Chapter 8 – Regression 2 Basic review, estimating the standard error of the estimate and short cut problems and solutions.

You can use the regression equation when:
1. the relationship between X and Y is linear, 2. r falls outside the CI.95 around and is therefore a statistically significant correlation, and 3. X is within the range of X scores observed in your sample,

df nonsignificant 1 2 3 4 5 6 7 8 9 10 11 12 . 100 200 300 500 1000 2000 10000 -.996 to .996 -.949 to .949 -.877 to .877 -.810 to .810 -.753 to .753 -.706 to .706 -.665 to .665 -.631 to .631 -.601 to .601 -.575 to .575 -.552 to .552 -.531 to .531 . -.194 to .194 -.137 to .137 -.112 to .112 -.087 to .087 -.061 to .061 -.043 to .043 -.019 to .019 .997 .950 .878 .811 .754 .707 .666 .632 .602 .576 .553 .532 . .195 .138 .113 .088 .062 .044 .020 .9999 .990 .959 .917 .874 .834 .798 .765 .735 .708 .684 .661 . .254 .181 .148 .115 .081 .058 .026

Simple problems using the regression equation
tY' = r * tX tY' = .150 * 0.40 = 0.06 tY' = .40 * -1.70 = -0.68 tY' = .40 * 1.70 = 0.68

Predictions from Raw Data
1. Calculate the t score for X. 2. Solve the regression equation. 3. Transform the estimated t score for Y into a raw score.

Problem: We look into the correlation between time spent studying and score on a midterm exam The mean study time in a random sample is 560 minutes (Range = ). The estimated standard deviation of the study times, sX, was The estimated mean on the midterm was and the sample’s estimated standard deviation, (sY) was 7.98. There were 10 pairs of tX,tY scores The estimated correlation coefficient is .851. John studied for 400 minutes for the midterm. Predict his midterm score.

Can you use the regression equation?

YES! You can use the regression equation.
r (8) = .851, p < .01 400 minutes is inside the range of X scores seen in the random sample ( minutes)

Predicting from and to raw scores
1. Translate raw X to tX score. X X-bar sX (X-X-bar) / sX = tX ( )/216.02= -0.74

Use regression equation
2. Find value of tY' r r * tX = tY' *-0.74=-0.63

Translate tY' to raw Y' Y sY Y + (tY' * sY) = Y'
(-0.63*7.98) = 70.97

Another reminder Never assume that a correlation will stay linear outside of the range you originally observed. Therefore, never use the regression equation to make predictions from X values outside of the range you found in your sample. Example: Basing a prediction of the height of a 83 year old adult based on a study examining the correlation of age and height in a sample composed only of children age 12 or less.

Correlation Characteristics: Which line best shows the relationship between age (X) and height (Y)
Linear vs Curvilinear

Reviewing the r table and reporting the results of calculating r from a random sample
.

How the r table is laid out: the important columns
Column 1 of the r table shows degrees of freedom for correlation and regression (dfREG) dfREG=nP-2 Column 2 shows the CI.95 for varying degrees of freedom Column 3 shows the absolute value of the r that falls just outside the CI.95. Any r this far or further from falsifies the hypothesis that rho=0.000 and can be used in the regression equation to make predictions of Y scores for people who were not in the original sample but who were part of the population from which the sample is drawn.

Notice how the interval in Column 2 gets narrower as df increase
Notice how the interval in Column 2 gets narrower as df increase. Why does this happen? Why is an r of .210 not significant with nP= 12 but is significant with nP=102?

This happens because r is a consistent estimate of rho, as well as a least-squares, unbiased estimate. As you increase df, r becomes a better and better estimate of rho. This is the same thing that happens when a sample mean gets closer to the population mean. Sample statistics get closer and closer to their population parameters as df increase.

AS dfREG increase, r becomes a better estimate of rho.
If rho were .400, r should get closer and closer to .400 as df increase. Why does this happen? Remember, r is really an index of how far from each other the ZXZY scores in the population are, on the average. An r of .400 means an estimated average squared difference between the Z scores of about squared units (r=1-.5(1.200) = .400

Each pair of scores tends to bring r closer to rho
Some of the pairs of tXtY scores will be further from each other than average, some will be closer. Each randomly added pair of scores tends to make the estimated average squared difference between the Z scores closer to the average found in the population. That makes r closer and closer to rho.

Test Scores correlation between X & Y about .500
F r e q u n c y Scores Mean Squared Distance of t scores score 1.2 1.6 Squared differences between tX & tY 2.2 1.2 1.0 1.3 .4 .7 Means: 1.7 1.5 1.4 1.3 1.3 1.2 1.1

According to the null, rho = 0.000
If the null is right, increasing df should result in r getting closer to the true value of rho, If we had 100 random samples, all would tend to get closer to 0.000 Therefore, the interval around in which we could expect 95 of the 100 samples to fall would get narrower and narrower. (That’s what you see in Column 2 of the r table.) Of course, that brings the critical values in Columns 3 and 4 similarly closer to So with larger samples, weaker correlations can be statistically significant.

Can we generalize to the population from the correlation in the sample?
A Type 1 error involves saying that there is a correlation in the population as a whole, when the correlation is actually (and the null is true). We carefully guard against Type 1 error by using significance tests to try to falsify the null hypothesis.

Why is it important to avoid Type 1 error.

Using the regression equation when rho=0
Using the regression equation when rho=0.000 increases error beyond that which we would get if we predicted everyone will score precisely at the mean of Y. The scientist’s rule is “First don’t increase error!” Such errors lead to unfair prejudgments rather than giving everyone an equal chance.

Example : Achovy pizza and horror films, rho=0.000
7 9 8 6 5 2 1 (scale 0-9) anchovies 7 3 8 4 1 H1: People who enjoy food with strong flavors also enjoy other strong sensations. H0: There is no relationship between enjoying food with strong flavors and enjoying other strong sensations. Can we reject the null hypothesis?

Can we reject the null hypothesis?
8 6 4 2 Pizza Horror films

Can we reject the null hypothesis?
We do the math and we find that: r = .352 df = 8

This finding falls within the CI.95 around 0.000
We call such findings “nonsignificant” Nonsignificant is abbreviated n.s. We would report these finding as follows r (8)=0.352, n.s. Given that it fell inside the CI.95, we must assume that rho actually equals zero and that our sample r is .352 instead of solely because of sampling fluctuation. We go back to predicting that everyone will score at the mean of Y.

In fact, the null hypothesis was correct; rho = 0.000
I made up that example using numbers randomly selected from a random number table. So there really was no relationship between the two sets of scores: rho really equaled 0.000 But samples don’t give you an r of zero, they fluctuate around 0.000 Significance testing is your protection against mistaking sampling fluctuation for a real correlation. Significance testing protects against Type 1 error.

We use significance testing to protect us from Type 1 error.
Our sample gave us an r of .352. Without the r table, we could have thought that far enough from zero to represent a true correlation in the population. 0.352 was the product only of sampling fluctuation Significance testing is your protection against mistaking sampling fluctuation for a real correlation. Significance testing protects against Type 1 error.

How to report a significant r
For example, let’s say that you had a sample (nP=30) and r = -.400 Looking under nP-2=28 dfREG, we find the interval consistent with the null is between and +.360 So we are outside the CI.95 for rho=0.000 We would write that result as r(28)=-.400, p<.05 That tells you the dfREG, the value of r, and that you can expect an r that far from five or fewer times in 100 when rho = 0.000

Then there is Column 4 Column 4 shows the values that lie outside a CI.99 (The CI.99 itself isn’t shown like the CI.95 in Column 2 because it isn’t important enough.) However, Column 4 gives you bragging rights. If your r is as far or further from as the number in Column 4, you can say there is 1 or fewer chance in 100 of an r being this far from zero (p<.01). For example, let’s say that you had a sample (nP=30) and r = The critical value at .01 is You are further from than that.So you can brag. You write that result as r(28)=-.525, p<.01.

To summarize If r falls inside the CI.95 around 0.000, it is nonsignificant (n.s.) and you can’t use the regression equation (e.g., r(28)=.300, n.s. If r falls outside the CI.95, but not as far from as the number in Column 4, you have a significant finding and can use the regression equation (e.g., r(28)=-.400,p<.05 If r is as far or further from zero as the number in Column 4, you can use the regression equation and brag while doing it (e.g., r(28)=-.525, p<.01

Here is one in exam format
Jack was comparing the degree of severity of a back injury as ascertained by an MRI and the amount of pain reported by the patient. Back injuries were rated in severity on a scale from 0-9. Jack randomly selected 26 patients for his sample. MRI Scores ranged from 1-8. The mean severity was 4.70 with a standard deviation of 2.00 points. Pain ratings were assessed on a scale that went from Mean pain rating in the sample was with a standard deviation of 3.00. The sum of the squared differences between the tX and tY scores was 1. Compute the correlation and report it in correct form. 2. Mrs. Green, who was part of the population but not part of your original sample, had a back injury rated as a 6. Predict her reported pain. 3. Mr. Jones, who was part of the population but not part of your original sample, had a back injury rated as a 9. Predict his reported pain.

Answer to Q. 1 r = 1-.5(30/25) =.400 r(24)=.400, p<.05

Q. 2 First translate to tX 1. Translate raw X to tX score.
X X-bar sX (X-X-bar)/sX = tX (6-4.70)/2.00= 0.65

Q.2-Use regression equation
2. Find value of tY' r r * tX = tY' *0.65= .26

Q.2 -Translate tY' to raw Y' YBAR sY YBAR + (tY' * sY) = Y'
(.26*3.00) = 11.78 Mrs. Green was predicted to score a little more than a quarter of a standard deviation about the mean, With a mean pain rating of and a standard deviation of 3.00 points, Mrs. Green’s pain rating is predicted to be

Q. 3 Mr. Jones’ pain rating It will turn out that Mr. Jones will have a lower predicted pain rating than Ms. Green, though his severity score is much higher. Mr. Jones should have a pain rating of 11.00 Why? Because his injury was rated 9. We had a range in our random sample of 1-8. He was not like anyone we had seen before. You can not use the regression equation. You must go back to predicting that he will score at your best estimate of mu, XBAR.

How much better than the mean can we guess?

Improved prediction If we can use the regression equation rather than the mean to make individualized estimates of Y scores, how much better are our estimates? We are making predictions about scores on the Y variable from our knowledge of the statistically significant correlation between X & Y and the fact that we know someone’s X score. The average unsquared error when we predict that everyone will score at the mean of Y equals sY, the ordinary standard deviation of Y. How much better than that can we do?

Estimating the standard error of the estimate the (very) long way.
Calculate the correlation of X and Y from the raw data (which includes calculating s for Y). If the correlation is significant, you can use the regression equation to make individualized predictions of scores on the Y variable. Then find the difference between each predicted and actual score in your sample, square the differences, add them up, divide by dfREG (nP-2), and take a square root. The average unsquared error of prediction when you do that is called the estimated standard error of the estimate. Its symbol is sEST.

Example for Prediction Error
A study was performed to investigate whether the quality of an image affects reading time. The experimental hypothesis was that reduced quality would slow down reading time. Quality was measured on a scale of 1 to 10. Reading time was in seconds.

Quality vs Reading Time data: Compute the correlation
(scale 1-10) 4.30 4.55 5.55 5.65 6.30 6.45 Reading time (seconds) 8.1 8.5 7.8 7.3 7.5 6.0 Is there a relationship? Check for linearity. Compute r.

Calculate t scores for X
tX = (X - X) / sX -1.48 -1.19 -0.07 0.05 0.78 0.95 X 4.30 4.55 5.55 5.65 6.30 6.45 X - X -1.31 -1.06 -0.06 0.04 0.69 0.84 (X - X)2 1.71 1.12 0.00 0.48 0.71 X=39.25 n= 7 X=5.61 MSW = 4.73/(7-1) = 0.79 sX = 0.89 SSW = 4.73

Calculate t scores for Y
tY = (Y - Y) / sY 0.76 1.26 0.38 -025 0.00 -0.25 -1.89 Y 8.1 8.5 7.8 7.3 7.5 6.0 Y - Y 0.60 1.00 0.30 -0.20 0.00 -1.50 (Y - Y)2 0.36 1.00 0.09 0.04 0.00 2.25 SSW = 3.78 Y=52.5 n= 7 Y=7.50 MSW = 3.78/(7-1) = 0.63 sY = 0.794

Plot t scores tX -1.48 -1.19 -0.07 0.05 0.78 0.95 tY 0.76 1.28 0.39 -0.25 0.00 -1.89

t score plot with best fitting line: linear? YES!

Calculate r tX tY tY -tX (tY -tX)2 -1.48 -1.19 -0.07 0.05 0.78 0.95
0.76 1.28 0.39 -0.25 0.00 -1.88 tY -tX -2.24 -2.47 -0.46 0.30 0.78 1.20 2.83 (tY -tX)2 5.02 6.10 0.21 0.09 0.61 1.44 8.01  (tX - tY)2 = 21.48  (tX - tY)2 / (nP - 1) = 3.580 r = 1 - (1/2 * 3.580) = =

Check whether r is significant
df = nP-2 = 5 Look in r table:With 5 dfREG, the CI.95 goes from to +.753  is .05 r(5)= -.790, p <.05 r is significant!

We can find the Y' for every raw X score by using the regression equation. Then we can compute how much error remains, subtracting the predicted score (Y' ) from the actual Y score obtained by each person. Y' 8.42 8.23 7.54 7.47 7.01 6.91 X 4.30 4.55 5.55 5.65 6.30 6.45

Can we show mathematically that regression estimates are better than saying everyone will score precisely at the mean of Y? Y' 8.42 8.23 7.54 7.47 7.01 6.91 Y 8.1 8.5 7.8 7.3 7.5 6.0 Y 7.5 To calculate the standard deviation we take deviations of Y from the mean of Y, square them, add them up, divide by degrees of freedom, and then take the square root. To calculate the estimated standard error of the estimate, sEST, we will take the deviations of each raw Y score from its regression equation estimate, square them, add them up, divide by degrees of freedom, and take the square root. We expect of course that there will be less error if we use regression.

Estimated standard error of the estimate
Y - Y' -0.32 0.27 0.26 -0.17 0.49 0.39 -0.91 (Y - Y')2 0.10 0.07 0.03 0.24 0.15 0.83 Y' 8.42 8.23 7.54 7.47 7.01 6.91 Y 8.1 8.5 7.8 7.3 7.5 6.0 SSRES = 1.49 MSRES = 1.49/(7-2) = 0.298 SEST = 0.546

How much better? MSY = 0. 630 MSRES = 0.298
53% less squared error when we use the regression equation instead of the mean to predict Y scores.

How much better is the estimated standard error of the estimate than the estimated standard deviation? SY = 0.794 SEST = 0.546 31% less error of prediction(using unsquared units) when we use the regression equation instead of the mean to predict.

Mathematical magic There is usually an alternative formula for calculating statistics that is easier to perform. We went through a lot of extra steps to calculate SEST = It is not necessary to calculate all of the estimated Y scores, find the difference between each actual Y score and Y', then square, sum, and divide by dfREG.

Another way to phrase it: How much error did we get rid of?
Treat it as a weight loss problem. If Jack is 30 pounds overweight and he loses 40% of it, how much is he still overweight. He lost .400 x 30 pounds = 12 pounds. He has 30 – 12 = 18 pounds left to lose.

How did we solve that problem?
First we found how much weight Jack had gotten rid of. That equaled the percent he lost (expressed as a proportion) times the amount overweight he started with He was 30 pounds overweight and lost 40% of it. 30 * .400 = He lost 12 pounds.

Then we found how much he was still overweight.
He started off 30 pounds overweight. He lost pounds. So he had 30-12=18 pounds of overweight left. To find what is left, subtract what you got rid of from the amount you started with.

So to compute how much of something is left after some is lost, you need to know how much there was to start with and what percentage was gotten rid of. Percentage gotten rid of times original quantity = amount gotten rid of. Original quantity minus amount gotten rid of = what’s left.

Calculate how much is left.
C. Munster was 100 pounds overweight. He lost 35% of it. How much is he still overweight? Mr. Hardy was 25.8 pounds overweight. He lost 60% of it. How much is he still overweight?

Cookie Muster is still 65 pounds overweight.
Mr. Hardy has earned some laurels. He only has pounds left to lose.

SSY= error to start r2=percent of error lost
SSY is the total amount of error we start with when prediction scores on Y. It is the amount of error when everyone is predicted to score at the mean. It is analogous to the total amount overweight. The proportion of error you get rid of using the regression equation as your predictor equals Pearson’s correlation coefficient squared (r2)! It is analogous to the proportion of weight lost.

To get the total error left find how much you got rid of, then subtract from what you started with
Amount you got rid of: SSY * r2 Amount you started with SSY Amount left: SSRES = SSY – (SSY * r2 )

Now you know how much error is left when you use the regression equation on your sample (SSRES). To estimate the average amount of squared error you will have left if you use the regression equation to make predictions for the whole population divide SSRES by dfREG to obtain MSRES. To compute the estimated standard error of the estimate, sEST, just take the square root of MSRES.

Computing sEST the easier way
Computing sEST the easier way! In the problem for which we computed sEST the long way, we already knew that SSY = 3.78 and r = Thus, r2 = (-0.790)2= Here is the computation: SSRES = SSY - (SSY * r2) = (3.78 * 0.624) = 1.42 MSRES = 1.42/(7-2) = 0.284 SEST = 0.533

Compare the two methods: Estimated standard error of the estimate
Y - Y' -0.32 0.27 0.26 -0.17 0.49 0.39 -0.91 (Y - Y')2 0.10 0.07 0.03 0.24 0.15 0.83 Y' 8.42 8.23 7.54 7.47 7.01 6.91 Y 8.1 8.5 7.8 7.3 7.5 6.0 SSRES = 1.49 MSRES = 1.49/(7-2) = 0.298 SEST = 0.546

The easier method is more accurate.
Why? Because there is less rounding.

Summary on estimating the standard error of the estimate.

In generic terms, as usual you divide the sum of squares (SSRES) by degrees of freedom (dfREG= nP-2) to estimate the average amount of squared error that will be left. Then you take a square root to get the average amount of unsquared error you will make when you use your best prediction to predict Y scores, which, in this case, is tY'

Only change is that you used the regression equation to get SSRES rather than using the mean to compute SSY. So you divide SSRES by dfREG=nP-2 to get MSRES, to estimate the average amount of squared error you will have left when you use the regression equation. Then take the square root of MSRES to get sEST. sEST is your best estimate of the average unsquared error of prediction when you properly use the regression equation to predict Y scores. Remember to properly use the regression equation, r must be significant and X within the range of X scores observed in your random sample.

Stating the (hopefully) obvious:
The estimated standard deviation (s) was the estimated average unsquared distance of scores in the population from mu. If we are looking at Y scores it is the average unsquared difference of Y scores from the mean of Y. When using the regression equation we are predicting Y scores other than the mean. The estimated standard error of the estimate (sEST) is the estimated average unsquared distance of Y scores in the population from the regression equation based predicted Y scores. Both s and sEST reflect error of prediction. Using the regression equation individualizes prediction. If r is significant, and prediction restricted to values of X within the range of X scores seen in the random sample, using the regression equation leads to less error.

How much better is using the regression equation rather than the mean as your predictor?
SEST = 0.533 In this case, there was 33% less unsquared error when the regression equation was used instead of the mean to predict scores on the Y variable. Note: the difference between 33% and 31% when we calculated using the long way is mostly due to rounding error in the long calculation. 33% is more accurate

Here are the formulae again:
Residual sum of squared error if the regression equation is used: SSRES = SSY - (SSY * r2) Estimated average amount of squared error left: MSRES = SSRES/dfREG = SSRES/(nP-2) Estimated average amount of unsquared error left: sEST = square root of MSRES

Chapter 8 – Regression 2 Basic review, estimating the standard error of the estimate and short cut problems and solutions.

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Chapter 8 – Regression 2 Basic review, estimating the standard error of the estimate and short cut problems and solutions.

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