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Exercises 35 and 36 refer to the following setting. Many chess masters and chess advocates

believe that chess play develops general intelligence, analytical skill, and the ability to

concentrate. According to such beliefs, improved reading skills should result from study to

improve chess-playing skills. To investigate this belief, researchers conducted a study. All the

subjects in the study participated in a comprehensive chess program, and their reading

performances were measured before and after the program. The graphs and numerical

summaries that follow provide information on the subjects’ pretest scores, posttest scores, and

the difference (Post − Pre) between these two scores.

Predicting posttest scores (3.2)

a. What is the equation of the least-squares regression line relating posttest and pretest

scores? Define any variables used.

b. Is a linear model appropriate for describing this relationship? Justify your answer.

c. If we use the least-squares regression line to predict students’ posttest scores from their

pretest scores, how far off will our predictions typically be?

Short Answer

Expert verified

a) y=17.897+0.78301xwhere x represents the pretest score and y represents the posttest score.

b) Yes, a linear model appropriate for describing this relationship.

c) The predicted posttest score deviates on average by12.55 points from the actual posttest score.

Step by step solution

01

Step 1. Given information  

We have given data.

02

a) Step 1. To find the equation of the least-squares regression line.

General equation of the least squares regression line is

y=b0+b1x

Eliminate,role="math" localid="1654232338347" b0=17.897andb1=0.78301

Thus,

y=17.897+0.78301x
03

b) Step 1  Justify a linear model is appropriate for describing this relationship.

A linear model is appropriate for describing the relationship in this case, because there is no strong curvature present in the scatterplot and there is no strong curvature present in the residual plot. Moreover, the points in the residual plot appear to be randomly scattered and there appear to be no strong outliers (points that deviate strongly from the pattern in the other points), thus the model does indeed appear to be appropriate.

04

c) Step 1. To find how far off posttest scores from their pretest scores,

The standard error of the estimate s is given after"S="in the computer output:S=12.55

The standard error of the estimate s represents the average error of predictions, thus the average deviation between actual y-values and the predicted y-values.

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Most popular questions from this chapter

Cranky mower To start her old lawn mower, Rita has to pull a cord and hope for some luck. On any particular pull, the mower has a 20%chance of starting.

a. Find the probability that it takes her exactly 3 pulls to start the mower.

b. Find the probability that it takes her more than 6 pulls to start the mower.

Ms. Hall gave her class a 10-question multiple-choice quiz.

LetX=the number of questions that a randomly selected student in the class answered correctly. The computer output gives information about the probability distribution of X. To determine each student’s grade on the quiz (out of 100), Ms. Hall will multiply his or her number of correct answers by 5and then add 50. Let G=the grade of a randomly chosen student in the class.

More easy quiz

a. Find the mean of G.

b. Find the range of G.

Bag check Refer to Exercise 91 .

a. Calculate and interpret the mean of R.

b. Calculate and interpret the standard deviation of R.

.Life insurance The risk of insuring one person’s life is reduced if we insure many people. Suppose that we insure two 21-year-old males, and that their ages at death are independent. If X1andX2are the insurer’s income from the two insurance policies, the insurer’s average income W on the two policies is

W=X1+X22=0.5X1+0.5X2

Find the mean and standard deviation of W. (You see that the mean income is the same as for a single policy but the standard deviation is less.)

Life insurance If four 21-year-old men are insured, the insurer’s average income is

V=X1+X2+X3+X44=0.25X1+0.25X2+0.25X3+0.25X4

where Xiis the income from insuring one man. Assuming that the amount of income earned on individual policies is independent, find the mean and standard deviation of V. (If you compare with the results of Exercise 57, you should see that averaging over more insured individuals reduces risk.)

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