Question

Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on monday versus wednesday. for 16 cars observed on monday, the average speed was 59.4 mph, with a standard deviation of 3.7 mph. for 20 cars observed on wednesday, the average speed was 56.3 mph, with a standard deviation of 4.4 mph. at the 0.05 level, and assuming normal populations, can we conclude that the average speed for all vehicles was higher on monday than on wednesday?

Answer 1

Hi! 

To compare this two sets of data, you need to use a t-student test:

You have the following data:

-Monday n1=16; x̄1=59,4 mph; s1=3,7 mph

-Wednesday n2=20;  x̄2=56,3 mph; s2=4,4 mph

You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

$t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510$

To calculate the degrees of freedom you need to use the following equation:

$df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89$≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.