Question

For studying the average pulse rates between three groups of people: smokers, ex-smokers, and non-smokers, three independent random samples of male subjects were selected from the three populations. The data, sitting pulse rates per minute measured in the morning, is listed below. Smokers: 88, 82, 80, 75 Ex-smokers: 70, 72, 73, 72 Non-smokers: 68, 70, 70, 75 Which of the followings are correct when one performed One-way ANOVA F-test for testing whether the difference in means of the sampled population is statistically significant? [Check all correct answers.] Question 15 options: A) At 5% level significance, the equal variances assumption is satisfied B) At 5% level significance, the difference among means of the sampled populations is statistically significant. C) At 5% level significance, the difference among means of the sampled populations is not statistically significant. D) At 5% level significance, the normality assumption is not satisfied.

Answer 1

The R code for doing ANOVA given below.

C1 = c(88, 82, 80, 75)
C2 = c( 70, 72, 73, 72 )
C3 = c(68, 70, 70, 75)
dati = c(C1,C2,C3)
groups = factor(rep(letters[1:3],c(4,4,4)))
fit = lm(formula = dati ~ groups)
fit
anova (fit)

The output is:

Analysis of Variance Table

Response: dati
Df Sum Sq Mean Sq F value Pr(>F)
groups 2 268.67 134.333 10.224 0.004823 **
Residuals 9 118.25 13.139

The p-value of the test is . Since the p-value , The null hypothesis   is rejected . The variances are not equal.

B) At 5% level significance, the difference among means of the sampled populations is statistically significant. Is correct.

For checking normality,

> by(dati,groups , shapiro.test)

Output is:

groups: a

Shapiro-Wilk normality test

data: dd[x, ]

W = 0.98996, p-value = 0.9573

------------------------------------------------------------------------

groups: b

Shapiro-Wilk normality test

data: dd[x, ]

W = 0.89495, p-value = 0.4064

------------------------------------------------------------------------

groups: c

Shapiro-Wilk normality test

data: dd[x, ]

W = 0.86521, p-value = 0.2793

The p-values from normality tests for all three samples are all greater 0.05. The normality assumption is acceptable.