Background

A study was done to determine whether or not the idea that shoes for boys tend to be wider than shoes for girls. A statistician collected data in a fourth grade classroom where the length and width of 39 children’s feet were measured

Below is a table showing the collected data. The width3 column is the 3x the width.

datatable(KidsFeet[,c("name", "length","width", "width3")],
options = list(lengthMenu = c(3,10,30)))

With this data I am hypothesizing that the length of a child’s foot is the same as 3x the width of their foot. Because I am going to compare the length and the width of the foot, this is a paired study as two measurements were obtained from each individual.

The null and alternative hypotheses are written as

\[ H_0: \mu_{\text{d}} = 0 \] \[ H_a: \mu_{\text{d}} \neq 0 \] where the differences \(d\) are defined as \[ d = \text{length} - 3\times\text{width} \] In other words, I am testing to see if the length of a foot is equal to three times the width.

The significance level for this study will be set at

\[ \alpha = 0.05 \]

Analysis

The dotplot below shoes the differences between the length and 3 times the width for each individual. Since 28 out of the 39 differences are negative, it shows that most individuals length is shorter than 3 times the width of their foot

with(KidsFeet, stripchart(length - width3, main = "Length vs Width", xlab = "Length - 3x the Width"), pch = 20)

The paired samples t test is only appropriate if the sampling distribution of the sample mean of the differences, \(\bar{d}\), is normally distributed

with(KidsFeet, qqPlot(length - width3, pch = 1, xlab = "Foot Length", ylab = "Difference", main = "Difference of Width \nGiven Length"))

Based on the QQ-Plot above, it appears that the data is normally distributed making it possible to continue on with the paired samples t test.

pander(with(KidsFeet, t.test(KidsFeet$length, KidsFeet$width3, paired = TRUE, mu = 0, alternative = "two.sided", conf.level = 0.95)), caption = "Paired Samples t Test")
Paired Samples t Test (continued below)
Test statistic df P value Alternative hypothesis
-11.53 38 5.668e-14 * * * two.sided
mean of the differences
-2.254

\[ p = 0.0000 < \alpha \] The p-value is less than our significance level, \(\alpha\), so we reject the null hypothesis

Interpretation

Due to the p-value being lower than the significance level, there is sufficient evidence to conclude that the length of the foot is not equivalent to 3 times the width of the foot. Because this sampling of data was a convinience sample we can’t not make any solid conclusions.

Because of this test we see that most of the data is actually close to 2 maybe 2.5.I would probably run this test one more time and see if 2.5 would work better than 3.