Question

homework problem

A random sample of the records of single births was selected from each of four populations. The weights (grams) of the babies at birth are shown below. Do these data provide sufficient evidence to indicate that the four populations differ with respect to average birth weight? Use a = 0.05. Sample А B с D 2946 3186 2300 2286 2913 2857 2903 2938 2280 3099 2572 2952 3685 2761 2584 2348 2310 3290 2675 2691 2582 2937 2571 2858 3002 3347 241 2408 2008 2850 2762

Answer 1

Sol:

Please see the R snippet

# read the data into R dataframe
data.df<- read.csv("singlebirths.csv",header=TRUE)
str(data.df)

# reshape the data to long format
library(reshape2)
melt(data.df)

#fit a lm model
melt.df<- na.omit(melt(data.df))
mod1 = lm(value ~ variable, data = melt.df)
# check anova for part 1 of the question
anova(mod1)

# the test is significant and they do differ
a1<-aov(value~variable,data=melt.df)

# part 2 , perform a post hoc test , such as the Tukey hsd
posthoc<-TukeyHSD(x=a1,conf.level=0.95)

#plot the results for visual reppresentation of the results
plot(posthoc)

# for part c we need to create a new dummy variable for rural vs urban
# midwest vs south

melt.df$var1 <- ifelse(melt.df$variable %in% c("MR","MU"),"Midwest", "South")
melt.df$var2 <- ifelse(melt.df$variable %in% c("SR","MR"),"Rural", "Urban")

# perform the anova again , you may also choose to do a t test
a2<-aov(value~var1,data=melt.df)
a3<-aov(value~var2,data=melt.df)

# summary of the results

summary(a2)
summary(a3)

Kindly change the path with the path of the file on your local machine. Enter the data exactly as shown in the table above

The results are

> summary(a1)
Df Sum Sq Mean Sq F value Pr(>F)
variable 3 943136 314379 4.652 0.0172 *
Residuals 15 1013628 67575   
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

> summary(a2)
Df Sum Sq Mean Sq F value Pr(>F)
var1 1 328280 328280 3.427 0.0816 .
Residuals 17 1628484 95793   
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> summary(a3)
Df Sum Sq Mean Sq F value Pr(>F)
var2 1 432305 432305 4.821 0.0423 *
Residuals 17 1524459 89674   
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

The result of the post hoc test is

> posthoc
Tukey multiple comparisons of means
95% family-wise confidence level

Fit: aov(formula = value ~ variable, data = melt.df)

$variable
diff lwr upr p adj
MU-MR 479.00 5.150837 952.84916 0.0471395
SR-MR -99.45 -602.042935 403.14293 0.9394047
SU-MR 36.80 -437.049163 510.64916 0.9958684
SR-MU -578.45 -1081.042935 -75.85707 0.0217381
SU-MU -442.20 -916.049163 31.64916 0.0714508
SU-SR 136.25 -366.342935 638.84293 0.8617146

The visual graph is

as we are trying to compare a quatitative variable against a categorical variable for with different levels of categories , a 1 way anova is a suitable choice to check whther the values differ across differetn categories. You may also chose to do a t test for the same