modeling Dongmei Li Department of Public Health Sciences Office of - - PowerPoint PPT Presentation

Dongmei Li Department of Public Health Sciences Office of Public Health Studies University of Hawai’i at Mānoa

Quantitative response variable modeling

Outline

2

 T-test  ANOVA  Correlation and simple linear regression

One-sample t test

3

One-Sample t Test: Example

Statement of the problem:

 Do Sudden Infant Death Syndrome (SIDS) babies have

lower than average birth weights?

 We know from prior research that the mean birth

weight of the non-SIDS babies in the population is 3300 grams.

 We study n = 10 SIDS babies, determine their birth

weights, and calculate x-bar = 2975.5 and s = 737.3.

 Do these data provide significant evidence that SIDS

babies have different birth weights than the rest of the population?

SIDS baby weights: 2229 2997 2314 3831 1788 2745 4151 2975 3463 3262

4

One-Sample t Test

• A. Hypotheses. H0: µ = µ0 vs. Ha: µ ≠ µ0 (two-sided) [ Ha: µ < µ0

(left-sided) or Ha: µ > µ0 (right-sided)]

• B. Test statistic.
• C. P-value. Convert tstat to P-value [software]. Small P  strong

evidence against H0

• D. Significance level α (compare P-value with α to determine

whether you will reject the null hypothesis or not).

1 with

stat

    n df n s x t 

5

A.

H0: µ = 3300 versus Ha: µ ≠ 3300 (two-sided)

• B. Test statistic
• C. P = 0.1974

Weak evidence against H0

• D. Data are not significant at α = .10. Fail to reject the null

hypothesis. 9 1 10 1 39 . 1 10 3 . 737 3300 5 . 2975

stat

           n df SE x t

x



One-Sample t Test: Example

6

Confidence Interval for µ

n s t x

n

   

 

2

1 , 1

for CI % 100 ) 1 (



 

• Typical point “estimate ± margin of error” formula
• tn-1,1-α/2 is from t table
• Alternative formula:

n s SE SE t x

x x n

  

 

where

2

1 , 1

 7

Confidence Interval: Example 1

grams 3502.9) to (2448.1 = 527.4 ± 5 . 2975 10 3 . 737 262 . 2 5 . 2975 for CI 95% 10 3 . 737 5 . 2975

2 05 .

1 , 1 10

         

 

n s t x n s x 

Let us calculate a 95% confidence interval for μ for the birth weight of SIDS babies.

8

How to do it in Excel?

Open the Presentation3_SIDS_BW .xlsx data set Use the AVRAGE functions in Excel to calculate mean Use the STDEV functions in Excel to calculate standard deviation Use the TDIST function to calculate p-value Use the TINV function to get critical value

9

How to do it in JMP?

 Open the Presentation3_SIDS_BW

.jmp data set

 Analyze---Distribution---Put SIDS_BW into

Y, Column box--

• click OK---click on the red arrow next to SIDS_BW---click
• n test mean---input 3300 for hypothesized mean---click OK

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Click confidence interval---95 to get 95% confidence interval

Paired-sample t test

11

Paired Samples

 Paired samples: Each point in one sample is

matched to a unique point in the other sample

 Pairs be achieved via sequential samples within

individuals (e.g., pre-test/post-test), cross-over trials, and match procedures

 Also called “matched-pairs” and “dependent

samples”

12

Example: Paired Samples

 A study addresses whether oat bran reduce LDL cholesterol

with a cross-over design.

 Subjects “cross-over” from a cornflake diet to an oat bran

diet.

 Half subjects start on CORNFLK, half on OATBRAN  Two weeks on diet 1  Measures LDL cholesterol  Washout period  Switch diet  Two weeks on diet 2  Measures LDL cholesterol

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Example, Data

Subject CORNFLK OATBRAN

• --- ------- -------

1 4.61 3.84 2 6.42 5.57 3 5.40 5.85 4 4.54 4.80 5 3.98 3.68 6 3.82 2.96 7 5.01 4.41 8 4.34 3.72 9 3.80 3.49 10 4.56 3.84 11 5.35 5.26 12 3.89 3.73

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Calculate Difference Variable “DELTA”

 Step 1 is to create difference variable “DELTA”  Let DELTA = CORNFLK - OATBRAN  Order of subtraction does not materially effect results (but does change sign

• f differences)

 Here are the first three observations:

Positive values represent lower LDL

• n oatbran

ID CORNFLK OATBRAN DELTA

• --- ------- ------- -----

1 4.61 3.84 0.77 2 6.42 5.57 0.85 3 5.40 5.85 -0.45 ↓ ↓ ↓ ↓

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Explore DELTA Values

Stemplot |-0|42 |+0|0133 |+0|667788 ×1

Here are all the twelve paired differences (DELTAs):

0.77, 0.85, −0.45, −0.26, 0.30, 0.86, 0.60, 0.62, 0.31, 0.72, 0.09, 0.16

EDA shows a slight negative skew, a median of about 0.45, with results varying from −0.4 to 0.8.

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Descriptive stats for DELTA

Data (DELTAs): 0.77, 0.85, −0.45, −0.26, 0.30, 0.86, 0.60, 0.62, 0.31, 0.72, 0.09, 0.16 0.4335 0.3808 12   

d d

s x n The subscript d will be used to denote statistics for difference variable DELTA

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95% Confidence Interval for µd

n s t x

d n d d

   

 

2

1 , 1

for CI % 100 ) 1 (



 

A t procedure directed toward the DELTA variable calculates the confidence interval for the mean difference.

) 656 . to 105 . ( 2754 . 0.3808 12 4335 . 201 . 2 0.3808 for CI % 95 Table) t (from 201 . 2 use confidence 95% For

975 ,. 11 1 1 12

2 05

       

  d ,

t t

.



“Oat bran” data:

18

Paired t Test

• Similar to one-sample t test



μ0 is usually set to 0, representing “no mean difference”, i.e., H0: μ = 0

• Test statistic:

n df n s x t

d d

1

stat

    

19

Paired t Test: Example

“Oat bran” data

• A. Hypotheses. H0: µd = 0 vs. Ha: µd  0
• B. Test statistic.
• C. P-value. P = 0.011 (via computer). The evidence against

H0 is statistically significant.

• D. Significance level. The evidence against H0 is significant at

α = .05 but is not significant at α = .01.

11 1 12 1 043 . 3 12 / 4335 . 38083 .

stat

          n df n s x t

d



20

How to do it in Excel?

 Open the Presentation3_cornflk.xlsx data set  Paired sample t test in Excel

21

How to do it in Excel?

22

Results and interpretation

 P-value = 0.0112: have evidence to show Oatbran can

significantly lower LDL cholesterol level compared to Cornflk.

23

How to do it in JMP?

24

How to do it in JMP?

25

P-value = 0.0112: have evidence to show Oatbran can significantly lower LDL cholesterol level compared to Cornflk. Notice the 95% confidence interval for their difference does not include 0.

Conditions for Inference

t procedures require these conditions:

 SRS (individual observations or DELTAs)  Valid information (no information bias)  Normal population or large sample (central limit theorem)

26

The Normality Condition

The Normality condition applies to the sampling distribution of the mean, not the population. Therefore, it is OK to use t procedures when:

 The population is Normal  Population is not Normal but is symmetrical and n is at

least 5 to 10

 The population is skewed and the n is at least 30 to 100

(depending on the extent of the skew)

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Can a t procedures be used?

 Dataset A is skewed and

small: avoid t procedures

 Dataset B has a mild skew

and is moderate in size: use t procedures

 Data set C is highly skewed

and is small: avoid t procedure

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Two-independent sample t test

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Example: Cholesterol and Type A & B

Personality

Group 1 (Type A personality): 233, 291, 312, 250, 246, 197, 268, 224, 239, 239, 254, 276, 234, 181, 248, 252, 202, 218, 212, 325 Group 2 (Type B personality): 344, 185, 263, 246, 224, 212, 188, 250, 148, 169, 226, 175, 242, 252, 153, 183, 137, 202, 194, 213

Do fasting cholesterol levels differ in Type A and Type B personality men? Data (mg/dl) are a subset from the Western Collaborative Group Study*

30

Exploratory & Descriptive Methods

 Start with EDA  Compare group shapes, locations

and spreads

 Examples of applicable techniques

Side-by-side stemplots (right) Side-by-side boxplots (next slide)

Group 1 | | Group 2

• |1t|3

|1f|45 |1s|67 98|1.|8889 110|2*|011 33332|2t|22 55544|2f|4455 76|2s|6 9|2.| 21|3*| |3t| |3f|4 (×100)

31

Side-by-Side Boxplots

20 20 N =

GROUP

2 1 400 300 200 100

21 20

Interpretation :

• Location:

group 1 > group 2

• Spreads:

group 1 < group 2

• Shapes: Both fairly

symmetrical, outside values in each; no major departures from Normality

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Summary Statistics

Group n mean std dev 1 20 245.05 36.64 2 20 210.30 48.34

33

Inference About Mean Difference (Notation)

Parameters (population) Group 1 N1 µ1 σ1 Group 2 N2 µ2 σ2 Statistics (sample) Group 1 n1 s1 Group 2 n2 s2

1

x

2

x

2 1 2 1

• f

estimator point the is     x x

34

Hypothesis Test

A.

Hypotheses. H0: μ1 = μ2 against Ha: μ1 ≠ μ2 (two-sided) [Ha: μ1 > μ2 (right-sided) Ha: μ1 < μ2 (left-sided) ]

B.

Test statistic.

• C. P-value. Convert the tstat to P-value with t table or software.

Interpret.

• D. Significance level (optional). Compare P to prior

specified α level.

Welch 2 2 2 1 2 1 2 1 stat

2 1 2 1

where ) ( df n s n s SE SE x x t

x x x x

   

 

35

Hypothesis Test – Example

• A. Hypotheses. H0: μ1 = μ2 vs. Ha: μ1 ≠ μ2
• B. Test stat. In prior analyses we calculated sample mean

difference = 34.75 mg/dL, SE = 13.563 and dfconserv = 19.

• C. P-value. P = 0.019 → good evidence against H0 (“significant

difference”).

• D. Significance level (optional). The evidence against H0 is

significant at α = 0.05 but not at α = 0.01. df SE x x t

x x

19 with 2.56 13.563 34.75 ) (

2 1

2 1 stat

   



36

Equal Variance t Procedure

 Also called pooled variance t procedure  Not as robust as prior method, but…  Historically important  Calculated by software programs  Leads to advanced ANOVA techniques

37

We start by calculating this pooled estimate of variance

1 and group in variance the is where ) )( ( ) )( (

2 2 1 2 2 2 2 1 1 2

    

i i i pooled

n df i s df df s df s df s

Pooled variance procedure

38

 The pooled variance is used to calculate this standard

error estimate:

 Confidence Interval  Test statistic  All with df = df1 + df2 = (n1−1) + (n2−1) 1 1

2 1 2

2 1

         



n n s SE

pooled x x

) )( ( ) (

2 1 2

1 , 2 1 x x df

SE t x x

 

 



) (

2 1

2 1 stat x x

SE x x t



 

39

Pooled Variance t Confidence Interval

38 ) 1 20 ( ) 1 20 ( 56 . 13 20 1 20 1 1839.623

2 1

             



df SE

x x

62.14) (7.36, 39 . 27 75 . 34 ) 13.56 )( 02 . 2 ( ) 30 . 210 05 . 245 ( ) )( ( ) ( for CI % 95

2 1

975 ,. 38 2 1 2 1

         

 x x

SE t x x  

Group ni si xbari 1 20 36.64 245.05 2 20 48.34 210.30

Data

40

Pooled Variance t Test

38 ) 1 20 ( ) 1 20 ( 56 . 13 20 1 20 1 1839.623

2 1

             



df SE

x x

014 . 38 2.56; 56 . 13 75 . 34 : :

2 1

2 1 stat 2 1 2 1

       



P df SE x x t H H

x x a

   

Data: Group ni si xbari 1 20 36.64 245.05 2 20 48.34 210.30

41

How to do it in Excel?

 Data set: Presentation3_FCL.xlsx  First do Levene’s test to see whether two group has equal

variance

42

How to do it in Excel?

 Levene’s test shows no significant difference in variance

between groups

43

How to do it in Excel?

 Next use t-Test: Two-Sample Assuming Equal Variances for

the test

44

How to do it in Excel?

 Click OK to get results

45

Excel results

 p-value = 0.014  Significant difference

in fasting cholesterol levels between Type A personality subjects and Type B personality subjects.

46

How to do it in JMP?

 Data set: Presentation3_FCL.jmp  Analyze --- Fit

Y by X

47

How to do it in JMP?

 Select

Means/ANOVA/ Pooled t for the equal variance t test

 Select t Test for

unequal variance t test

48

Results from JMP

 p-value <0.05  Significant

difference in fasting cholesterol levels between Type A personality subjects and Type B personality subjects.

49

Conditions for Inference

Conditions required for t procedures: “Validity conditions”

• a. Good information (no information bias)
• b. Good sample (“no selection bias”)
• c. “No confounding”

“Sampling conditions”

• a. Independence
• b. Normal sampling distribution

50

ANOVA

51

52

Illustrative Example: Data

Pets as moderators of a stress response. This chapter follows the analysis of data from a study in which heart rates (bpm) of participants were monitored after being exposed to a psychological stressor. Participants were randomized to one of three groups:

 Group 1 - monitored in presence of pet dog  Group 2 - monitored in the presence of human friend  Group 3 - monitored with neither dog nor human friend present

53

Illustrative Example: Data

54

Descriptive Statistics

 Data are described and explored before moving to inferential

calculations

 Here are summary statistics by group:

55

Side-by-Side Boxplots

56

Analysis of Variance

 One-way ANalysis Of VAriance (ANOVA)

 Categorical explanatory variable  Quantitative response variable  Test group means for a significant difference

 Statistical hypotheses

 H0: μ1 = μ2 = … = μk  Ha: at least one of the μis differ

 Method: compare variability between groups to variability within

groups (F statistic)

57

Analysis of Variance, cont.

• R. A. Fisher

(1890-1962) The F in the F statistic stands for “Fisher”

58

Mean Square Between: Graphically

59

Mean Square Between: Example

60

Mean Square Within: Graphically

61

Mean Square Within: Example

62

The e F F sta stati tisti stic a c and nd AN ANOVA A ta table

 Data are arranged to form an ANOVA table  F statistic is the ratio of the MSB to MSW

08 . 14 793 . 84 843 . 1193    MSW MSB Fstat

Fstat “signal-to- noise” ratio

63

Fstat and P-value

 The Fstat has numerator and denominator degrees of

freedom: df1 and df2 respectively (corresponding to dfB and dfW)

 Convert Fstat to P-value with a computer program  The P-value corresponds to the area in the right tail

beyond

64

Fstat and P-value

P < 0.001

How to do one-way ANOVA in EXCEL?

65

 Data set: Presentation3_pet.xlsx

How to do one-way ANOVA in EXCEL?

66

Analysis results from Excel

 One-way ANOVA shows significant difference in mean FEV

values among the four different smoker groups.

67

How to do one-way ANOVA in JMP?

 Presentation3_pet.jmp file  Analyze---Fit

Y by X

68

How to do one-way ANOVA in JMP?

69

Analysis results from JMP

 Pairwise comparisons

from Tukey’s method shows the signficant difference among all three groups.

70

Correlation and simple linear regression

71

72

Data type for correlation and regression

 Quantitative response variable Y (“dependent variable”)  Quantitative explanatory variable X (“independent variable”)  Historically important public health data set used to

illustrate techniques (Doll, 1955)

 n = 11 countries  Explanatory variable = per capita cigarette consumption in 1930

(CIG1930)

 Response variable = lung cancer mortality per 100,000

(LUNGCA)

73

Data, cont.

74

Scatterplot

Bivariate (xi, yi) points plotted as scatter plot.

75

Inspect scatterplot’s

 Form: Can the relation be described with a straight or some

• ther type of line?

 Direction: Do points tend trend upward or downward?  Strength of association: Do point adhere closely to an

imaginary trend line?

 Outliers (in any): Are there any striking deviations from

the overall pattern?

Judging Correlational Strength

 Correlational strength refers

to the degree to which points adhere to a trend line

 The eye is not a good judge of

strength.

 The top plot appears to show a

weaker correlation than the bottom plot. However, these are plots of the same data sets. (The perception of a difference is an artifact of axes scaling.)

76

Correlation



Correlation coefficient r quantifies linear relationship with a number between −1 and 1.



When all points fall on a line with an upward slope, r = 1. When all data points fall on a line with a downward slope, r = −1



When data points trend upward, r is positive; when data points trend downward, r is negative.



The closer r is to 1 or −1, the stronger the correlation.

77

Examples of correlations

78

Calculating r (Pearson Correlation)

 Formula

Correlation coefficient tracks the degree to which X and Y “go together.”

 Recall that z scores quantify the amount a value lies above or

below its mean in standard deviations units.

 When z scores for X and

Y track in the same direction, their products are positive and r is positive (and vice versa).

79

Calculating r, Example

80

Scatter plot and r in Excel

 Data set: Presentation3_CIGLungCA.xlsx  Scatter plot: select the data in column B and C---Insert ---

Scatter Plot---Add x and y axis labels

 Correlation:  Data analysis---correlation

81

5 10 15 20 25 30 35 40 45 50 200 400 600 800 1000 1200 1400 Lung cancer mortality CIG1930

Lung cancer mortality vs. cigarette consumption

Scatter plot and r in JMP

 Data set: Presentation3_CIGLungCA.jmp  Scatter plot: Graph---Scatter plot matrix

82

Interpretation of r

1.

• Direction. The sign of r indicates the direction of the

association: positive (r > 0), negative (r < 0), or no association (r ≈ 0).

2.

• Strength. The closer r is to 1 or −1, the stronger the

association.

3.

Coefficient of determination. The square of the correlation coefficient (r2) is called the coefficient of

• determination. This statistic quantifies the proportion of

the variance in Y [mathematically] “explained” by X. For the illustrative data, r = 0.737 and r2 = 0.54. Therefore, 54% of the variance in Y is explained by X.

83

Notes, cont.

• 4. Reversible relationship. With correlation, it does not

matter whether variable X or Y is specified as the explanatory variable; calculations come out the same either

• way. [This will not be true for regression.]
• 5. Outliers. Outliers can have

a profound effect on r. This figure has an r of 0.82 that is fully accounted for by the single outlier.

84

Notes, cont.

• 6. Linear relations only. Correlation applies only to linear

relationships This figure shows a strong non-linear relationship, yet r = 0.00.

• 7. Correlation does not

necessarily mean causation. Beware lurking variables (next slide).

85

Confounded Correlation

A near perfect negative correlation (r = −.987) was seen between cholera mortality and elevation above sea level during a 19th century epidemic.

We now know that cholera is transmitted by water. The

• bserved relationship

between cholera and elevation was confounded by the lurking variable proximity to polluted water.

86

Hypothesis Test

We conduct the hypothesis test to guard against identifying too many random correlations. Random selection from a random scatter can result in an apparent correlation

87

Hypothesis Test

• A. Hypotheses. Let ρ represent the population

correlation coefficient. H0: ρ = 0 vs. Ha: ρ ≠ 0 (two-sided)

[or Ha: ρ > 0 (right-sided) or Ha: ρ < 0 (left-sided)]

• B. Test statistic
• C. P-value. Convert tstat to P-value with software or t

table.

2 2 1 where

2 stat

      n df n r SE SE r t

r r

88

Hypothesis Test – Illustrative Example

A. H0: ρ = 0 vs. Ha: ρ ≠ 0 (two-sided) B. Test stat

• C. .005 < P < .01 by Table C. P = .0097 by computer.

The evidence against H0 is highly significant.

9 2 11 3.27 0.2253 737 . 0.2253 2 11 737 . 1

stat 2

         df t SE r

89

Exercise (True/False)

1. Correlation coefficient r quantifies the

relationship between quantitative variables X and Y.

2. The closer r is to 1, the stronger the

linear relation between X and Y.

3. If r is close to zero, X and Y are unrelated. 4. The value of r changes when the units of

measure are changed.

90

Regression

 Regression describes the

relationship in the data with a line that predicts the average change in Y per unit X.

 The best fitting line is

found by minimizing the sum of squared residuals, as shown in this figure.

91

Regression Line, cont.

The regression line equation is:

where ŷ ≡ predicted value of Y, a ≡ the intercept of the line, and b ≡ the slope of the line

Equations to calculate a and b

SLOPE: INTERCEPT:

92

Regression Line, cont.

Slope b is the key statistic produced by the regression

93

Calculate regression line in Excel

 Data analysis---regression

94

Calculate regression line in JMP

 Analyze---fit

Y by X

95

Conditions for Inference

Inference about the regression line requires these conditions

 Linearity  Independent observations  Normality at each level of X  Equal variance at each level of X

96

Conditions for Inference

This figure illustrates Normal and equal variation around the regression line at all levels of X

97

Assessing Conditions

 The scatterplot should be visually inspected for linearity,

Normality, and equal variance

 Plotting the residuals from the model can be helpful in this

regard.

 The table lists residuals for the illustrative data

98

Assessing Conditions, cont.

 A stemplot of the residuals show

no major departures from Normality

 This residual plot shows more

variability at higher X values (but the data is very sparse)

|-1|6 |-0|2336 | 0|01366 | 1|4 x10

99

Residual Plots

With a little experience, you can get good at reading residual plots. Here’s an example of linearity with equal variance.

100

Residual Plots

Example of linearity with unequal variance

101

Example of Residual Plots

Example of non-linearity with equal variance

102

103