Part 1)
A physical therapists wishes to determine whether an exercise program increases flexibility. He measures the flexibility (in inches) of 12 randomly selected subjects both before and after an intensive eight-week training program and obtains the following data:
Before: 18.5, 21.5, 16.5, 21, 20, 15, 19.75, 15.75, 18, 22, 15, 20.5
After: 19.25, 21.75, 16.5, 20.5, 22.25, 16, 19.5, 17, 19.25, 19.5, 16.5, 20
Test the claim that the flexibility before the exercise program is less than the flexibility after the exercise program at a significance level of 0.01. Note: the data is approximately normal with no outliers.
Part 2)
Using the data from the previous problem, construct a 95% confidence interval about the population mean difference. Interpret your result.
Any help would be greatly appreciated. Thank you for your time.
Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
Use this data with the information in my response to your previous post.
Part 1: Hypothesis Testing
Step 1: State the null and alternative hypotheses
Null hypothesis (H0): The flexibility before the program is not less than the flexibility after the program.
Alternative hypothesis (Ha): The flexibility before the program is less than the flexibility after the program.
Step 2: Determine the significance level
The significance level, represented by α, is given as 0.01.
Step 3: Identify the test statistic
Since we have small sample sizes and no population standard deviation, we will use the t-test for paired samples.
Step 4: Formulate the decision rule
We will compare the calculated test statistic with the critical value from the t-distribution at the given significance level.
Step 5: Calculate the test statistic
To calculate the test statistic, we will need to compute the paired differences and then find the mean difference (d bar), the sample standard deviation (s), and the standard error of the mean difference (SE).
Before: 18.5, 21.5, 16.5, 21, 20, 15, 19.75, 15.75, 18, 22, 15, 20.5
After: 19.25, 21.75, 16.5, 20.5, 22.25, 16, 19.5, 17, 19.25, 19.5, 16.5, 20
Differences: (After - Before)
0.75, 0.25, 0, -0.5, 2.25, 1, -0.25, 1.25, 1.25, -2.5, 1.5, -0.5
Mean difference (d bar): 0.625
Sample standard deviation (s): 1.271
Standard error (SE) = s/√n
n = 12 (since there are 12 pairs)
SE = 1.271/√12 = 0.3677
Step 6: Calculate the test statistic t
t = (d bar - 0)/SE
t = (0.625 - 0)/0.3677 = 1.699
Step 7: Determine the critical value
Since our alternative hypothesis states that the flexibility before the program is less than the flexibility after, it is a one-tailed test. We need to find the critical value for a one-tailed test with a significance level of 0.01 and 11 degrees of freedom (n-1).
Using a t-table or calculator, the critical value is approximately -2.718 (df = 11, α = 0.01, one-tailed test).
Step 8: Make a decision
If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Since the absolute value of the test statistic (1.699) is greater than the critical value (-2.718), we fail to reject the null hypothesis.
Conclusion: Based on the given data, we fail to provide sufficient evidence to support the claim that the flexibility before the exercise program is less than the flexibility after the exercise program at a significance level of 0.01.
Part 2: Confidence Interval
Step 1: Identify the sample mean difference and standard error (already calculated)
Mean difference (d bar): 0.625
Standard error (SE): 0.3677
Step 2: Determine the confidence level
The confidence level, represented by (1 - α), is 95%. Therefore, α = 0.05.
Step 3: Calculate the margin of error
The margin of error is given by multiplying the critical value from the t-distribution with the standard error.
Using the t-table or calculator, the critical value for a 95% confidence level with 11 degrees of freedom (n-1) is approximately 2.201.
Margin of error = critical value * SE = 2.201 * 0.3677 = 0.8086
Step 4: Construct the confidence interval
The confidence interval can be calculated as:
Lower bound = d bar - margin of error
Upper bound = d bar + margin of error
Lower bound = 0.625 - 0.8086 = -0.1836
Upper bound = 0.625 + 0.8086 = 1.4336
Step 5: Interpret the result
We are 95% confident that the true mean difference in flexibility before and after the exercise program lies between -0.1836 and 1.4336 inches. This means that, on average, the exercise program may lead to an increase in flexibility by 0.1836 to 1.4336 inches.
Part 1: To test the claim that the flexibility before the exercise program is less than the flexibility after the exercise program at a significance level of 0.01, we can use a paired t-test. Here are the steps to follow:
Step 1: State the hypotheses:
- Null hypothesis (H0): The mean flexibility before the exercise program is not less than the mean flexibility after the exercise program.
- Alternative hypothesis (H1): The mean flexibility before the exercise program is less than the mean flexibility after the exercise program.
Step 2: Calculate the differences between the before and after measurements for each subject.
Differences: 0.75, 0.25, 0, -0.5, 2.25, 1, 0.75, 1.25, 1.25, -0.5, 1.5, -0.5
Step 3: Calculate the mean difference, sample standard deviation of the differences, and the standard error.
Mean difference (d̄): sum of differences / number of differences = 1 / 12 = 0.0833
Sample standard deviation of the differences (s): √(sum of (each difference - mean difference)² / (number of differences - 1))
s = √[(0.75-0.0833)²+(0.25-0.0833)²+(0-0.0833)²+(-0.5-0.0833)²+(2.25-0.0833)²+(1-0.0833)²+(0.75-0.0833)²+(1.25-0.0833)²+(1.25-0.0833)²+(-0.5-0.0833)²+(1.5-0.0833)²+(-0.5-0.0833)²] / (12-1) = 0.9101
Standard error (SE): s / √(number of differences) = 0.9101 / √12 = 0.2627
Step 4: Calculate the t-value.
t = (mean difference - hypothesized mean difference) / SE = (0.0833 - 0) / 0.2627 = 0.3167
Step 5: Determine the critical value and compare it with the calculated t-value.
Since our alternative hypothesis is that the mean flexibility before the exercise program is less than the mean flexibility after the exercise program, we will perform a one-tailed test.
At a significance level of 0.01, with 11 degrees of freedom (12 subjects - 1), the critical t-value is -2.718.
Since the calculated t-value (0.3167) is greater than the critical t-value (-2.718), we fail to reject the null hypothesis. There is not enough evidence to support the claim that the flexibility before the exercise program is less than the flexibility after the exercise program.
Part 2: To construct a 95% confidence interval about the population mean difference, we can use the same set of differences calculated in Part 1. Here are the steps to follow:
Step 1: Calculate the mean difference and the standard error of the mean difference.
Mean difference (d̄): 0.0833 (calculated in Part 1)
Standard error of the mean difference (SE): s / √(number of differences) = 0.9101 / √12 = 0.2627 (calculated in Part 1)
Step 2: Calculate the margin of error (ME).
ME = t-value * SE
For a 95% confidence interval, with 11 degrees of freedom (12 subjects - 1), the t-value is 2.718.
ME = 2.718 * 0.2627 = 0.7140
Step 3: Calculate the lower and upper bounds of the confidence interval.
Lower bound = Mean difference - ME = 0.0833 - 0.7140 = -0.6307
Upper bound = Mean difference + ME = 0.0833 + 0.7140 = 0.7973
Step 4: Interpret the result.
The 95% confidence interval for the population mean difference between the flexibility before and after the exercise program is (-0.6307, 0.7973). This means that we are 95% confident that the true population mean difference falls between these two values. Since the interval includes zero, it suggests that the exercise program may not have had a statistically significant effect on flexibility.