Problem : A swimming instructor wants to prove that the swimming speed of an athlete increases if the athelete performs some specific exercises before the swim. He undertakes an experiment with 16 participants and randomly assignes 8 participants to each team. For team A he recommends some common exercises and for team B he recommends some specific exercises. The results of the experiment are below.
Team A (Speed) – 10 12 11 16 13 9 15 6
Team B (Speed) – 11 12 13 11 8 14 7 7
Use the t test to find out if the average speed of the two teams is significantly different. Use alpha = 0.05.
Null Hypothesis => A(speed) = B(speed)
Alternate Hypothesis => A(speed) ne B(speed)
We assume unequal variances and therefore use the t-test for unequal variances.
The calculation in R is given below.
We see that the p-value is greater than 0.05 and hence the null hypthesis is not rejected.
What this means is that the special exercises are not so special after all!
Observe the difference in the degree of freedom and the t-value in all three cases.
look at http://janda.org/c10/Lectures/topic07/pairedt-test.htm for an example of the significance of using paired t-test.
ANOVA – one way:
Problem : An independent magazine wants to find out if the tyre life differs for four different tyre manufacturers. (for the same kind of tyres). It randomly selects 10 participants for each tyre make type and compares the mean of the tyre life.
Make A(tyre life in months) – 12 17 23 20 18 10 30 12 23 20
Make B(tyre life in months) – 10 23 19 30 13 15 18 17 20 22
Make C(tyre life in months) – 11 34 12 20 33 20 18 19 12 17
Make D(tyre life in months) – 10 12 34 23 27 18 15 17 10 15
Null Hypothesis – All means are equal.
Alternate Hypothesis – Atleast one of the means is different.
The results of the analysis are as given below.
The p-value is greater than 0.05 and hence the null hypthesis is not rejected.
What this means is the the shelf life of the tyre is not dependent on the make.
More Detailed examples of ANOVA can be found here: