# Continuous Outcome, Several Independent Groups

## Background

In a previous chapter we considered the analysis of a continuous outcome from two independent groups. In this chapter we extendt his method to consider 2 or more groups.

Illustrative Data. Let us consider ages of subjects from three centers. Data are:

Center 1: 60, 66, 65, 55, 62, 70, 51, 72, 58, 61, 71, 41, 70, 57, 55, 63, 64, 76, 74, 54, 58, 73
Center 2: 56, 65, 65, 63, 57, 47, 72, 56, 52, 75, 66, 62, 68, 75, 60, 73, 63, 64
Center 3: 67, 56, 65, 61, 63, 59, 42, 53, 63, 65, 60, 57, 62, 70, 73, 63, 55, 52, 58, 68, 70, 72, 45

The data set has 63 records (N = 63) and a grand mean of 62.1 years (standard deviation = 8.1 years).

For these data to be analyzed in Epi Info, they must be structured with two variables -- one for the dependent (outcome) variable and one for the independent (group) variable. Data for the illustrative data set are stored in AGEBYCEN.ZIP as the file AGEBYCEN.REC with variables AGE (dependent variable) and CENTER (independent variable). The first three records and last record of this data set are:

REC      AGE       CENTER
---  --------- -----------
1         60           1
2         66           1
3         65           1
etc.
63         45           3

## Descriptive Statistics

Summary statistics are computed with the MEANS command:

EPI6> MEANS <DV> <IV>

where <DV> represents the dependent variable and <IV> represents the independent variable.

To compute statistics for the illustrative data, issue the commands:

EPI6> MEANS AGE CENTER

This produces the following output:

MEANS of AGE for each category of CENTER

CENTER            Obs      Total       Mean   Variance    Std Dev
1                  22       1376     62.545     75.212      8.672
2                  18       1139     63.278     60.683      7.790
3                  23       1399     60.826     64.059      8.004

CENTER        Minimum     25%ile     Median     75%ile    Maximum       Mode
1              41.000     57.000     62.500     70.000     76.000     55.000
2              47.000     57.000     63.500     68.000     75.000     56.000
3              42.000     56.000     62.000     67.000     73.000     63.000

Thus, n1 = 22, n2 = 18, and n3 = 23. We also see that groups have similar means (means are 62.5, 63.3, and 60.8, respectively) and standard deviations are not too dissimilar.

Side-by-side quartile plots can be drawn (by hand) with minimal effort by graphing each group median as a dot and whiskers from the group's minimum to 25 percentile (bottom whisker) and 75 percentile to the maximum (top whisker). Click here for an example.

# Inferential Statistics

## Confidence Intervals

We assume the Mean Square Within (MSW) in the ANOVA table is an pooled estimate of the common within group variance:

ANOVA
Variation          SS   df          MS  F statistic    p-value
Between        66.614    2      33.307        0.497   0.616421
Within       4020.370   60      67.006
Total        4086.984   62

The standard error of the mean for group i (sei) is thus:

sei =sqrt(MSW/ ni)

with dfw = N-k, where N represents the total sample size (all groups combined) and k represents the number of groups.

For the illustrative example, se1 = sqrt(67.006 / 22) = 1.745.

A 95% confidence interval for the mean of group ii) is given by:

(sample mean of group i) ± (tdfw,.975)(sei)

For example, a 95% confidence for the mean of group 1 = 62.5 ± (t60,.025)(1.745) = 62.5 ± (2.00)(1.745) = (58.9, 66.1).

## ANOVA

The objective of ANOVA is to determine whether one or more population means of the k groups differs. The null and alternative hypotheses are:

H0: µ1 = µ2 = ... = µk
H1: at least one population mean differs

where µi represents the population mean of group i {i: 1, 2, . . . k}.

Briefly, ANOVA partitions the variance in the data into the variance or mea square between (MSB) and the variance or mean square (MSW). The ratio of these means squares is the F statistic:

Fstat = (MSB) / (MSW)

Under the null hypothesis, this test statistic has an F distribution with dfB = k-1 and dfW = N-k. The test is one-tailed focusing on the upper extent of the FdfB, dfw distribution. For the illustrative data set, the ANOVA table and F statistic are:

ANOVA
Variation          SS   df          MS  F statistic    p-value
Between        66.614    2      33.307        0.497   0.616421
Within       4020.370   60      67.006
Total        4086.984   62

Thus, p =.62.

Assumptions: The ANOVA tests has several hidden assumptions. Traditionally, we speak of the assumption of independence, normality, and equal variance. In addition, statistical inferences assume validity of the data (i.e., freedom from selection bias and information bias) and minimal confounding.

### The Kruskal-Wallis Test

The Kruskal-Wallis procedure is a non-parametric analogue of ANOVA. The null and alternative are:

H0: the population medians are equal
H
1: at least one population median differs

Statistics are provided in the two-variable MEANS command output:

Kruskal-Wallis One Way Analysis of Variance
Kruskal-Wallis H (equivalent to Chi square) =       0.916
Degrees of freedom =           2
p value =    0.632634

The results of this test (p = .63) provide no reason to reject H0.

### Bartlett's Test

Bartlett's test address whether population variances differ. The null and alternative hypotheses are:

H0: s²1 = s²2 = . . . = s²k
H1: at least one population variance differs

where s²i represents the population variance in group i. Results are provided in the output of the two-variable MEANS command. Output for the illustrative example is:

Bartlett's test for homogeneity of variance
Bartlett's chi square = 0.243  deg freedom = 2  p-value = 0.885608

Thus, for this example, c2 = 0.24 with 2 df (p = .89), providing no real evidence against the null hypothesis.

Comment: Because Bartlett's test performs poorly in non-normal populations and has poor power some statisticians advise against its routine use (Box, 1953, Biometrika, pp. 318 - 335).

## Sample Size Requirements

We frequently want to know how large a sample is needed when testing k means. Although there is no simple answer to this question, a reasonable sample size can be determined if certain assumptions are made. Let us concern ourselves with trying to establish a significant difference among k means (via ANOVA) by asking how big a sample size is needed to (a) detect a difference between two means of D, (b) at a type I error rate of a, (c) with probability (power) 1-b. It is necessary to have a prior estimate of variability s of the outcome variable, with such estimates coming from a pilot study, prior published results, a preliminary analysis, or intuition. Computational solutions are possible once these underlying assumptions are made clear, with formulas are available in Sokal & Rohlf, 1996, pp. 263-264 (for instance). Calculations have been programmed into a the Dept of OB/GYN at the University of Hong Kong  website via the URL http://department.obg.cuhk.edu.hk/ResearchSupport/Sample_size_CompMean.asp. (If this link does not take you directly to the sample size calculator, click Sample Size > Comparing Means.)

Illustrative example. Suppose we test H0: µ1 = µ2 = µ3. Prior study suggests the measurement has s @ 8. To find a mean difference of 5, the University of Hong Kong  website derives the following results.

Type I error=0.05 Type I error=0.01 Type I error=0.001 41 60 87 54 76 107 67 91 125

Notice that the output provides samples sizes per group (ni) at various power and alpha levels. For example, under the stated assumptions, we need n = 54 for 90% power at a = .05.

## Exercises

(1) ALCOHOL.ZIP: Alcohol Consumption by Income Level (Data from Monder, 1986). Data come from a survey of alcohol consumption and socioeconomic status. Data, in ALCOHOL.REC, are coded as follows:
 Variable Name Type Description and codes ALCS ## Alcohol consumption score. Codes are as follows:  00 = non-drinker  01 = 1 drink per week  02 = 1-2 drinks per week  03 = 2 drinks per week  04 = 2-3 drinks per week  05 = 3 drinks per week  06 = 3-4 drinks per week  07 = 4 drinks per week  08 = 4-5 drinks per week  09 = 5 drinks per week  10 = 5-6 drinks per week  11 = 6 drinks per week  12 = 7-11 drinks per week  13 = 12+ drinks per week AGE ## Age (in years) INC # Income level: 1 = low, 5 = high

(A) Univariate description of ALCS. Before performing ANOVA, describe the distribution of alcohol scores for all groups combined (MEANS ALCS). Show the distribution of this variable in the form of a histogram (HISTOGRAM ALCS). Are data skewed? What percentage of people in the sample are non-drinkers?
(B) ALCS by INC. Assess alcohol consumption by income.

(2) DEERMICE: Weight Gain in White-Footed Deer Mice (Hampton, 1994, p. 118, modified). Fifteen deermice are randomly assigned to one of three groups. Group A receives a standard diet, Group B receives a diet of junk food, and Group C receives a diet of health food. The research question is to determine whether WTGAIN differs by DIET. Weight gains (gms.) are as follows:

Group A: 11.8, 12.0, 10.7, 9.1, 12.1
Group B: 13.6, 14.4, 12.8, 13.0, 13.4
Group C: 9.2, 9.6, 8.6, 8.5, 9.8

(3) ROOSTER: Testosterone Levels in Roosters (Data from Hampton, 1994, p. 147). A chicken pathologist believes testosterone levels differ by rooster strain. To test this hypothesis, testosterone levels are measured in 3 strains of roosters. The research question is to determine whether testosterone levels differ by rooster strain. Data are:

REC  TESTOSTERO STRAIN
---  ---------- ------
1         439 A
2         568 A
3         134 A
4         897 A
5         229 A
6         329 A
7         103 B
8         115 B
9          98 B
10         126 B
11         115 B
12         120 B
13         107 C
14          99 C
15         102 C
16         105 C
17          89 C
18         110 C