Lesson 12.7 TI-83

Linear Regression and Correlation Problems

Experiment - View a Tutorial:

Now, it is your turn. To understand the ideas of linear regression and correlation, try the above interactive example. You can use your own data or use the data from a homework problem in Chapter 112 in Introductory Statistics (pick any problem). Close the window when you are finished and you will return to this portion of the lesson to try the interactive example.

Note: You will need to have "shockwave" installed on your browser to try the above example.

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Example Showing TI-83 Steps

Kim, a personal trainer, was interested in knowing if there was a linear relationship between the number of visits her clients made to the gym each week (independent variable) and the average amount of time her clients exercised per visit (dependent variable).

She took the following data.

The line of best fit or regression equation and correlation coefficient are:

 

r = -0.9381

The graph shows the scatter plot and line of best fit.

TI-83 Steps:

• Do this step only once. When you turn the calculator off, it will stay as you set it. Press 2nd CATALOG. Arrow down to DiagnosticOn. Press ENTER twice.
• Clear lists L1 and L2.
• Enter the data into L1 (x values) and L2 (y values).

• Press 2nd QUIT to go to the home screen.
• Press Y = and clear any equations using CLEAR.
• Press 2nd STATPLOT.
• Press 4 and ENTER.
• Press 2nd STATPLOT.
• Press 1 and ENTER.Arrow down to the scatter plot picture (the first picture) and press ENTER.
• Arrow down to Xlist: and enter L1.
• Arrow down to Ylist: and enter L2.
• Arrow down to Mark: and press ENTER.
• Press ZOOM and 9. You should see the scatter plot.
• Press the TRACE key and the arrow keys to move from point to point.
• Press STAT and arrow over to CALC.
• Press 8. Enter L1, L2 and press ENTER. You should see:

• Press Y =.
• Press VARS.
• Press 5.
• Arrow over to EQ and press 1. You should see \Y1 = 2.62 + - .44X.
• Press GRAPH. You should see the scatterplot together with the line of best fit.

Is the correlation coefficient significant?

• Go to the 95% critical values chart to find the critical value. n - 2 = 6 - 2 = 4. Since r = - 0.9381, we will compare r to - 0.811. (Close the window when you are finished with the chart.)

Using the line of best fit, we will estimate the average time spent exercising per visit for 4 visits per week.

TI-83 Steps (You should already have done linear regression):

• Go to the home screen and clear it.
• Press VARS.
• Press 5.
• Arrow over to EQ and press 1.
• Arrow onto the X and press the multiplication key.
• Press 4 and ENTER. You should see .86. This means that the average time spent exercising per visit for 4 visits is 0.86 hours.

Redoing the above example using LinRegTTest in TI-83/84 calculator

Data is entered into L1 and L2 as before. (See page 643 in Introductory Statistics for more explicit details.)

• Go to STAT, TESTS in the TI-83/84 calculator and scroll up to LinRegTTest, press ENTER
  
• L1 should be in Xlist, L2 in Ylist.  The Freq is 1.  The not equal sign should be highlighted. 
  
• Leave RegEQ blank
  
• Highlight Calculate and press ENTER.

The calculator will give you the test statistics, the pvalue, the degrees of freedom, a, b, s (the standard deviation), r^2 (the coefficent of determination) and r.  Everything you need to decide if the correlation coefficient is significant. 

Note that the pvalue is 0.0056.  Assuming an alpha of 0.05, this means we reject the null hypothesis.  Rejecting the null means that the correlation coefficient IS significant and therefore the least squares line can be used for prediction.

The other thing to note is that the calculator gives you the standard deviation so you don't need to go through the steps described in 12.6 Outliers.  Just mulitiply 0.26 by 1.9 to get 0.49.  Any |y - y-hat| > 0.49 is a potential outlier.  Where can you find those y - y-hat values without calculating each of them? Your calculator did the work, you just need to find the RESID list, which is where the calulator put them.  Press 2nd, LIST and scroll down to RESID and press ENTER twice.  The numbers -0.18, 0.20,0.14, 0.26, -.030, -0.12 will be visible (use the right scoll button to move to the right).  Taking the absolute value of these numbers we see that none of them are above 0.49, which means there are no potential outliers for this problem.

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Examples Showing the Keypad and Keystrokes of the TI-83

Example

Is there a linear relationship between the age of a backpacker and the number of days the backpacker backpacks on one backpack trip?

The following linear regression and correlation example attempts to answer that question.  This example will also show you a way answer this question using a LinRegTTest Close the window when you are finished viewing the example. You will return here.

Think About It

Runners consider stride rate to be one of the most important measures of form. Stride rate is the number of steps taken per second and should increase as running speed increases. In a study of some of the best American female runners, researchers measured the stride rate for different speeds. The following table gives speed in feet per second (x) and the stride rate (y) for these runners.

• Verify the following graph by creating a scatter plot and then plotting the line of best fit. Before you graph the line of best fit, make sure you do linear regression. Does the scatter plot indicate a linear relationship between speed and stride rate?

• Verify that the line of best fit and the correlation coefficient are:

• Is the correlation coefficient r significant? Go to the 95% critical values chart to find the critical value. Close the window when you are finished with the chart.

• If the speed is 18 feet per second, what is the estimated stride rate?

This is the last section of this lesson.

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Up » 12.1 Linear Equations » 12.2 Scatter Plots » 12.3 The Regression Equation » 12.4 The Correlation Coefficient » 12.5 Prediction » 12.6 Outliers » 12.7 TI-83