter plot:

Part (a) Step 2: Calculation

The least square regression line is, $\hat{y}=3.42+0.02x$

The y-intercept is $3.42$ which explain the average grade is $3.42$ when $0$ books read. Because the slope is $0.02$ every book read, the average English grade rises by $0.02$ per book read.

Part (b) Step 1: Calculation

First, we need to put value of x = 17 to calculate predicted value.

$\hat{y}=3.42+0.02\times 17=3.8331$

Therefore, residual is,

$Residual=2.85-3.8331=-0.9831$

This means that while using the lease square regression equation to forecast the average grade, we overestimated it by $0.9831$

Part (c) Step 1: Calculation

The $r^2=0.083=8.3\%$This value is very small. Therefore, The model employing the number of books read explains $8.3$percent of the difference in the average English grade.