inference for regression parameters, correlation & analysis...
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Inference for Regression Parameters, Correlation &Analysis of Variance
9.4, 9.5 & 11
Cathy Poliak, [email protected] Fleming 11c
Department of MathematicsUniversity of Houston
Lecture 13 - 3339
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 1 / 41
Outline
1 Inference for Regression Parameters
2 F-test
3 Estimating Correlation
4 Comparing More Than Two Means
5 ANOVA
6 Pairwise Tests
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 2 / 41
Least-Squares regression
The least-squares regression line (LSRL) of Y on X is the linethat makes the sum of the squares of the vertical distances of thedata points from the line as small as possible.The linear regression model is: Y = β0 + β1x + ε
I Y is dependent variable (response).I x is the independent variable (explanatory).I β0 is the population intercept of the line.I β1 is the population slope of the line.I ε is the error term which is assumed to have mean value 0. This is
a random variable that incorporates all variation in the dependentvariable due to factors other than x .
I The variability: σ of the response y about this line. More precisely,σ is the standard deviation of the deviations of the errors, εi in theregression model.
We will gather information from a sample so we will have the leastsquares estimates model: Y = β0 + β1x .
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 3 / 41
Principle of Least Squares
The vertical deviation of the point (xi , yi) from the line y = b0 + b1x is
hieght of point − height of line = yi − (b0 + b1xi)
The sum of the square vertical deviations from the points(x1, y1), (x2, y2), . . . , (xn, yn) to the line is then
f (b0,b1) =n∑
i=1
[yi − (b0 + b1xi)]2
The point estimates of β0 and β1, denoted by β0 and β1 and called theleast squares estimates, are those values that minimize f (b0,b1).
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 4 / 41
Estimating the Regression Parameters
In the simple linear regression setting, we use the slope b1 andintercept b0 of the least-squares regression line to estimate theslope β1 and intercept β0 of the population regression line.The standard deviation, σ, in the model is estimated by theregression standard error
s =
√∑(yi − yi)2
n − 2=
√∑all residuals2
n − 2
Recall that yi is the observed value from the data set and yi is thepredicted value from the equation.In R s is the called the Residual Standard Error in the lastparagraph of the summary.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 5 / 41
The Least - Squares Estimates
Recall ei = observed Y - predicted Y is the i th residual. Think of itas an estimate of the unobservable true random error εi .
The method of least squares selects estimators β0 and β1 thatminimizes the residual sum of squares:
SS(resid) = SSE =n∑
i=1
e2i =
n∑i=1
(Yi − Yi
)Where the estimate of the slope coefficient β1 is:
β1 =
∑(xi − x)(yi − y)∑
(xi − x)2 =Sxy
Sxx
The estimate for the intercept β0 is:
β0 = y − β1x
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 6 / 41
t Test for Significance of β1
HypothesisH0 : β1 = β10 versus Ha : β1 6= β10
Usually β10 = 0Test statistic
t =observed− hypothesized
standard deviation of observedobserved = b1
hypothesized = 0
standard error = SEb1 =s√∑
(xi − x)2
With degrees of freedom df = n − 2.P-value: based on a t distribution with n − 2 degrees of freedom.Decision: Reject H0 if p-value ≤ α.Conclusion: If H0 is rejected we conclude that the explanatoryvariable x can be used to predict the response variable y .
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 7 / 41
Conditions for regression inference
The sample is an SRS from the population.
There is a linear relationship in the population.
The standard deviation of the responses about the population lineis the same for all values of the explanatory variable.
The response varies Normally about the population regressionline.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 8 / 41
Example: Real Estate
Can the size of the house help us predict the list price?Explanatory Variable: Price (List Price of a home per $1,000)
Response Variable: Size (Size of the house in square feet)
Data: 55 houses for sale in a certain neighborhood https://www.math.uh.edu/~cathy/Math3339/data/homes.xlsx.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 9 / 41
R output
> attach(homes)> plot(Size,Price)> homes.lm=lm(Price~Size)> summary(homes.lm)
Call:lm(formula = Price ~ Size)
Residuals:Min 1Q Median 3Q Max
-99.157 -31.898 4.873 36.389 105.515
Coefficients:Estimate Std. Error t value Pr(>|t|)
(Intercept) 92.529120 25.701237 3.60 7e-04 ***Size 0.099106 0.008001 12.39 <2e-16 ***---Signif. codes:0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 46.69 on 53 degrees of freedomMultiple R-squared: 0.7432,Adjusted R-squared: 0.7384F-statistic: 153.4 on 1 and 53 DF, p-value: < 2.2e-16
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 10 / 41
Height
Because elderly people may have difficulty standing to have theirheights measured, a study looked at predicting overall height fromheight to the knee. Here are data (in centimeters, cm) for five elderlymen:
Knee Height (cm) 57.7 47.4 43.5 44.8 55.2Overall Height(cm) 192.1 153.3 146.4 162.7 169.1
1. We want to test if the relationship of the measurement of theheight is more than double that of the measurement from the floorto the knee. Test H0 : β1 = 2 versus Ha : β1 > 2.
2. Give a conclusion for the relationship between using knee lengthto predict overall height.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 11 / 41
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 12 / 41
Confidence Intervals for β1
If we want to know a range of possible values for the slope we can usea confidence interval.
Remember confidence intervals are
estimate± t∗ × standard error of the estimate
Confidence interval for β1 is
b1 ± tα/2,n−2 × SEb1
Where t∗ is from table D with degrees of freedom n − 2 where n =number of observations.In R we can get this by confint(name.lm,level = 0.9) For a 90%confidence interval.> confint(shelf.lm,level = 0.9)
5 % 95 %(Intercept) 105.519121 184.48088space 4.516724 10.28328
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 13 / 41
Determine a 90% Confidence Interval for CoffeeExample
From the output in R:
Coefficients:Estimate Std. Error t value Pr(>|t|)
(Intercept) 145.000 21.783 6.657 5.66e-05 ***space 7.400 1.591 4.652 0.000906 ***
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 14 / 41
Definitions of Regression Output
1. The error sum of squares, denoted by SSE is
SSE =∑
(yi − yi)2
2. A quantitative measure of the total amount of variation in observedvalues is given by the total sum of squares, denoted by SST .
SST =∑
(yi − y)2
3. The regression sum of squares, denoted SSR is the amount oftotal variation that is explained by the model
SSR =∑
(yi − y)2
4. The coefficient of determination, r2 is given by
r2 =SSRSST
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 15 / 41
Finding these values using R
> anova(shelf.lm)Analysis of Variance Table
Response: soldDf Sum Sq Mean Sq F value Pr(>F)
space 1 20535 20535 21.639 0.0009057 ***Residuals 10 9490 949---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 16 / 41
F-distribution
The F distribution with ν1 degrees of freedom in the numeratorand ν2 degrees of freedom in the denominator is the distribution ofa random variable
F =U/ν1
V/ν2,
where U ∼ χ2(df = ν1) and V ∼ χ2(df = ν2) are independent.That F has this distribution is indicated by F ∼ F (ν1, ν2).Notice U = SSR
σ2 ∼ χ2(df = 1) and V = SSEσ2 ∼ χ2(df = n − 2) are
independent.Let MSE = SSE/(n − 2) and MSR = SSR/1. Then
F =MSRMSE
=SSR/1
SSE/(n − 2)∼ F (1,n − 2)
Then we can use the F-distribution to test the hypothesisH0 : β1 = 0 versus Ha : β1 6= 0.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 17 / 41
F-test for Shelf Space
Analysis of Variance Table
Response: soldDf Sum Sq Mean Sq F value Pr(>F)
space 1 20535 20535 21.639 0.0009057 ***Residuals 10 9490 949---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Note: F = t2 and the p-value is the same.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 18 / 41
Estimating the Correlation Coefficient
Notice that β1 = ρσyσx
Where ρ is the population correlation, and σis the population standard deviation.
This means that the regression null hypothesis H0 : β1 = 0 isequivalent to H0 : ρ = 0.
Recall that we can use the sample correlation, to estimate theslope,
β1 = Rsy
sxThus this can be rewritten as:
R = β1sx
sy
Recall the student t statistic for testing H0 : β1 = 0,
t =β1s√∑(xi − x)2
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 19 / 41
Test Statistic for H0 : ρ = 0
From the test statistic of H0 : β1 = 0 we get the test statistic for testingH0 : ρ = 0.
T =R√
n − 2√1− R2
Therefore, if X and Y have a bivariate normal distribution, a test ofH0 : ρ = 0 against Ha : ρ 6= 0 is to reject H0 when |T | > tα/2(n − 2)or when the p-value based on the the student-t distribution TDist(df = n − 2) is smaller than the chosen significance level.
Note: We have three equivalent tests in the case of a bivariatenormal distribution:
1. the T test for H0 : β1 = 02. the T test for H0 : ρ = 03. the F test for significance of regression
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 20 / 41
Test H0 : ρ = 0 for Shelf Space Example
Given: R = 0.827 and n = 12 find the test statistic T and the p-value.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 21 / 41
R output> cor.test(sold,space)
Pearson’s product-moment correlation
data: sold and spacet = 4.6517, df = 10, p-value = 0.0009057alternative hypothesis: true correlation is not equal to 095 percent confidence interval:0.4817419 0.9500109sample estimates:cor0.8270006
> summary(shelf.lm)
Call:lm(formula = sold ~ space)
Residuals:Min 1Q Median 3Q Max
-42.00 -26.75 5.50 21.75 41.00
Coefficients:Estimate Std. Error t value Pr(>|t|)
(Intercept) 145.000 21.783 6.657 5.66e-05 ***space 7.400 1.591 4.652 0.000906 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 30.81 on 10 degrees of freedomMultiple R-squared: 0.6839,Adjusted R-squared: 0.6523F-statistic: 21.64 on 1 and 10 DF, p-value: 0.0009057
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 22 / 41
Weight Loss
From: http://sphweb.bumc.bu.edu/otlt/MPH-Modules/BS/BS704_HypothesisTesting-ANOVA/BS704_HypothesisTesting-Anova_print.html
Is there a difference in the mean weight loss among differentprograms?A clinical trial is run to compare weight loss programs andparticipants are randomly assigned to one of the comparisonprograms and are counseled on the details of the assignedprogram.Participants follow the assigned program for 8 weeks.Three popular weight loss programs are considered.
I Low calorie diet.I Low fat dietI Low carbohydrate dietI Control group
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 23 / 41
Results
Response variable = weight loss = weight at the end of 8 weeks -weight at beginning of the studyThe observed weight losses of twenty people in this study are asfollows:
Low Calorie Low Fat Low Carbohydrate Control8 2 3 29 4 5 26 3 4 -17 5 2 03 1 3 3
Is there a statistically significant difference in the mean weight lossamong the four diets?
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 24 / 41
Box Plots of Weight Loss
calorie carb control fat
02
46
8
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 25 / 41
Hypotheses
We want to know if there is a "statistically significant difference" inthe mean weight loss among the four diets.
Null hypothesis: mean weight loss is the same among the fourdiets
H0 : µcalorie = µcarb = µcontrol = µfat
Alternative hypothesis is that at least one of the mean weight lossamong the four diets is different.
Rejecting H0 is evidence that the mean of at least one group isdifferent from the other means.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 26 / 41
Hypotheses
We want to know if there is a "statistically significant difference" inthe mean weight loss among the four diets.
Null hypothesis: mean weight loss is the same among the fourdiets
H0 : µcalorie = µcarb = µcontrol = µfat
Alternative hypothesis is that at least one of the mean weight lossamong the four diets is different.
Rejecting H0 is evidence that the mean of at least one group isdifferent from the other means.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 26 / 41
Hypotheses
We want to know if there is a "statistically significant difference" inthe mean weight loss among the four diets.
Null hypothesis: mean weight loss is the same among the fourdiets
H0 : µcalorie = µcarb = µcontrol = µfat
Alternative hypothesis is that at least one of the mean weight lossamong the four diets is different.
Rejecting H0 is evidence that the mean of at least one group isdifferent from the other means.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 26 / 41
Hypotheses
We want to know if there is a "statistically significant difference" inthe mean weight loss among the four diets.
Null hypothesis: mean weight loss is the same among the fourdiets
H0 : µcalorie = µcarb = µcontrol = µfat
Alternative hypothesis is that at least one of the mean weight lossamong the four diets is different.
Rejecting H0 is evidence that the mean of at least one group isdifferent from the other means.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 26 / 41
Assumptions
The assumptions of analysis of variance are the same as those of thetwo sample t-test, but they must hold for all k groups.
The measurements in every group is a SRS.
We have a Normal distribution for each of the k populations.
The variance is the same in all k populations.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 27 / 41
Analysis: ANOVA
ANalysis Of VArianceWe can estimate how much variation among group means oughtto be present from sampling error alone if the null hypothesis istrue.ANOVA lets us determine whether there is more variance amongthe sample means than we would expect by chance alone.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 28 / 41
The Formulas
Let Xi. =∑ni
j=1 Xij
nidenote the average of the observations in the i th
group.
Let N =∑M
i=1 ni , be the total number of observations in all the Mgroups.
Let X.. =∑M
i=1 ni Xi.N be the average of all the observations (the
grand average)
The treatment sum of squares (between groups) isSS(betw) =
∑Mi=1 ni(Xi. − X..)2.
The error sum of squares (residual) isSSE =
∑Mi=i∑ni
n=1(Xij − Xi.)2 =
∑Mi=1(n − 1)S2
i
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 29 / 41
Diets Example
Low Calorie Low Fat Low Carbohydrate Control8 2 3 29 4 5 26 3 4 -17 5 2 03 1 3 3
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 30 / 41
The F Test
The mean square for treatments is MSTr = SSTrM−1 .
The mean square for error is MSE = SSEN−M .
The test statistic is F = MSTrMSE .
This test statistic has an F distribution with parameters "numeratordegrees of freedom" = M - 1 and "denominator degrees offreedom" = N - M. Where N is the total number of observationsand M is the number of groups.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 31 / 41
The ANOVA Table
Source of degrees of Sum of Mean FVariation freedom Squares SquareTreatments M - 1 SSTr MSTr MSTr
MSEError N - M SSE MSETotal N - 1 SST
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 32 / 41
ANOVA for Diets
Source of degrees of Sum of Mean FVariation freedom Squares SquareDiets
Error
Total
p-value = 1 - pf(f,M - 1, N - M)
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 33 / 41
R Code
> diet.lm=lm(Loss~Diet,data=diet)> anova(diet.lm)Analysis of Variance Table
Response: LossDf Sum Sq Mean Sq F value Pr(>F)
Diet 3 75.75 25.25 8.5593 0.001278 **Residuals 16 47.20 2.95---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 34 / 41
Tukey’s Method (The T Method)
The T-method is used to determine which pair (or pairs) of meansdiffers significanctly.
1. Select α, determine Qα,k ,N−k in R it is qtukey(1− α, k ,N − k )where, k = number of groups and N = total sample size.
2. Calculate w = Qα,k ,N−k√
MSE/j . Where j = the number ofelements in each group.
3. List the sample means in increasing order and underline thosepairs that differ by less than w .
4. Any pair of sample means not underscored by the same linecorresponds to a pair of population or treatment means that arejudged significantly different.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 35 / 41
Tukey’s Method for Diet Example
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 36 / 41
Multiple Comparisons
If our p-value is small for the ANOVA F test, this implies that atleast one of the means is different from the other.
Which one(s) are different?
We could do a t-test for each pair of means.
Problem: when we do multiple t-tests our P(Type 1 error) becomesgreater than α.
Solution: There are methods of adjustments to reduce thesignificance level of the pairwise test enough so that theprobability of one or more type I errors in the whole set ofcomparisons is less than α.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 37 / 41
Multiple Comparisons
If our p-value is small for the ANOVA F test, this implies that atleast one of the means is different from the other.
Which one(s) are different?
We could do a t-test for each pair of means.
Problem: when we do multiple t-tests our P(Type 1 error) becomesgreater than α.
Solution: There are methods of adjustments to reduce thesignificance level of the pairwise test enough so that theprobability of one or more type I errors in the whole set ofcomparisons is less than α.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 37 / 41
Multiple Comparisons
If our p-value is small for the ANOVA F test, this implies that atleast one of the means is different from the other.
Which one(s) are different?
We could do a t-test for each pair of means.
Problem: when we do multiple t-tests our P(Type 1 error) becomesgreater than α.
Solution: There are methods of adjustments to reduce thesignificance level of the pairwise test enough so that theprobability of one or more type I errors in the whole set ofcomparisons is less than α.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 37 / 41
Multiple Comparisons
If our p-value is small for the ANOVA F test, this implies that atleast one of the means is different from the other.
Which one(s) are different?
We could do a t-test for each pair of means.
Problem: when we do multiple t-tests our P(Type 1 error) becomesgreater than α.
Solution: There are methods of adjustments to reduce thesignificance level of the pairwise test enough so that theprobability of one or more type I errors in the whole set ofcomparisons is less than α.
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 37 / 41
The Bonferroni Method
The Bonferroni Method of adjustments reduces the significancelevel for the pairwise test to α/k , where k is the number ofcomparisons.R Code:> attach(diet)> pairwise.t.test(Loss,Diet,"bonferroni")
Pairwise comparisons using t tests with pooled SD
data: Loss and Diet
calorie carb controlcarb 0.05695 - -control 0.00083 0.35914 -fat 0.02632 1.00000 0.70193
P value adjustment method: bonferroni
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 38 / 41
The Bonferroni Method
The Bonferroni Method of adjustments reduces the significancelevel for the pairwise test to α/k , where k is the number ofcomparisons.R Code:> attach(diet)> pairwise.t.test(Loss,Diet,"bonferroni")
Pairwise comparisons using t tests with pooled SD
data: Loss and Diet
calorie carb controlcarb 0.05695 - -control 0.00083 0.35914 -fat 0.02632 1.00000 0.70193
P value adjustment method: bonferroni
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 38 / 41
Example MPG
Is there a difference in the average miles per gallon for different makesof automobiles? The following table shows the mean mpg of threedifferent makes of automobiles. The data is on the data sets list calledmpg.
Make n X SHonda 5 29.9 1.468Toyota 6 33.04 2.1173Nissan 4 29.3 1.3115
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 39 / 41
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 40 / 41
Cathy Poliak, Ph.D. [email protected] Office Fleming 11c (Department of Mathematics University of Houston )Sections 9.4, 9.5 & Chapter 11 Lecture 13 - 3339 41 / 41