Chapter 23. Inferences for Regression

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Chapter 23. Inferences for Regression

Topics covered in this chapter:

• Simple Linear Regression

Simple Linear Regression

Example 23.1: Crying and IQ

The Problem: Infants who cry easily may be more easily stimulated than others. This may be a sign of higher IQ. Child development researchers explored the relationship between the crying of infants four to ten days old and their later IQ test scores. A snap of a rubber band on the sole of the foot caused the infants to cry. The researchers recorded the crying and measured its intensity by the number of peaks in the most active 20 seconds. They later measured the children’s IQ at age three years using the Stanford- Binet IQ test. Do children with higher crying counts tend to have higher IQ?

1. Create a scatterplot:

a. Open data set ta23-01.por.

b. Click Graphs, scroll to Legacy Dialogs then Scatter/Dot.

c. Click on Simple Scatter, then click on Define.

d. Move IQ into the Y Axis box since IQ is the response variable

e. Move Crycount into the X Axis box since Crycount is the explanatory variable.

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f. Click OK. The scatterplot will appear in the output window.

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2. Find the least-squares regression line.

a. Click Analyze. Scroll to Regression then Linear.

b. Move IQ to the Dependent box.

c. Move Crycount to the Independent box.

d. Click OK.

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The least squares regression line is given by yˆ=91.268+1.493x. The slope of the least squares regression line, 1.493, is found in the Coefficients table under the B column in the row for Crycount. The y-intercept of the least squares regression line, 91.268, is also found in the Coefficients table under the B column in the Constant row.

Example 23.7: Beer and blood alcohol

The Problem: The EESEE story “Blood Alcohol Content” describes a study in which 16 student volunteers at the Ohio State University drank a randomly assigned number of cans of beer. Thirty minutes later, a police officer measured their blood alcohol content (BAC) in grams of alcohol per deciliter of blood. The students were equally divided between men and women and differed in weight and usual drinking habits. Because of this variation, many students don’t believe that number of drinks predicts blood alcohol well. Steven thinks he can drive legally 30 minutes after he finishes 5 beers. The legal limit for driving is BAC 0.08 in all states. We want to predict Steve’s blood alcohol content, using no information except that he drinks 5 beers.

1. Regress BAC on number of beers.

a. Open the data set eg23-07.por.

b. Click Analyze, scroll down to Regression, then click Linear.

c. Move BAC to the Dependent box.

d. Move Beers to the Independent box.

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2. Display predicted values and residuals.

a. Click the Save button at the right of the window.

b. Under Predicted Values select Unstandardized.

c. Under Residuals select Unstandardized.

d. Click Continue.

3. Create a 95% Confidence Interval.

a. Click the Statistics button.

b. Under Regression Coefficients select Confidence Intervals.

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c. Click Continue.

d. Click OK. A new window will pop up with the output.

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e. The predicted values and residuals can be seen on the data sheet as 2 new columns PRE_1 and RES_1.

Notice that the predicted BAC when a person drinks 5 beers is 0.07712.

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Chapter 23 Exercises

23.1 Ebola and gorillas.

23.3 Great Arctic rivers.

23.5 Great Arctic rivers: testing.

23.7 Ebola and gorillas: testing correlation.

23.9 Ebola and gorillas: estimating slope.

23.11 Great Arctic rivers: estimating slope.

23.29 Manatees: conditions for inference.

23.33 Predicting tropical storms.

23.41 Squirrels and their food supply.

23.43 Beavers and beetles.

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Chapter 23 SPSS Solutions

23.1 Use Graphs, Legacy Dialogs, Scatter/Dot to create a plot of the data.

The plot is strongly linear and increasing. We could use Analyze, Correlate, Bivariate to find the correlation, but we also want to find the regression equation, so use Analyze, Regression to compute the regression equation (we’ll use the square root of r² as the correlation). We can have SPSS find the residuals for us by clicking Save and checking the box for Unstandardized residuals.

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Model Summary^b

Model R R Square

Adjusted R Square

Std. Error of the Estimate

1 .962^a .926 .908 4.903

a. Predictors: (Constant), Distance b. Dependent Variable: Days

Coefficients^a Unstandardized Coefficients

Standardized Coefficients

Model B Std. Error Beta t Sig.

(Constant) -8.088 5.917 -1.367 .243

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Distance 11.263 1.591 .962 7.080 .002

a. Dependent Variable: Days

The correlation is r = 0.962 – a very strong relationship. Our estimated slope of b = 11.263 says the virus takes about 11.263 days for each additional home range it must travel. The estimated intercept is a = –8.088. The standard deviation around the regression line is s = 4.90345 (labeled as Std. Error of the Estimate). To sum (or find the mean of) the residuals (created as variable RES_1) use Analyze, Descriptive Statistics, Descriptives. With a mean of 0.0000000, the sum must be 0.

Descriptive Statistics

N Minimum Maximum Mean Std. Deviation Unstandardized Residual 6 -4.70175 6.03509 .0000000 4.38578245

Valid N (listwise) 6

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23.3 We define a scatterplot of the data in ta23-02.

To add the regression line in the graph, double-click for the Chart Editor, then click Elements, Fit line at total. There is an increasing trend in the graph, with lots of scatter. SPSS gives r² = 0.112; the relationship is fairly weak. Only 11% of the variation in discharge is explained by time (year); there certainly are other factors involved.

Use Analyze, Regression, Linear to fit the regression. Looking ahead, we have asked for confidence intervals for the coefficients using the Statistics button.

Coefficients^a Unstandardized

Coefficients

95% Confidence Interval for B

Model B Std. Error Beta t Sig. Lower

Bound

Upper Bound

(Constant) -2056.769 1384.687 -1.485 .143 -4824.720 711.181

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Year 1.966 .704 .334 2.794 .007 .559 3.373

a. Dependent Variable:

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Model Summary

Model R R Square

Adjusted R Square

1 .334^a .112 .097 104.003

a. Predictors: (Constant),

The regression equation is Discharge = –2057 + 1.97*Year. The regression standard error is s = 104.003.

23.5 From the SPSS output in Exericise 23.3, the test statistic is t = 2.794 with (two- sided) P-value 0.007. The one-sided P-value is 0.0035. Since this P-value is less than any standard α, we reject a null hypothesis of no relationship and conclude that these data do show an increase in Arctic river discharge (supporting the global warming hypothesis).

23.7 Refer to the solution for Exercise 23.1 In the SPSS results, we were given t = 7.08 and P = 0.002. Since this is a two-sided P-value, divide by 2. The one-sided P-value is 0.001. Minitab gives the same (two-sided) P-value for the correlation if you use Stat, Basic Statistics, Correlation. If we use Analyze, Correlate, Bivariate, and ask for the one-sided P-value, we have the same result.

Correlations

Distance Days Pearson Correlation 1.000 .962^**

Sig. (1-tailed) .001

Distance

N 6.000 6

Pearson Correlation .962^** 1.000

Sig. (1-tailed) .001

Days

N 6 6.000

**. Correlation is significant at the 0.01 level (1-tailed).

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23.9 SPSS will find the 95% confidence interval if you redo the regression and click Statistics, then check the box to ask for confidence intervals for the coefficients.

Coefficients

Bound

Upper Bound

(Constant) -8.088 5.917 -1.367 .243 -24.516 8.341

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Distance 11.263 1.591 .962 7.080 .002 6.846 15.680

a. Dependent Variable: Days

We have the same output as in Exercise 23.1, with the addition of confidence bounds at the right side. Based on this data, Ebola takes between 6.85 and 15.68 days to travel one home range, with 95% confidence. However, the problem asked for 90% confidence.

For this, we use Transform, Compute Variable to find t*, then compute the interval “by hand.”

The interval is 11.263 2.132*1.591 (7.871, 14.655).± = Based on this data, Ebola takes between 7.87 and 14.66 days to travel one home range, with 90% confidence.

23.11 SPSS gives only 95% confidence intervals for regression parameters. We saw in Exercise 23.3 that a 95% confidence interval for the slope is from 0.559 to 3.373.

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However, this question asks for a 90% confidence interval. We use Transform, Compute Variable to find t* (degrees of freedom are n – 2), then compute the interval

“by hand.”

The confidence interval is calculated as b t SE b± * ( ), giving 1.9662 ±1.670*0.7037, or (0.791, 3.141). We are 90% confident that arctic river discharge increases between 0.791 and 3.141 cubic kilometers per year. Since the low end is positive, we’re convinced that discharge is increasing over time.

23.29 To make the stemplot, use Analyze, Descriptive Statistics, Explore.

Stem-and-Leaf Plot

Frequency Stem & Leaf 1.00 -1 . 5 1.00 -1 . 0 6.00 -0 . 566899 7.00 -0 . 0111223 8.00 0 . 01112244 5.00 0 . 55788 1.00 1 . 3 1.00 1 . 7 Stem width: 10.00 Each leaf: 1 case(s)

This plot is pretty symmetric and bell-shaped with no outliers. The Normal assumption is reasonable for these residuals. To make the scatterplot, use Graphs, Legacy Dialogs, Scatter/Dot. Use Residual on the y axis and Boats on the x axis. To add the “residual = 0” line, double click in the graph for the Chart Editor, then click Options, Y axis reference line. Close the properties window and the Chart editor.

The plot is random (no discernable pattern), so the regression model is reasonable. While pollution may have caused some manatee deaths, the data are labeled as manatees killed by boats, so pollution would not explain more of these deaths.

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23.33 We create a scatterplot of Dr. Gray’s predictions against actual storms and compute the regression.

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There is a positive relationship seen in the graph; however, there are a couple of years in which he predicted a large number of storms and the actual number was much less. Part (b) asks for a 95% confidence interval for the mean number of storms when Dr. Gray predicts 16 storms. To do this, add a forecast value of 16 in the spreadsheet (SPSS will only create prediction and confidence intervals for values in the spreadsheet), then in the Analyze, Regression, Linear dialog box, click Save and check the box for Mean Prediction Intervals.

Model Summary^b

Model R R Square

Adjusted R Square

1 .529^a .280 .247 4.086

a. Predictors: (Constant), b. Dependent Variable:

(Constant) 1.803 3.587 .503 .620

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Forecast .903 .309 .529 2.923 .008

The regression equation is ActualStorms = 1.80 + 0.90*Predicted. With a t statistic of 2.923 and (two-sided) P-value of 0.008 (so, the one-sided P-value is 0.004), the relationship is significantly positive. Return to the data spreadsheet. SPSS has created 95% confidence intervals for each value of Forecast. At the bottom, we see the values for the interval of interest.

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We predict the mean number of actual storms for years when Professor Gray predicts 16 will be between 12.78 and 19.73, with 95% confidence. If you wanted values for a particular year, you would have checked the box for Individual Prediction Intervals in the Save dialog box.

23.41 We create a scatterplot of the Cones as the X axis variable and Offspring as the Y axis variable using Graphs, Legacy Dialogs, Scatter/Dot. The pattern is roughly linear (there is a fair amount of scatter) and increasing – more cones seem to be associated with more offspring.

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Use Analyze, Regression, Linear to find linear regression and measures of association.

We will want to examine the residuals for adequacy of the regression, so click Save and check the box for Unstandardized Residuals. You can also ask for a histogram of the standardized residuals – these are z-scores – (and a Normal plot of them) in the Plots box.

The regression equation is Offspring = 1.415 + 0.44*Cones. The relationship is fairly strong – r = 0.756; the cone index explains r² = 57.2% of the variation in offspring.

Model Summary

Model R R Square

Adjusted R Square

1 .756^a .572 .542 .60031

a. Predictors: (Constant), Cones

(Constant) 1.415 .252 5.619 .000

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Cones .440 .102 .756 4.328 .001

a. Dependent Variable: Offspring

The relationship is indeed statistically significant; we have t = 4.328 with (two-sided) P = 0.001, so the one-sided P-value is 0.0005.

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There are some gaps in the histogram, but with the imposed density curve, the Normal assumption seems reasonable. Note the mean of these is (essentially) 0. Create a scatterplot of the saved residuals against the cone index using Graphs, Legacy Dialogs, Scatter/Dot. This plot shows no definite pattern, so our inference is reliable.

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23.43 Open data file ex05-51. First, create a scatterplot of the data using Graphs, Legacy Dialogs, Scatter/Dot. Enter Stumps as the X variable and Larvae as the Y variable. Give your graph an appropriate title using Titles.

We see that these data indicate that there are more beetle larvae with more stumps. Use Analyze, Regression, Linear to fit the line using Stumps as the Independent and Larvae as the Dependent. We’d like a 95% confidence interval for the slope (how many more clusters accompany each additional stump), so click Statistics, and check the box for Confidence Intervals.

Model Summary

Model R R Square

Adjusted R Square

1 .918^a .843 .835 6.455

a. Predictors: (Constant), Stumps

Coefficients

Bound

Upper Bound

(Constant) -1.286 2.853 -.451 .657 -7.220 4.647

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Stumps 11.894 1.136 .916 10.467 .000 9.531 14.257

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The regression equation is Larvae= −1.29 11.89 *+ Stumps. The relationship is strong;

the regression model explains 84.3% of the variability in larvae (the correlation is r = .843

r= = 0.918). We are confident that more stumps lead to more larvae because the 95% confidence for the slope is between 9.53 and 14.26 which is well above 0.

Our scatterplot of the residuals against Stumps (the predictor variable) indicates no discernable pattern, so this regression model is reasonable.

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