Section 12.1

HAWKES LEARNING SYSTEMS

Students Matter. Success Counts.

Copyright © 2013 by Hawkes Learning

Systems/Quant Systems, Inc.

All rights reserved.

Section 12.1

Scatter Plots and Correlation






Objectives

o Construct and interpret scatter plots. o Calculate and interpret the correlation between two

variables.






Table

Table 12.2: Sample of NFL Quarterbacks (2011–2012 Season)

Number of Passing

Touchdowns

2012 Base Salary

(in Millions of Dollars)

Quarterback Rating

Drew Brees 46 3.0 110.6Michael Vick 18 12.5 84.9Philip Rivers 27 10.2 88.7Tony Romo 31 0.825 102.5

Aaron Rodgers 45 8.0 122.5Jay Cutler 13 7.7 85.7Alex Smith 17 5.0 90.7






Table (cont.)

Table 12.2: Sample of NFL Quarterbacks (2011–2012 Season)Eli Manning 29 1.75 92.9Tim Tebow 12 2.1 72.9Tom Brady 39 0.95 105.6

Source: Yahoo! Sports. “NFL - Statistics by Position.” http://sports.yahoo.com/nfl/stats/byposition?pos=QB&conference=NFL&year=season_20 11&sort=49&timeframe=All (20 May 2012). Source: Spotrac.com. “NFL Player Contracts, Salaries, and Transactions.” http://www.spotrac.com/nfl/ (2 Oct. 2012).






Example 12.1: Creating a Scatter Plot to Identify Trends in Data

Use the data from Table 12.2 to produce a scatter plot that shows the relationship between the base salary of an NFL quarterback and the number of touchdowns the quarterback has thrown in one season. Solution We might expect for the number of touchdowns a quarterback throws in one season to influence his salary. Taking this into consideration, we will place the number of touchdowns on the x-axis and the base salary on the y-axis.






Example 12.1: Creating a Scatter Plot to Identify Trends in Data (cont.)







Looking at this scatter plot, we do not see a linear pattern. Actually, no pattern is evident. This probably indicates that these two variables do not have a relationship after all.






Example 12.2: Creating a Scatter Plot to Identify Trends in Data

Use the data in Table 12.2 to produce a scatter plot that shows the relationship between the number of touchdowns thrown in one season and the corresponding quarterback rating for the given sample of NFL quarterbacks. Solution In this case, we would expect that the number of touchdowns thrown by a quarterback does influence that quarterback’s rating, since number of touchdowns is one of many factors used to determine the quarterback rating.







Hence, the logical way to label the axes is to place the number of passing touchdowns on the x-axis and the quarterback rating on the y-axis.







Notice that the points tend to go up from left to right, and fall close to a straight line. This pattern can be described as a linear pattern with a positive slope.






Example 12.3: Determining Whether a Scatter Plot Would Have a Positive Slope, Negative Slope, or Not Follow a Straight-Line Pattern

Determine whether the points in a scatter plot for the two variables are likely to have a positive slope, negative slope, or not follow a straight-line pattern. a. The number of hours you study for an exam and the

score you make on that exam b. The price of a used car and the number of miles on

the odometer c. The pressure on a gas pedal and the speed of the car d. Shoe size and IQ for adults






Example 12.3: Determining Whether a Scatter Plot Would Have a Positive Slope, Negative Slope, or Not Follow a Straight-Line Pattern (cont.)

Solution a. As the number of hours you study for an exam

increases, the score you receive on that exam is usually higher. Thus, the scatter plot would have a positive slope.

b. As the number of miles on the odometer of a used car increases, the price usually decreases. Thus, the scatter plot would have a negative slope.






Example 12.3: Determining Whether a Scatter Plot Would Have a Positive Slope, Negative Slope, or Not Follow a Straight-Line Pattern (cont.)

c. The more you push on the gas pedal, the faster the car will go. Thus, the scatter plot would have a positive slope.

d. Common sense suggests that there is not a relationship, linear or otherwise, between a person’s IQ and his or her shoe size.







The Pearson correlation coefficient, , is the parameter that measures the strength of a linear relationship between two quantitative variables in a population. The correlation coefficient for a sample is denoted by r. It always takes a value between −1 and 1, inclusive.

1 1r







Pearson Correlation Coefficient The Pearson correlation coefficient for paired data from a sample is given by

where n is the number of data pairs in the sample, xi is the ith value of the explanatory variable, and yi is the ith value of the response variable.

2 22 2

i i i i

i i i i

n x y x yr

n x x n y y






Example 12.4: Calculating the Correlation Coefficient Using a TI-83/84 Plus Calculator

Calculate the correlation coefficient, r, for the data from Table 12.2 relating touchdowns thrown and base salaries. Solution The data we need from Table 12.2 are reproduced in the following table.






Example 12.4: Calculating the Correlation Coefficient Using a TI-83/84 Plus Calculator (cont.)

NFL Quarterbacks Number of Passing

Touchdowns Base Salary (in

Millions of Dollars) 46 3.0 18 12.5 27 10.2 31 0.825 45 8.0 13 7.7 17 5.0 29 1.75 12 2.1 39 0.95







Let’s enter these data into our calculator. • Press . • Select option 1:Edit. • Enter the values for number of touchdowns (x) in

L1 and the values for base salary (y) in L2. • Press . • Choose CALC. • Choose option 4:LinReg(ax+b). • Press twice.







From the scatter plot, we would expect r to be close to 0. The calculator confirms that the correlation coefficient for these two variables is r ≈ 0.251, indicating a weak negative relationship, if any relationship exists at all.






Example 12.5: Calculating the Correlation Coefficient Using a TI-83/84 Plus Calculator

Calculate the correlation coefficient, r, for the data from Table 12.2 relating touchdowns thrown and quarterback ratings. Solution The data we need from Table 12.2 are reproduced in the following table.







NFL Quarterbacks Number of Passing

Touchdowns Quarterback

Rating

46 110.6 18 84.927 88.731 102.545 122.513 85.717 90.729 92.912 72.939 105.6







Let’s enter these data into our calculator. • Press . • Select option 1:Edit. • Enter the values for number of touchdowns (x) in

L1 and the values for base salary (y) in L2. • Press . • Choose CALC. • Choose option 4:LinReg(ax+b). • Press twice.







From the scatter plot, we would expect r to be close to 1. The calculator confirms that the correlation coefficient for these two variables is r ≈ 0.925. Since the value is close to 1, this indicates a very strong positive correlation between the variables.






Testing the Correlation Coefficient for Significance

Using Critical Values of the Pearson Correlation Coefficient to Determine the Significance of a Linear

Relationship A sample correlation coefficient, r, is statistically significant if .r r






Example 12.6: Using a Table of Critical Values to Determine Significance of a Linear Relationship

Use the critical values in Table I to determine if the correlation between the number of passing touchdowns and base salary from Example 12.4 is statistically significant. Use a 0.05 level of significance. Solution Begin by finding the critical value for = 0.05 with n = 10 in Table I. Find the value in the table where the row for n = 10 intersects the column for = 0.05.






Example 12.6: Using a Table of Critical Values to Determine Significance of a Linear Relationship (cont.)

n = 0.05 = 0.016 0.811 0.9177 0.754 0.8758 0.707 0.8349 0.666 0.798

10 0.632 0.76511 0.602 0.73512 0.576 0.708






Example 12.6: Using a Table of Critical Values to Determine Significance of a Linear Relationship (cont.)

Thus, r = 0.632. Comparing this critical value to the absolute value of the correlation coefficient we found for the data in Example 12.4, we have 0.251 < 0.632, and thus r < r. Therefore, the linear relationship between the variables is not statistically significant at the 0.05 level of significance. Thus, we do not have sufficient evidence, at the 0.05 level of significance, to conclude that a linear relationship exists between the number of passing touchdowns during the 2011–2012 season and the 2012 base salary of an NFL quarterback.






Testing the Correlation Coefficient for Significance Using Hypothesis Testing

Testing Linear Relationships for Significance Significant Linear Relationship (Two-Tailed Test)

H0: = 0 (Implies that there is no significant linear relationship)

Ha: ≠ 0 (Implies that there is a significant linear relationship)







Testing Linear Relationships for Significance (cont.)Significant Negative Linear Relationship (Left-Tailed Test)

H0: ≥ 0 (Implies that there is no significant negative linear relationship)

Ha: < 0 (Implies that there is a significant negative linear relationship)







Testing Linear Relationships for Significance (cont.) Significant Positive Linear Relationship (Right-Tailed Test)

H0: ≤ 0 (Implies that there is no significant positive linear relationship)

Ha: > 0 (Implies that there is a significant positive linear relationship)







Test Statistic for a Hypothesis Test for a Correlation Coefficient

The test statistic for testing the significance of the correlation coefficient is given by

212

rt

rn







Test Statistic for a Hypothesis Test for a Correlation Coefficient (cont.)

where r is the sample correlation coefficient and n is the number of data pairs in the sample. The number of degrees of freedom for the t-distribution of the test statistic is given by n 2.







Rejection Regions for Testing Linear Relationships Significant Linear Relationship (Two-Tailed Test)

Reject the null hypothesis, H0 , if Significant Negative Linear Relationship (Left-Tailed Test)

Reject the null hypothesis, H0 , ifSignificant Positive Linear Relationship (Right-Tailed Test)

Reject the null hypothesis, H0 , if

2.tt

.tt

.tt






Example 12.7: Performing a Hypothesis Test to Determine if the Linear Relationship between Two Variables Is Significant

Use a hypothesis test to determine if the linear relationship between the number of parking tickets a student receives during a semester and his or her GPA during the same semester is statistically significant at the 0.05 level of significance. Refer to the data presented in the following table.

GPA and Number of Parking Tickets

Number of Tickets 0 0 0 0 1 1 1 2 2 2 3 3 5 7 8

GPA 3.6 3.9 2.4 3.1 3.5 4.0 3.6 2.8 3.0 2.2 3.9 3.1 2.1 2.8 1.7






Example 12.7: Performing a Hypothesis Test to Determine if the Linear Relationship between Two Variables Is Significant (cont.)

Solution Step 1: State the null and alternative hypotheses. We wish to test the claim that a significant linear relationship exists between the number of parking tickets a student receives during a semester and his or her GPA during the same semester. Thus, the hypotheses are stated as follows.

0: 0: 0a

HH







Step 2: Determine which distribution to use for the test statistic, and state the level of significance. We will use the t-test statistic presented previously in this section along with a significance level of = 0.05 to perform this hypothesis test. Step 3: Gather data and calculate the necessary sample statistics. We need to begin by calculating the correlation coefficient, r.







Since it is possible to argue for either of these two variables affecting the other, let’s assign the number of tickets to be our explanatory variable (x), and thus the GPA as the response variable (y). Using a TI-83/84 Plus calculator, enter the values for the numbers of tickets (x) in L1 and the values for the GPAs (y) in L2. Then press and choose CALC and option 4:LinReg(ax+b). Press twice. We get r ≈ 0.586619 from the calculator and we know that n = 15.







Note that we rounded r to six decimal places, rather than three decimal places, to avoid additional rounding error in the following calculation of the test statistic. Substituting these values into the formula for the t-test statistic yields the following.

2 2

0.586619

1 1 0.58

2.612

66192 15 2

r

rt

n







Step 4: Draw a conclusion and interpret the decision. We will use rejection regions in this example to draw the conclusion. Since the sample size for this example is 15, the number of degrees of freedom is n 2 = 15 2 = 13. Using the t distribution table or ‑appropriate technology, we find the critical value for this test, So we will reject the null hypothesis, H0, if t ≥ 2.160.

2 0.05 2 0.025 2.160.tt t







Since t ≈ 2.612 and 2.612 ≥ 2.160, the test statistic falls in the rejection region. Thus, we reject the null hypothesis. Therefore, there is sufficient evidence at the 0.05 level of significance to support the claim that there is a significant linear relationship between the number of parking tickets a student receives during a semester and his or her GPA during the same semester.






Example 12.8: Performing a Hypothesis Test to Determine if the Linear Relationship between Two Variables Is Significant

An online retailer wants to research the effectiveness of its mail-out catalogs. The company collects data from its eight largest markets with respect to the number of catalogs (in thousands) that were mailed out one fiscal year versus sales (in thousands of dollars) for that year. The results are as follows.

Number of Catalogs Mailed and SalesNumber of Catalogs

(in Thousands) 2 3 3 3 4 4 5 6

Sales (in Thousands) $126 $98 $255 $394 $107 $122 $334 $403







Use a hypothesis test to determine if the linear relationship between the number of catalogs mailed out and sales is statistically significant at the 0.01 level of significance. Solution Step 1: State the null and alternative hypotheses. We wish to test the claim that a significant linear relationship exists between the number of catalogs mailed out and the corresponding sales for that area.







Step 2: Determine which distribution to use for the test statistic and state the level of significance. We will use the t-test statistic with the given level of significance, = 0.01.

0: 0: 0a

HH







Step 3: Gather data and calculate the necessary sample statistics. We first need to calculate the correlation coefficient, r. It is possible to infer that mailing a larger number of catalogs to a region will influence the number of sales in that region. Thus, the explanatory variable (x) will be the number of catalogs and the response variable (y) will be the sales. Using a TI-83/84 Plus calculator, enter the values for the numbers of catalogs mailed (x) in L1 and the sales values (y) in L2.







Then press and choose CALC and option 4:LinReg(ax+b). Press twice. From the calculator we see that r ≈ 0.504505, and we know that n = 8. Note that we rounded r to six decimal places, rather than three decimal places, to avoid additional rounding error in the following calculation of the test statistic. Substituting these values into the equation for the test statistic, we have the following.







2

2

12

0.504505

1 0.5045

1.4

0

3

58 2

1

r

rn

t







Step 4: Draw a conclusion and interpret the decision. We will use rejection regions to draw the conclusion. Since the sample size for this example is 8, the number of degrees of freedom is n 2 = 8 2 = 6. Using the t-distribution table or appropriate technology, we find the critical value for this test, So, we will reject the null hypothesis, H0 , if t ≥ 3.707.

2 0.01 2 0.005 3.707.tt t







Since the value of the test statistic, t ≈ 1.431, is less than the critical value, we fail to reject the null hypothesis. Hence, there is not enough evidence at the 0.01 level of significance to support the claim that there is a significant linear relationship between the number of catalogs distributed in a particular area and the corresponding sales in that area.

2 3.707,t






Coefficient of Determination

The coefficient of determination, r2 , is a measure of the proportion of the variation in the response variable (y) that can be associated with the variation in the explanatory variable (x).






Example 12.9: Calculating and Interpreting the Coefficient of Determination

If the correlation coefficient for the relationship between the numbers of rooms in houses and their prices is r = 0.65, how much of the variation in house prices can be associated with the variation in the numbers of rooms in the houses? Solution Recall that the coefficient of determination tells us the amount of variation in the response variable (house price) that is associated with the variation in the explanatory variable (number of rooms).






Example 12.9: Calculating and Interpreting the Coefficient of Determination (cont.)

Thus, the coefficient of determination for the relationship between the numbers of rooms in houses and their prices will tell us the proportion or percentage of the variation in house prices that can be associated with the variation in the numbers of rooms in the houses. Also, recall that the coefficient of determination is equal to the square of the correlation coefficient.






Example 12.9: Calculating and Interpreting the Coefficient of Determination (cont.)

Since we know that the correlation coefficient for these data is r = 0.65, we can calculate the coefficient of determination as Thus, approximately 42.3% of the variation in house prices can be associated with the variation in the numbers of rooms in the houses.

22 0.65 .0.4225r

Section 12.1

Documents

Transcript of Section 12.1