Section 12.1

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Section 12.1

Scatter Plots and Correlation

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Types of Relationships:


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Regression, Inference, and Model Building

12.1 Scatter Plots and Correlation

Strong Linear

Relationship

Non-LinearRelationship

NoRelationship

Weak LinearRelationship

Plot (x,y) data points and think about whether x and y are somehow related






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

More






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.

Scatter Plot of (touchdowns, salary) on TI-84

• Touchdowns in list L1, Salary in list L2

• Y= old algebra plots should be cleared out of there• 2nd STAT PLOT all should be “Off” to start with• 1:Plot 1: On, choose Type, Lists L1 and L2, Mark

• Remember 2ND 1, 2ND 2 to put in list names?• ZOOM 9:ZoomStat• If unexplainable error, 2ND MEM 7 1 2 to clear all

and then retype the lists of data.• TRACE and Left Arrow and Right Arrow to explore it






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.

Scatter Plot of (touchdowns, rating) on TI-84

• Ratings in list L3, Touchdowns still in L1, Salary in L2

• 2nd STAT PLOT• 1:Plot 1: Change to Lists L1 and L3

• ZOOM 9:ZoomStat• TRACE and Left Arrow and Right Arrow to explore it







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






Question: “Are x and y related?”

ρ (Greek letter rho) is the population parameter for the Correlation Coefficient

r (our alphabet’s letter r) is the sample statistic for the Correlation Coefficient

We use our sample r to estimate the population’s parameter ρ







math courseware specialists

Regression, Inference, and Model Building

12.1 Scatter Plots and Correlation

• –1 ≤ r ≤ 1

• Close to –1 means a strong negative correlation.

• Close to 0 means no correlation.

• Close to 1 means a strong positive correlation.







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.

But we will not dig into the details of that awful formula! The TI-84 has built-in goodies.






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

Use TI-84 LinRegTTest

• The next few slides describe the use of LinRegTTest.• It’s STAT, TESTS, ALPHA F (ALPHA E on the 83/Plus)• This description is about the full hypothesis test to

determine “Is the relationship significant?”• The outputs include the value of r, the correlation

coefficient, which is of greatest interest at this early point in our study.

The Hawkes materials talk about the LinReg feature but I’m recommending the LinRegTTest instead because you get more information for about the same effort.

Hypothesis Test for significant

Null Hypothesis: “No relationship”Alternative: “But there IS a significant relationship!”There’s some level of significance specified in advance, like or It involves calculating a value and finding “what is the -value of this ?”And if -value < , reject the null hypothesis

– If so, then we say “Yes, significant relationship!”

Hypothesis Test for significant

Usually we do this two-tailed test:– Null Hypothesis : “No relationship”– Alternative Hypothesis: , “There is a significant

linear relationship.”Be aware of a couple one-tailed variations:

– Test for significant POSITIVE correlation only:using and

– Test for significant NEGATIVE correlation only:using and

LinRegTTest inputs (not identical to the quarterback example!)

• Here are the inputs:

• Xlist and Ylist – where you put the data– Shortcut: 2ND 2 puts L2

• Freq: 1 (unless…)

• β & ρ: ≠ 0– This is the Alternative

Hypothesis• RegEq: VARS, right

arrow to Y-VARS, 1, 1– Just put it in for later

• Highlight “Calculate”• Press ENTER

LinRegTTest Outputs, first screen(from a different problem)

•

• t= the t statistic value for this test (the formula is in the book)

• p = the p-value for this t test statistic

• in this kind of a test• later – for regression

LinRegTTest Outputs, second screen (from a different problem)

• b later, for Regression• s much later, for

advanced Regression

• r2 = how much of the output variable (weight) is explained by the input variable (girth)

• r = the correlation coefficient for the sample– Close to – strong

positive relationship– Or – strong negative






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

(Why is this discussion here? Sometimes they give you a shred of a problem that gives some summary results and you have to use a printed table to make the determination. That’s the only time you’ll need to do this, for a few of those kinds of problems. In “real life”, in large problems, the LinRegTTest p-value is compared to alpha.)






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

INTERPRETATION: “If my sample’s correlation coefficient, r, is at least as big as the value you look up in this table, then YES, significant linear relationship. Otherwise, no, no significant linear relationship.”






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)

(Now they’re getting into the Hypothesis Testing we saw a brief preview of earlier in this set of slides.)







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

TI-84 LinRegTTest will calculate this value for us.







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

Use the TI-84 LinRegTTest to perform the hypothesis test. Use the p-value method:• The LinRegTTest gives you a p-value.• If the p-value is < the given Level of Significance α =

0.05, then REJECT the null hypothesis; conclude that there IS a significant linear relationship.

• Otherwise, Fail To Reject – no significant relationship.And you can disregard most or all of the by-hand detail that follows in the next few slides.






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.

Example 12.8

And use the TI-84 LinRegTTest to complete the hypothesis test. You can then disregard most or all of the by-hand detail that follows in the next few slides.

Disregard also their use of LinReg; the LinRegTTest is bigger and better.







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).

This too is reported to you in the LinRegTTest outputs.






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

Correlation Coefficient in Excel

More with Excel

That’s about all that can be done with basic Excel.There is an advanced feature on Data tab, then the Data Analysis add-in.

It gets intothe Regressiontopic in thenext lesson.

Section 12.1

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Transcript of Section 12.1