Question

Braking Reaction Times: Boxplots Use the same data from Exercise 6 and use the same scale to construct a boxplot of the braking reaction times of males and another boxplot for the braking reaction times of females. What do the boxplots suggest?

Short answer

The two box plots are shown below:

The data for the male and female braking reaction times appear to be similar. In general, the male braking reaction times are lower than the female braking reaction times.

Step-by-step solution

Given information

Two samples of braking reaction times of males and females are considered.

Computation of summary statistics for male braking reaction times

The summary statistics for male braking reaction times are computed below:

The minimum value in the sample is equal to is 28.

The maximum value in the sample is 63.

The first quartile is equal to the 25^th percentile.

\({Q_1} = {P_{25}}\)

The formula to calculate the percentile is equal to:

\(L = \frac{k}{{100}} \times n\)

Here, L is the locator of the value

k is the percentile of the value

n is the total number of values

So, for k=25 and n=36, the value of L is equal to:

\(\begin{array}{c}L = \frac{k}{{100}} \times n\\ = \frac{{25}}{{100}} \times 36\\ = 9\end{array}\)

Since L is a whole number, compute the sum of the 9^thand the 10^th observation and divide by 2.

\(\begin{array}{c}{P_{25}} = \frac{{{9^{th}}obs + {{10}^{th}}obs}}{2}\\ = \frac{{36 + 38}}{2}\\ = 37\end{array}\)

Thus, the first quartile of the sample is equal to 37.

Similarly, the second quartile is the 50^th percentile.

\({Q_2} = {P_{50}}\)

So, for k=50 and n=36, the value of L is equal to:

\(\begin{array}{c}L = \frac{k}{{100}} \times n\\ = \frac{{50}}{{100}} \times 36\\ = 18\end{array}\)

Since L is a whole number, compute the sum of the 18^th and the 19^th observation and divide by 2.

\(\begin{array}{c}{P_{50}} = \frac{{{{18}^{th}}obs + {{19}^{th}}obs}}{2}\\ = \frac{{41 + 42}}{2}\\ = 41.5\end{array}\)

Thus, the second quartile/median is equal to 41.5.

Similarly, the third quartile is the 75^th percentile.

\({Q_3} = {P_{75}}\)

So, for k=75 and n=36, the value of L is equal to:

\(\begin{array}{c}L = \frac{k}{{100}} \times n\\ = \frac{{75}}{{100}} \times 36\\ = 37\end{array}\)

Since L is a whole number, compute the sum of the 27^th and the 28^th observation and divide by 2.

\(\begin{array}{c}{P_{75}} = \frac{{{{27}^{th}}obs + {{28}^{th}}obs}}{2}\\ = \frac{{51 + 53}}{2}\\ = 52\end{array}\)

Thus, the third quartile is equal to 52.

Thus, the 5 number summary for the male braking reaction time is as follows:

Minimum Value: 28

First Quartile: 37

Median: 41.5

Third Quartile: 52

Maximum Value: 63

Computation of summary statistics for female braking reaction times

The summary statistics for female braking reaction times are computed below:

The minimum value in the sample is equal to is 22.

The maximum value in the sample is 80.

Note that the value of the locator for each percentile will remain the same as above because the value of n is the same (36).

The first quartile is equal to the 25^th percentile.

\({Q_1} = {P_{25}}\)

\(\begin{array}{c}{P_{25}} = \frac{{{9^{th}}obs + {{10}^{th}}obs}}{2}\\ = \frac{{41 + 43}}{2}\\ = 42\end{array}\)

Thus, the first quartile of the sample is equal to 42.

Similarly, the second quartile is the 50^th percentile.

\({Q_2} = {P_{50}}\)

\(\begin{array}{c}{P_{50}} = \frac{{{{18}^{th}}obs + {{19}^{th}}obs}}{2}\\ = \frac{{55 + 56}}{2}\\ = 55.5\end{array}\)

Thus, the second quartile/median is equal to 55.5

Similarly, the third quartile is the 75^th percentile.

\({Q_3} = {P_{75}}\)

\(\begin{array}{c}{P_{75}} = \frac{{{{27}^{th}}obs + {{28}^{th}}obs}}{2}\\ = \frac{{66 + 67}}{2}\\ = 66.5\end{array}\)

Thus, the third quartile is equal to 66.5

Thus, the 5 number summary for the female braking reaction time is as follows:

Minimum Value: 22

First Quartile: 42

Median: 55.5

Third Quartile: 66.5

Maximum Value: 80

Construction of Box Plots

Now using the five-number summary statistics, construct the box plots for female braking reaction times and male braking reaction times on the same scale.

Take the values of the braking reaction times on the vertical scale

The boxplots are shown below:

Analysis of box plots

The following features can be observed:

• The minimum value of the braking reaction time is ower for women than men.
• The maximum value of the braking reaction time is higher for women than men.
• The width of the boxplot corresponding to women is greater than the width of the boxplot corresponding to men. This implies that there is a greater variability present in the breaking times of women as compared to that of men.
• For males, the median of the braking reaction time lies towards the bottom of the boxplot. Thus, the data appears to be positively skewed (skewed to the right).

• For females, the median of the braking reaction time lies in the centre. Thus, the distribution is symmetric.

Therefore, overall there are no significant differences between the two samples.

The braking reaction times of males are generally lower than the braking reaction times of females.