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A physical trainer has four workouts that he recommends for his clients. The workouts have been designed so that the average maximum heart rate achieved is the same for each workout. To test this design he randomly selects twenty people and randomly assigns five of them to use each of the workouts. During each workout, he measures the maximum heart rate in beats per minute with the following results.

Maximum Heart Rates (Beats per Minute)
Workout #1 Workout #2 Workout #3 Workout #4
172 184 178 170
182 169 166 194
179 185 160 196
154 165 174 192
155 178 179 182

Requird:
a. Can the physical trainer conclude that there is a difference among the average maximum heart rates which are achieved during the four workouts?
b. Find the value of the test statistic to test for a difference between the workouts.

Answer :

Solution :

ANOVA : Single factor

Groups         [tex]$\text{count}$[/tex]    [tex]$\text{Sum}$[/tex]     [tex]$\text{Average}$[/tex]   [tex]$\text{Variance}$[/tex]

Workout 1        5       842      168.4         174.3

Workout 2       5       881       176.2         79.7

Workout 3       5       857       171.4          66.8

Workout 4       5       934       186.8         117.2

ANNOVA

Source of Variance    SS       df      MS           F            P-value        F crit

Between groups      976.2     3    325.4  2.971689   0.063098     3.238872

Within groups           1752     16    109.5

Total                        2728.2    19

The null hypothesis is : [tex]$u_1=u_2=u_3=u_4$[/tex]

Alternative hypothesis : At least one of the [tex]$u$[/tex] is not equal.

The significance level = 0.05

Analyzing the sample data:

F statistics is given by :

[tex]$F=\frac{MSB}{MSE}$[/tex]

  = 2.97

[tex]$F_{critical}=3.24$[/tex]

The P-value= 0.06

The result : Since the P-value of 0.06 is greater than the significance level (0.05), hence we fail to reject the null hypothesis.

So we cannot conclude that there is no difference between the average maximum heart rates that are achieved during the four workouts.

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