Statistics in medicine - Yale Universitytbl.med.yale.edu/files/bivariate_association_continuous_variables/...9 0.703 0.883 1.100 1.383 1.833 2.262 2.821 3.250 4.781 10 0.700 ... •

S L I D E 0

Statistics in medicine

Lecture 3: Bivariate association : continuous variables

Fatma Shebl, MD, MS, MPH, PhD Assistant Professor Chronic Disease Epidemiology Department Yale School of Public Health [email protected]

mailto:[email protected]

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Outline • Mean in one group: Normal distribution • Mean in the same group repeated twice: Normal

distribution • Mean in one group: Not normal distribution • Mean in the same group repeated twice: Not normal

distribution • Mean in two groups: Normal distribution • Mean in two groups: Not normal distribution • Mean in three or more groups: Normal distribution • Mean in three or more groups : Not normal

distribution

S L I D E 2

Bivariate analysis Bivariate

One-sample

Mean

T-test Paired t-test Sign Wilcoxon

sign rank test

Proportion

Z test McNemar

> One-sample

Continuous and

categorical

variable

Two categories

Pooled t-test

Welch’s t-test

Wilcoxon rank sum

test

>= two categories

ANOVA Kruskal-

Wallis test

Continuous and

continuous

variable

Correlation

Categorical and

categorical

variable

Chi-square

test Fisher’s

exact test

S L I D E 3

Statistical tests of one group

One group

Mean

Normal distribution

Single time

T test

Two times/paired

Paired-t test

Not normal distribution

Transformation Nonparametric test

Single time

Sign test

Two times/paired

Wilcoxon sign rank test

Proportion

Single times

Z test

Two times/paired

McNemar test

S L I D E 4

Mean in one group: Normal distribution • Single time point:

–One sample Student’s t-test • Is a test to compare mean of a sample to

a known value • Can be used for small sample size (<30) • Use the t-distribution • Violations of assumptions of t-test

– Nonnormality – Skeweness – Outliers – Dependence between units of the sample

S L I D E 5

Mean in one group: Normal distribution • Single time point:

–One sample Student’s t-test • E.g., is the mean energy intake of 2-

year-olds (i.e. mean of the sample) significantly different from the value reported in NHANES III (i.e. known value)?

S L I D E 6

Steps of the statistical significance testing, t-test

• 1- Calculate the test statistic (critical ratio)

• 2- Calculate degrees of freedom • 3- Determine the critical value of

significance • 4-Compare the test statistic with the

critical value • 5- Calculate the p value • 6- Calculate the confidence interval • 7- Draw a conclusion

S L I D E 7

Mean in one group: Normal distribution, t-test 1- Calculate the critical value • Critical ratio (critical value)

¾General formula= 𝑃𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟𝑆𝐸 𝑜𝑓 𝑡ℎ𝑎𝑡 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟

¾t= 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑚𝑒𝑎𝑛−𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑚𝑒𝑎𝑛𝑆𝐸 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑒𝑎𝑛

¾t= 𝑥 −𝜇𝑆𝐷 𝑛

2- Calculate degrees of freedom • One sample t test df= N-1

S L I D E 8

Mean in one group: Normal distribution

3-Determine the critical value of significance Critical value:

– The value that a test statistic must exceed (in absolute value sense) for the null hypothesis to be rejected

– Depends on: • Alpha level • Degrees of freedom • Two-tail vs. one-tail test – E.g., for an alpha of .05,

two-tailed test, 8 df, the value of the t that defines the central 95% area (acceptance area) is between -2.306 and 2.306

Source: http://elvers.us/psy216/tables/tvalues.htm

df

Levels of Significance for a One-Tailed Test

0.2500 0.2000 0.1500 0.1000 0.0500 0.0250 0.0100 0.0050 0.0005

Levels of Signficance for a Two-Tailed Test

0.5000 0.4000 0.3000 0.2000 0.1000 0.0500 0.0200 0.0100 0.0010

1 1.000 1.376 1.963 3.078 6.314 12.706 31.821 63.657 636.619

2 0.816 1.061 1.386 1.886 2.920 4.303 6.964 9.925 31.599

3 0.765 0.978 1.250 1.638 2.353 3.182 4.541 5.841 12.924

4 0.741 0.941 1.189 1.533 2.132 2.776 3.747 4.604 8.610

5 0.727 0.920 1.156 1.476 2.015 2.570 3.365 4.032 6.869

6 0.718 0.906 1.134 1.440 1.943 2.447 3.143 3.707 5.959

7 0.711 0.896 1.119 1.415 1.895 2.365 2.998 3.499 5.408

8 0.706 0.889 1.108 1.397 1.859 2.306 2.896 3.355 5.041

9 0.703 0.883 1.100 1.383 1.833 2.262 2.821 3.250 4.781

10 0.700 0.879 1.093 1.372 1.812 2.228 2.764 3.169 4.587

S L I D E 9

Mean in one group: Normal distribution 4-Compare the test statistic with the critical value

–The test statistic is < the critical value (in absolute sense)Æ acceptance area

–The test statistic is > the critical value (in absolute sense)Æ rejection area

S L I D E 10

Mean in one group: Normal distribution

5- Calculate the p value • From the t-table

– Locate the row of the relevant df

– Find the t value that is < test statistic

– Find the p value at the top of the column

– E.g., for a two-tailed test, 8 df, t=3.1

– Æ .02 > p value > .01

df


0.2500 0.2000 0.1500 0.1000 0.0500 0.0250 0.0100 0.0050 0.0005


0.5000 0.4000 0.3000 0.2000 0.1000 0.0500 0.0200 0.0100 0.0010

1 1.000 1.376 1.963 3.078 6.314 12.706 31.821 63.657 636.619

2 0.816 1.061 1.386 1.886 2.920 4.303 6.964 9.925 31.599

3 0.765 0.978 1.250 1.638 2.353 3.182 4.541 5.841 12.924

4 0.741 0.941 1.189 1.533 2.132 2.776 3.747 4.604 8.610

5 0.727 0.920 1.156 1.476 2.015 2.570 3.365 4.032 6.869

6 0.718 0.906 1.134 1.440 1.943 2.447 3.143 3.707 5.959

7 0.711 0.896 1.119 1.415 1.895 2.365 2.998 3.499 5.408

8 0.706 0.889 1.108 1.397 1.859 2.306 2.896 3.355 5.041

9 0.703 0.883 1.100 1.383 1.833 2.262 2.821 3.250 4.781

10 0.700 0.879 1.093 1.372 1.812 2.228 2.764 3.169 4.587

S L I D E 11

Mean in one group: Normal distribution 6- Calculate the confidence interval • CI of mean in one group:

– 𝑥 ± 𝑡 × SE • E.g. if the test has 8 df

–95% CI= 𝑥 ± 2.306× SE

Obtain this value from the t table

S L I D E 12

Mean in one group: Normal distribution 7- Draw a conclusion • Reject or fail to reject the null hypothesis

– Fail to reject the null • If the p value > alpha level • If the confidence interval crosses the 0

– Reject the null • If the p value < alpha level • If the confidence interval does not include

0

S L I D E 13

One-sample t-test example • A study of 94 2-year old

energy intake compared to NHANES III mean energy intake value were conducted. The calculated t value was -1.67. What is the corresponding p value?

• Answer: – df=94-1=93 – 0.1>P value>0.05 – Conclusion: we fail to

reject the null hypothesis i.e. the 2-year old intake is not significantly different from NHANES III


df


0.2500 0.2000 0.1500 0.1000 0.0500 0.0250 0.0100 0.0050 0.0005


0.5000 0.4000 0.3000 0.2000 0.1000 0.0500 0.0200 0.0100 0.0010

91 0.677 0.846 1.042 1.291 1.662 1.986 2.368 2.631 3.401

92 0.677 0.846 1.042 1.291 1.662 1.986 2.367 2.630 3.399

93 0.677 0.846 1.042 1.291 1.661 1.986 2.367 2.630 3.398

94 0.677 0.845 1.042 1.291 1.661 1.986 2.367 2.629 3.397

95 0.677 0.845 1.042 1.291 1.661 1.985 2.366 2.628 3.396

96 0.677 0.845 1.042 1.290 1.661 1.985 2.366 2.628 3.395

97 0.677 0.845 1.042 1.290 1.661 1.985 2.365 2.627 3.394

98 0.677 0.845 1.042 1.290 1.661 1.984 2.365 2.627 3.392

99 0.677 0.845 1.042 1.290 1.660 1.984 2.365 2.626 3.391

100 0.677 0.845 1.042 1.290 1.660 1.984 2.364 2.626 3.390

S L I D E 14

Mean in one group, paired/matched: Normal distribution • One group is measured twice (or matched

design) – Paired Student’s t-test

• Is a test to compare the difference (change) in the mean of two paired (matched) groups.

– Can be used for designs where individuals’ characteristics are measured before and after intervention

• Use the t-distribution • Violations of assumptions of t-test

– Nonnormality – Skeweness – Outliers – Dependence between units of the sample

S L I D E 15

Mean in one group, paired/matched: Normal distribution, paired t-test

• One group is measured twice (or matched design): –Paired Student’s t-test

• E.g., Is there a change in serum LDL levels after cholecystectomy?

S L I D E 16

Steps of the statistical significance testing, paired t-test

• 1- Calculate the test statistic (critical ratio)

• 2- Calculate degrees of freedom • 3- Determine the critical value of



S L I D E 17


1- Calculate the critical value • Critical ratio (critical value)

¾t= 𝑑 −0𝑆𝐷𝑑 𝑛

¾Where, ¾ 𝑑 = mean difference

¾𝑆𝐷𝑑 = 𝑑−𝑑 2

𝑛−1

¾2- Calculate degrees of freedom • Paired-t test df= N-1

S L I D E 18


3-Determine the critical value of significance Critical value: – Depends on: • Alpha level • Degrees of freedom • Two-tail vs. one-tail

test – E.g., for an alpha of

.01, two-tailed test, 50 df, the value of the t that defines the central 99% area (acceptance area) is between -2.678 and 2.678


df


0.2500 0.2000 0.1500 0.1000 0.0500 0.0250 0.0100 0.0050 0.0005


0.5000 0.4000 0.3000 0.2000 0.1000 0.0500 0.0200 0.0100 0.0010

41 0.680 0.850 1.050 1.303 1.683 2.020 2.421 2.701 3.544

42 0.680 0.850 1.049 1.302 1.682 2.018 2.418 2.698 3.538

43 0.680 0.850 1.049 1.302 1.681 2.017 2.416 2.695 3.532

44 0.680 0.850 1.049 1.301 1.680 2.015 2.414 2.692 3.526

45 0.680 0.850 1.048 1.301 1.679 2.014 2.412 2.689 3.520

46 0.680 0.850 1.048 1.300 1.679 2.013 2.410 2.687 3.515

47 0.680 0.849 1.048 1.300 1.678 2.012 2.408 2.684 3.510

48 0.680 0.849 1.048 1.299 1.677 2.011 2.406 2.682 3.505

49 0.679 0.849 1.047 1.299 1.677 2.010 2.405 2.680 3.500

50 0.679 0.849 1.047 1.299 1.676 2.009 2.403 2.678 3.496

S L I D E 19


4-Compare the test statistic with the critical value



S L I D E 20


5- Calculate the p value • From the t-table

– Locate the row of the relevant df

– Find the t value that is < test statistic

– Find the p value at the top of the column

– E.g., for a two-tailed test, 50 df, t=3.1

–Æ .01 > p value > .001

df


0.2500 0.2000 0.1500 0.1000 0.0500 0.0250 0.0100 0.0050 0.0005


0.5000 0.4000 0.3000 0.2000 0.1000 0.0500 0.0200 0.0100 0.0010

41 0.680 0.850 1.050 1.303 1.683 2.020 2.421 2.701 3.544

42 0.680 0.850 1.049 1.302 1.682 2.018 2.418 2.698 3.538

43 0.680 0.850 1.049 1.302 1.681 2.017 2.416 2.695 3.532

44 0.680 0.850 1.049 1.301 1.680 2.015 2.414 2.692 3.526

45 0.680 0.850 1.048 1.301 1.679 2.014 2.412 2.689 3.520

46 0.680 0.850 1.048 1.300 1.679 2.013 2.410 2.687 3.515

47 0.680 0.849 1.048 1.300 1.678 2.012 2.408 2.684 3.510

48 0.680 0.849 1.048 1.299 1.677 2.011 2.406 2.682 3.505

49 0.679 0.849 1.047 1.299 1.677 2.010 2.405 2.680 3.500

50 0.679 0.849 1.047 1.299 1.676 2.009 2.403 2.678 3.496

S L I D E 21


6- Calculate the confidence interval • CI of mean difference in paired design:

– 𝑑 ± 𝑡𝑛−1 × 𝑆𝐷𝑑𝑛

• E.g. if the test has 50 df

–95% CI= 𝑑 ± 2.009× 𝑆𝐷𝑑𝑛

S L I D E 22


7- Draw a conclusion • Reject or fail to reject the null hypothesis



the 0

S L I D E 23

Paired t-test example

• A study of LDL in 51 patients who had cholecystectomy were conducted. The calculated t value was 5.68. What is the corresponding p value?

• Answer: – df=51-1=50 – P value<0.001 – Conclusion: we

reject the null hypothesis i.e. the post operative LDL is significantly different from the pre operative LDL

df


0.2500 0.2000 0.1500 0.1000 0.0500 0.0250 0.0100 0.0050 0.0005


0.5000 0.4000 0.3000 0.2000 0.1000 0.0500 0.0200 0.0100 0.0010

41 0.680 0.850 1.050 1.303 1.683 2.020 2.421 2.701 3.544

42 0.680 0.850 1.049 1.302 1.682 2.018 2.418 2.698 3.538

43 0.680 0.850 1.049 1.302 1.681 2.017 2.416 2.695 3.532

44 0.680 0.850 1.049 1.301 1.680 2.015 2.414 2.692 3.526

45 0.680 0.850 1.048 1.301 1.679 2.014 2.412 2.689 3.520

46 0.680 0.850 1.048 1.300 1.679 2.013 2.410 2.687 3.515

47 0.680 0.849 1.048 1.300 1.678 2.012 2.408 2.684 3.510

48 0.680 0.849 1.048 1.299 1.677 2.011 2.406 2.682 3.505

49 0.679 0.849 1.047 1.299 1.677 2.010 2.405 2.680 3.500

50 0.679 0.849 1.047 1.299 1.676 2.009 2.403 2.678 3.496

S L I D E 24

Median in one group: Not normal distribution • Single time point:

–Sign test • Is a nonparametric (distribution free)

test to compare median of a sample to a known value

• Indications: single sample with –Abnormal distribution –Skewed data –Outliers

S L I D E 25

Median in one group: Not normal distribution • Single time point:

–Sign test • Because the median is used, binomial

distribution with π=.5 could be used • If sample size is large, z distribution

could be used • E.g., is the median energy intake of 2-

year-olds significantly different from the value reported in NHANES III?

S L I D E 26

Steps of the statistical significance testing, sign test

• 1- Calculate the test statistic • 2- Determine the critical value of


critical value • 4- Calculate the p value • 5- Draw a conclusion

S L I D E 27

Median in one group: Not normal distribution, sign test 1- Calculate the critical value • Critical ratio (critical value) ¾Binomial distribution: p(x)=

𝑛!𝑥! 𝑛−𝑥 ! 𝜋

𝑥(1 − 𝜋)𝑛−𝑥

¾Z approximation: z= 𝑋−𝑛𝜋 −(12)

𝑛𝜋(1−𝜋)

S L I D E 28

Median in one group: Not normal distribution, sign test

2-Determine the critical value of significance Critical value identification • Find the area

• For +z – one-tailed test: 1-alpha – two-tailed test: 1-

(alpha/2) • For -z

– one-tailed test: alpha – two-tailed test: (alpha/2)

• Look in the table for the area value and find the corresponding z critical value

• E.g., for an alpha of .05, two-tailed test, the value of the z that defines the central 95% area (acceptance area) is between -1.96 and 1.96

Source of table: http://image.slidesharecdn.com/copyofz-table-130515110049-phpapp02/95/copy-of-z-table-1-638.jpg?cb=1368615687

S L I D E 29

Median in one group: Not normal distribution, sign test 3-Compare the test statistic with the critical value



S L I D E 30


4- Calculate the p value • Using the z-table –Find the z value that is =test statistic by scrolling down the

z column then scrolling to the right in the row –The entries of the tables are used to calculate the p values

• For negative z values • One-tailed, the p value is the entry of the z table • Two-tailed, the p value is the entry of the z table multiplied by

2 • For positive z values

• One-tailed, the p value = 1- the entry of the z table • Two-tailed,: p value = (1- the entry of the z table )multiplied

by 2

S L I D E 31


4- Calculate the p value

– E.g., for a z=3.1 • one-tailed test,

p=(1-.999) =.001 • two-tailed test,

p=(1-.999) *2=.002

Source of table : http://image.slidesharecdn.com/copyofz-table-130515110049-phpapp02/95/copy-of-z-table-1-638.jpg?cb=1368615687

S L I D E 32

Median in one group: N0t normal distribution, sign test 5- Draw a conclusion • Reject or fail to reject the null hypothesis

– Fail to reject the null • If the p value > alpha level

– Reject the null • If the p value < alpha level

S L I D E 33

One-sample sign test example

• A study of 94 2-year old energy intake compared to NHANES III median energy intake value were conducted. The calculated z value was 1.92. Using a two-tail test and .05 alpha level, what is the critical value and the corresponding p value?

• Answer: – Critical value=+1.96 – p=(1-.9726) *2=.0548 – Conclusion: we fail to reject

the null hypothesis i.e. the 2-year old intake is not significantly different from NHANES III

Source: http://www.had2know.com/academics/normal-distribution-table-z-scores.html

S L I D E 34

Mean in one group, paired/matched: Not normal distribution • One group is measured twice (or matched

design) – Sign test

–Used for the difference of means – Wilcoxon signed rank test

–A non parametric test for two DEPENDENT samples with ordinal data or with numerical observations that are NOT NORMALLY distributed

–As powerful as t-test –More powerful than sign test –Requires extensive computation, therefore

statistical programs are needed

S L I D E 35

Statistical tests of two independent groups

Two groups

Mean

Normal distribution

Equal variance

Pooled t test

Unequal variance

Welch’s t test

Not normal distribution

Transformation Nonparametric test

Wilcoxon rank sum test

Proportion

Z test Chi-

square test

Fisher’s exact test

Risk ratios

S L I D E 36

Mean in two groups: Normal distribution

–Two-sample t-test • Is a test to compare means of two small

samples • Can be used for small sample size (<30) • Use the t-distribution • Violations of assumptions of t-test

– Nonnormality in either group: if sample size>30 it might be okay

– Equal variance in the two groups – Dependence between units of the sample

S L I D E 37

Mean in two groups: Normal distribution, t-test • Two-sample t-test:

• E.g., Did women who receive paracervical block prior to the cryosurgery have less severe total cramping than women who did not?

S L I D E 38

Steps of the statistical significance testing, Two sample t-test

• 1- Determine the type of t test • 2- Calculate the test statistic • 3- Calculate degrees of freedom • 4- Determine the critical value of



S L I D E 39

Steps of the statistical significance testing, Two sample t-test 1- Determine the type of t test • Examine the equality of the variance in the two

groups ¾F test for equal variance ¾Sensitive for data normality i.e. may be

significant because data is not normal, not because the unequal variance ¾If significant Æ reject the hypothesis of equal

variance ¾Levene test for equal variances ¾If significant Æ reject the hypothesis of equal

variance

S L I D E 40

Steps of the statistical significance testing, Two sample t-test 1- Determine the type of t test • Equal variance in the two groups

– Pooled t test • Unequal variance in the two groups

– Welch’s t test

S L I D E 41

Steps of the statistical significance testing, Two sample t-test 2- Calculate the critical value Equal variances: Pooled t test

tn1+n2-2= 𝑥 1−𝑥 2𝑆𝐷𝑝 [(1/𝑛1)+(1/𝑛2)]

Where, pooled SDp =

¾SDp= 𝑛1−1 𝑆𝐷12+ 𝑛2−1 𝑆𝐷2

2

𝑛1+ 𝑛2−2

S L I D E 42

Steps of the statistical significance testing, Two sample t-test 2- Calculate the critical value Unequal variances: Welch’s t test

t= 𝑥 1−𝑥 2𝑆𝐷

Where, SD =

¾SD= 𝑆𝐷12

𝑛1+ 𝑆𝐷2

2

𝑛2

S L I D E 43

Steps of the statistical significance testing, Two sample t-test 3- Calculate degrees of freedom • Pooled t test df= n1 + n2 – 2

• Welch’s t test df=(𝑆𝐷1

2

𝑛1+𝑆𝐷2

2

𝑛2)2

(𝑆𝐷12

𝑛1)2

𝑛1−1 +(𝑆𝐷2

2

𝑛2)2

𝑛2−1

S L I D E 44

Steps of the statistical significance testing, Two sample t-test 4-Determine the critical value of significance, using t-distribution table based on

– Alpha level – df – Two-tail or one-tail

5-Compare the test statistic with the critical value – The test statistic is < the critical value (in

absolute sense)Æ acceptance area – The test statistic is > the critical value (in

absolute sense)Æ rejection area

S L I D E 45

Steps of the statistical significance testing, Two sample t-test

6- Calculate the p value from the t-distribution table 7- Calculate the confidence interval • CI of mean difference between two groups:

– 𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 ± 𝑡1−𝛼, 𝑑𝑓 ∗ 𝑆𝐸𝑥 1−𝑥 2 • Where,

–Equal variance: 𝑆𝐸𝑥 1−𝑥 2 =1𝑛1

+ 1𝑛2

–Unequal variance: 𝑆𝐸𝑥 1−𝑥 2 =𝑆𝐷1

2

𝑛1+ 𝑆𝐷2

2

𝑛2

S L I D E 46

Steps of the statistical significance testing, Two sample t-test 8- Draw a conclusion • Reject or fail to reject the null hypothesis



0

S L I D E 47

Steps of the statistical significance testing, Two sample t-test, example

•A study of cramping score in two groups of women (group1: 45 who received paracervical block , and group2: 39 who did not receive paracervical block) who underwent cryosurgery were conducted. The mean and SD were (35.6,and 28.45), and (51.41, and 8.11) in group1 and group2 respectively. The calculated t value was -2.56. Assuming an alpha level of 0.01, and the alternative hypothesis is that “women who receive paracervical block prior to the cryosurgery have less severe total cramping than women who did not”. Calculate df, critical value, and p value?


S L I D E 48

Steps of the statistical significance testing, Two sample t-test, example

• Answer: –df=45+39-2=82 –Critical value=2.373 –0.005<P value <0.01 –Conclusion: we reject

the null hypothesis i.e. women who received the block had significantly lower total cramping than women who did not receive the block


df


0.2500 0.2000 0.1500 0.1000 0.0500 0.0250 0.0100 0.0050 0.0005


0.5000 0.4000 0.3000 0.2000 0.1000 0.0500 0.0200 0.0100 0.0010

81 0.677 0.846 1.043 1.292 1.664 1.990 2.373 2.638 3.415

82 0.677 0.846 1.043 1.292 1.664 1.989 2.373 2.637 3.413

83 0.677 0.846 1.043 1.292 1.663 1.989 2.372 2.636 3.412

84 0.677 0.846 1.043 1.292 1.663 1.989 2.371 2.636 3.410

85 0.677 0.846 1.043 1.292 1.663 1.988 2.371 2.635 3.409

86 0.677 0.846 1.043 1.291 1.663 1.988 2.370 2.634 3.407

87 0.677 0.846 1.043 1.291 1.663 1.988 2.370 2.633 3.406

88 0.677 0.846 1.042 1.291 1.662 1.987 2.369 2.633 3.404

89 0.677 0.846 1.042 1.291 1.662 1.987 2.369 2.632 3.403

90 0.677 0.846 1.042 1.291 1.662 1.987 2.368 2.631 3.402

S L I D E 49

Mean in two groups: Not normal distribution, Wilcoxon rank sum test

• A nonparametric test for comparing two independent samples with ordinal data or with numerical observations that are not normally distributes

• It tests the hypothesis that the means of the ranks are equal

• Steps: 1.Rank or observations regardless of the group 2.Sum the ranks of each group 3.Calculate the mean and SD of the ranks in each group 4.Calculate pooled SD of the ranks

Documents

Statistics in medicine - Yale Universitytbl.med.yale.edu/files/bivariate_association_continuous_variables/...9 0.703 0.883 1.100 1.383 1.833 2.262 2.821 3.250 4.781 10 0.700 ... •