In: Statistics and Probability
A coach uses a new technique to train gymnasts. 7
gymnasts were randomly selected and their competition scores were
recorded before and after the training. The results are shown
below
Subject A B
C D E F
G
Before 9.6 9.7 9.7 9.4 9.7 9.5 9.5
After 9.7 9.9 9.7 9.3 9.8 9.8
9.3
Using a 0.01 level of significance, test the claim that the
training technique is effective in raising the gymnasts’
score.
Use the traditional method of hypothesis testing with critical
value t= -3.143.
Solution:
Given: A coach uses a new technique to train gymnasts. 7
gymnasts were randomly selected and their competition scores were
recorded before and after the training. The results are shown
below
Subject A B
C D E F
G
Before 9.6 9.7 9.7 9.4 9.7 9.5 9.5
After 9.7 9.9 9.7 9.3 9.8 9.8
9.3
We have to use a 0.01 level of significance, test the claim that the training technique is effective in raising the gymnasts’ score.
We are also given that: critical value t= -3.143
Thus we use following steps:
Step 1) State H0 and H1:
Vs
where d = before - After
Step 2) Find test statistic
Since the data is taken on same subjects before and after , the data is paired data set. Thus we use paired t test.
Where
Thus we need to make following table:
Subject | Before | After | di = Before -After | di2 |
A | 9.6 | 9.7 | -0.1 | 0.01 |
B | 9.7 | 9.9 | -0.2 | 0.04 |
C | 9.7 | 9.7 | 0 | 0 |
D | 9.4 | 9.3 | 0.1 | 0.01 |
E | 9.7 | 9.8 | -0.1 | 0.01 |
F | 9.5 | 9.8 | -0.3 | 0.09 |
G | 9.5 | 9.3 | 0.2 | 0.04 |
Thus we get:
and
Thus t test statistic is:
Step 3) t critical value = -3.143
Step 4) Decision Rule:
Reject H0, if t test statistic value < t critical value , otherwise we fail to reject H0.
Since t test statistic value = t = -0.880 > t critical value = -3.143, we failed to reject H0.
Step 5) Conclusion:
Since we failed to reject H0, at 0.01 level of significance, there is not sufficient evidence to support the claim that the training technique is effective in raising the gymnasts’ score.