Lab 5:Analysis of Variance

Parametric Analysis of Variance

Exercise: Forced expiratory volume in one second for patients with coronary heart disease sampled at three different medical centers is given in pulmonary function data.

a) State the null/alternative hypothesis that the mean forced expiratory volume in one second for patients with coronary heart disease is the same for each of the threemedical centers.

Ho:

Ha:

b) What assumptions about the data must be true for you to use the one-way analysis of variance technique? State whether they are satisfied for the data.

1) Normality of data (given below, tested by Kolmogorov Smirnov orShapiro Wilks and also by looking at the Normal probability plot)

2) Homogenous variance (Levene test, given as an output of ANOVA below)

Normality test:

Analyze→ Descriptive Statistics→ Explore

→Plots

Tests of Normality
center / Kolmogorov-Smirnova / Shapiro-Wilk
Statistic / df / Sig. / Statistic / df / Sig.
Forcedexpiratoryvolume in onesecond / Johns Hopkns / ,133 / 21 / ,200* / ,966 / 21 / ,652
Rancho Los Amigos / ,155 / 16 / ,200* / ,953 / 16 / ,532
St Louis / ,109 / 23 / ,200* / ,980 / 23 / ,906
*. This is a lowerbound of thetruesignificance.
a. LillieforsSignificanceCorrection

Each group is normally distributed (p-values>0.05).

b) Test the null hypothesis that the mean forced expiratory volume in one second for patients with coronary heart disease are the same for each of the four medical centers.

To conduct One-way ANOVA to compare means:

Analyze→Compare Means→One-way ANOVA

→Options

Complete the following descriptive statistics table fromthe output and state which medical center has the highest mean.

Descriptives
Forcedexpiratoryvolume in onesecond
N / Mean / Std. Deviation / Std. Error / 95% ConfidenceIntervalforMean / Minimum / Maximum
LowerBound / UpperBound
Johns Hopkns / 21 / 2,84333 / ,325413 / ,071011 / 2,69521 / 2,99146 / 2,100 / 3,470
Rancho Los Amigos / 16 / 3,18631 / ,384130 / ,096033 / 2,98162 / 3,39100 / 2,610 / 3,860
St Louis / 23 / 2,87870 / ,497716 / ,103781 / 2,66347 / 3,09392 / 1,980 / 4,060
Total / 60 / 2,94835 / ,432382 / ,055820 / 2,83665 / 3,06005 / 1,980 / 4,060
Test of Homogeneity of Variances
Forcedexpiratoryvolume in onesecond
LeveneStatistic / df1 / df2 / Sig.
1,646 / 2 / 57 / ,202

Are the variances homogeneous?

ANOVA
Forcedexpiratoryvolume in onesecond
Sum of Squares / df / MeanSquare / F / Sig.
BetweenGroups / 1,249 / 2 / ,625 / 3,640 / ,033
WithinGroups / 9,781 / 57 / ,172
Total / 11,030 / 59

What do you conclude from the above ANOVA table?

c) If you have found a significant difference continue with post-hoc tests.

What do you conclude? Which pairs are significantly different than each other?

According to Tukey HSD test the mean of Rancho Los Amigos is significantly higher than JohnsHopkins (p-value=0.41).

Nonparametric analysis of Variance(Kruskal Wallis Test)

When the assumption of normality or homogeneity of variance is not satisfied we can use a nonparametric version of ANOVA which is called Kruskal Wallis test.

Exercise:

Three teaching methods were tested on a group of 38 students with homogeneous backgrounds in statistics and comparable aptitudes. Each student was randomly assigned to a method and at the end of a 6-week program was given a standardized exam. Because of classroom space and group size, the students were not equally allocated to each method. The data is called teaching. Test for a difference in distributions (medians) of the test scores for the different teaching methods using the Kruskal-Wallis test.

Analyze→ Nonparametric Tests→ Independent Samples

Customize Tests→ Kruskal-Wallis →All Pairwise

What do you conclude?

1