Question

In: Statistics and Probability

A coach uses a new technique to train gymnasts. 7 gymnasts were randomly selected and their...

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.

Solutions

Expert Solution

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.


Related Solutions

A coach uses a new technique to train gymnasts. 7 gymnasts were randomly selected and their...
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...
A coach uses a new technique to train gymnasts. 7 gymnasts were randomly selected and their...
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. Use a 0.01 significance level to test the claim that the training technique is effective in rising the gymnasts’ scores. Subject A B C D E F G Before 9.6 9.7 9.7 9.1 9.7 9.2 9.1 After 9.7 9.9 9.7 9.3 9.8 9.8 9.3
Question #4 A coach uses a new technique to train gymnasts. 7 gymnasts were randomly selected...
Question #4 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. Before 9.4 9.0 9.2 9.4 9.2 9.3 9.7 After 9.5 9.3 9.5 9.6 9.4 9.5 9.9 Using a 0.10 level of significance and Critical Value method to test the claim that the training technique is effective in raising the gymnasts' scores. (Mean difference is less than zero)
A coach uses a new technique in training middle distance runners. The times for 9 randomly...
A coach uses a new technique in training middle distance runners. The times for 9 randomly selected, independent athletes to run 800 meters before and after this training are shown below. Assume the distribution of differences are normally distributed Before 115.2 120.9 108.0 112.4 107.5 119.1 121.3 110.8 122.3 After 116.0 119.1 105.1 111.9 109.1 115.1 118.5 110.7 120.9 Build and Interpret 90% confidence interval for the true mean difference in 800-meter times.
A coach times 6 randomly selected runners to measure their times early in the week and...
A coach times 6 randomly selected runners to measure their times early in the week and at the end of the week (in minutes). Using the parts below, help the coach run a paired t-test to determine if the runners were faster (had lower times) at the end of the week. Use α = 0.05. Runner Monday time Fridays time 1 26.2 25.8 2 25.2 24.9 3 24.8 24.9 4 24.1 24.6 5 25.9 25.2 6 25.1 24.5 A. State...
For a new study conducted by a fitness magazine, 240 females were randomly selected. For each,...
For a new study conducted by a fitness magazine, 240 females were randomly selected. For each, the mean daily calorie consumption was calculated for a September-February period. A second sample of 230 females was chosen independently of the first. For each of them, the mean daily calorie consumption was calculated for a March-August period. During the September-February period, participants consumed a mean of 2385.3 calories daily with a standard deviation of 206 . During the March-August period, participants consumed a...
For a new study conducted by a fitness magazine,285 females were randomly selected. For each, the...
For a new study conducted by a fitness magazine,285 females were randomly selected. For each, the mean daily calorie consumption was calculated for a September-February period. A second sample of 275 females was chosen independently of the first. For each of them, the mean daily calorie consumption was calculated for a March-August period. During the September-February period, participants consumed a mean of 2383.5 calories daily with a standard deviation of 208. During the March-August period, participants consumed a mean of...
For a new study conducted by a fitness magazine, 270 females were randomly selected. For each,...
For a new study conducted by a fitness magazine, 270 females were randomly selected. For each, the mean daily calorie consumption was calculated for a September-February period. A second sample of 285 females was chosen independently of the first. For each of them, the mean daily calorie consumption was calculated for a March-August period. During the September-February period, participants consumed a mean of 2387.2 calories daily with a standard deviation of 216. During the March-August period, participants consumed a mean...
For a new study conducted by a fitness magazine, 240 females were randomly selected. For each,...
For a new study conducted by a fitness magazine, 240 females were randomly selected. For each, the mean daily calorie consumption was calculated for a September-February period. A second sample of 210 females was chosen independently of the first. For each of them, the mean daily calorie consumption was calculated for a March-August period. During the September-February period, participants consumed a mean of 2385.5 calories daily with a standard deviation of 222 . During the March-August period, participants consumed a...
The following sample observations were randomly selected:
The following sample observations were randomly selected:12345X:213832Y:18132249            a. Not available in Connect.b. Determine the regression equation.(Negative answer should be indicated by a minus sign. Do not round intermediate calculations. Round the final answers to 4 decimal places.)                                           b =  a =Y' =  +  X           c. Determine the value of Y' whenX is 11. (Do not round intermediate calculations. Round the final answer to 4 decimal places.)
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT