Pairs of Scores: How to Find the Perfect Pair of Scores for Your Needs

Chapter Six 

6.7 (a) Estimate whether the following pairs of scores for X and Y reflect a positive relationship, a negative relationship, or no relationship. Hint: Note any tendency for pairs of X and Y scores to occupy similar or dissimilar relative locations. 

The data of X and Y scores showcase a negative relationship, since in most instances, except one, X and Y are on opposite spectrums of their mean - X < X mean, Y > Y mean; or X > X mean, Y < Y mean – implying dissimilar relative locations. 

(b) Construct a scatterplot for X and Y. Verify that the scatterplot does not describe a pronounced curvilinear trend. 

(c) Calculate r using the computation formula 

X Values 

Summation (Σ) = 451 

Mean= 50.111 

Σ(X - Mx)^2= SSx = 1756.889 

Y Values 

Σ = 687 

Mean = 76.333 

Σ(Y - My)^2 = SSy = 858 

X and Y Values 

N = 9 

Σ(X - Mx)(Y - My) = Sxy = -1122.333 

R 

r = Σ((X - Mx)(Y - My)) / √((SSx)(SSy)) 

r = -1122.333 / √((1756.889)(858)) = -0.9141 

6.10 On the basis of an extensive survey, the California Department of Education reported an r of -.32 for the relationship between the amount of time spent watching TV and the achievement test scores of school children. Each of the following statements represents a possible interpretation of this finding. 

Indicate whether each is true or false. 

(a) Every child who watches a lot of TV will perform poorly on the achievement tests. 

False. For the statement to be considered true, it must be r = -1. 

(b) Extensive TV viewing causes a decline in test scores. 

False 

(c) Children who watch little TV will tend to perform well on the tests. 

True 

(d) Children who perform well on the tests will tend to watch little TV. 

Tue 

(e) If Gretchen’s TV-viewing time is reduced by one-half, we can expect a substantial improvement in her test scores. 

False 

(f) TV viewing could not possibly cause a decline in test scores. 

False 

6.11 Assume that an r of .80 describes the relationship between daily food intakes, measured in ounces, and body weight, measured in pounds, for a group of adults. Would a shift in the units of measurement from ounces to grams and from pounds to kilograms change the value of r? Justify your answer. No. A change of unit will not change the value of r. 

r = Σ((X - Mx)(Y - My)) / √((SSx)(SSy)) = Σ((X - Mx)(Y - My)) / √((Σ(X - Mx )^2)(Σ(Y - My)^2)) 

Assume a change units of x to x’ = mx, and units of y to y’ = ny, then 

r’ = Σ((X’ - Mx’)(Y’ - My’)) / √((SSx’)(SSy’)) = 

Σ((X’ - Mx’)(Y’ - My’)) / √((Σ(X’ - Mx’)^2)(Σ(Y’ - My’)^2)) = 

Σ((mX - M(mx))(nY - M(ny)) / √((Σ(mX - M(mx))^2)(Σ(nY - M(ny))^2)) = 

(both numerator and denominator multiply by mn, which can be cancelled with each other. 

= Σ((X - Mx)(Y - My)) / √((Σ(X - Mx)^2)(Σ(Y - My)^2)) = r 

The value of correlation value does not change with change in units since r is a numerical value that measure association between two variables. 

Chapter Seven 

7.8 Each of the following pairs represents the number of licensed drivers ( X ) and the number of cars ( Y) for seven houses in my neighborhood: 

(a) Construct a scatterplot to verify a lack of pronounced curvilinearity. 

(b) Determine the least squares equation for these data. (Remember, you will first have to calculate r, SS y, and SS x .) 

X Values 

5,5,2,2,3,1,2 

Σ = 20 

Mean = 2.85714 

Σ(X - Mx)^2= SSx = 14.8571 

Y Values 

4,3,2,2,2,1,2, 

Σ = 16 

Mean = 2.28571. 

Σ(Y - My)^2 = SSy = 5.42857 

X and Y Values 

N = 7. 

Σ(X - Mx)(Y - My) = Sxy = 8.28571 

R 

r = Σ((X - Mx)(Y - My)) / √((SSx)(SSy)) = Σ((X - Mx)(Y - My)) / √((Σ(X - Mx)^2)(Σ(Y - My)^2)) = 

0.922613 

Least square equation 

b1 = Sxy / SSx = 8.28571 / 14.8571 = 0.557694 

b0 = My - b1 Mx = 2.28571 - 0.557694 * 2.85714 = 0.6923 

y = b0 + b1 x = 0.6923 + 0.558445 x 

Fit curve is in scatterplot. 

(c) Determine the standard error of estimate, s y | x, given that n 5 7. 

se = sqrt((SSy - b1 *Sxy)/(n - 2)) = sqrt((5.42857 - 0.557694*8.28571)/(7 - 2)) = 0.401915 

( d) Predict the number of cars for each of two new families with two and five drivers. 

y =a+b.x= 

y(2) = 0.6923 + 0.558445*2 = 1.80919 

y(5) = 0.6923 + 0.558445*5 = 3.48453 

7.10 Assume that r 2 equals .50 for the relationship between height and weight for adults. 

Indicate whether the following statements are true or false. 

(a) Fifty percent of the variability in heights can be explained by variability in weights. 

True 

(b) There is a cause-effect relationship between height and weight. 

False. 

(c) The heights of 50 percent of adults can be predicted exactly from their weights. 

False 

(d) Fifty percent of the variability in weights is predictable from heights. 

True 

7.13 In the original study of regression toward the mean, Sir Francis Galton noted a tendency for offspring of both tall and short parents to drift toward the mean height for offspring and referred to this tendency as “regression toward mediocrity.” What is wrong with the conclusion that eventually all heights will be close to their mean? 

The issue with the Galton’s conclusion is eventually everybody will be the same height, which is not the case despite centuries of human existence. The conclusion to no small extent implies convergence over time. It ignores the fact that regression works backwards as well as forward. Regression to the mean occurs when enduring features combine casually with transient elements. Convergence overtime, and as such, Galton’s conclusion ignores transient factors.