   Chapter 14, Problem 7P Essentials of Statistics for the B...

8th Edition
Frederick J Gravetter + 1 other
ISBN: 9781133956570

Solutions

Chapter
Section Essentials of Statistics for the B...

8th Edition
Frederick J Gravetter + 1 other
ISBN: 9781133956570
Textbook Problem

With a small sample, a single point can have a large effect on the magnitude of the correlation. To create the following data, we started with the scores from problem 8 and changed the first X value from X = 1 to X = 6. X Y 6 6 4 1 1 4 1 3 3 1 a. Sketch a scatter plot and estimate the value of the Pearson correlation.b. Compute the Pearson correlation.

a.

To determine

To Construct: A scatter plot.

To determine: The estimated value of the Pearson correlation.

Output using the SPSS software is given below: The estimated value of the Pearson correlation is around 0.25 to 0.30.

Explanation

Given info:

The scores of X and Y.

Software procedure:

Step-by-step procedure to obtain the scatter plot using the SPSS software:

• Enter the data under X and Y column.
• Go to Graphs > select Chart builder.
• From Gallery option drag Simple scatter and drop under chart preview box. Similarly drag X and Y variables from Variables box then drop X variable in X-axis box under chart preview box and drop Y variable in Y-axis box under chart preview box.
• Click OK.

Justification:

The scatter plot shows points clustered around a line sloping up to the right.

Thus, the estimated value of the Pearson correlation, just by looking at the scatter plot, is around 0.25 to 0.30.

b.

To determine

To Calculate: The value of Pearson correlation.

The Pearson correlation is 0.277.

Explanation

Calculation:

The Pearson correlation (r) is calculated as:

r=SP(SSx)×(SSy)

Here SP is calculated as:

SP=(XMx)(YMy)

Squared deviations SSx and SSy are calculated as:

SSx=(XMx)2 and SSy=(YMy)2

The below table showing the calculations required for calculating correlation:

 S.No. Scores Deviations squared deviation Products X Y (X-Mx) (Y-My) (X-Mx)2 (Y-My)2 (X-Mx)(Y-My) 1 6 6 3 3 9 9 9 2 4 1 1 -2 1 4 -2 3 1 4 -2 1 4 1 -2 4 1 3 -2 0 4 0 0 5 3 1 0 -2 0 4 0 Sum 3 3 18 18 5

So the sum of product of deviation (SP) is 5, SSx is 18 and SSy is 18, substitute these values in Pearson correlation formula then:

r=SP(SSx)×(SSy)=5(18)×(18)=0.277

Thus, the Pearson correlation is 0.277.

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