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Chapter 9: Problem 17

Sports and Achilles Tendon Injuries Some sports that involve a significant amount of running, jumping, or hopping put participants at risk for Achilles tendinopathy (AT), an inflammation and thickening of the Achilles tendon. A study in The American Journal of Sports Medicine looked at the diameter (in \(\mathrm{mm}\) ) of the affected tendons for patients who participated in these types of sports activities. Suppose that the Achilles tendon diameters in the general population have a mean of 5.97 millimeters (mm). When the diameters of the affected tendon were measured for a random sample of 31 patients, the average diameter was 9.80 with a standard deviation of \(1.95 \mathrm{~mm}\). Is there sufficient evidence to indicate that the average diameter of the tendon for patients with AT is greater than \(5.97 \mathrm{~mm}\) ? Test at the \(5 \%\) level of significance.

Short Answer

Expert verified
Yes, there is enough evidence to support the claim since the test statistic (t = 9.32) is greater than the critical value (t伪 = 1.697), leading to the rejection of the null hypothesis and the conclusion that the average diameter of the affected tendon for patients with AT is greater than 5.97 mm at a 5% level of significance.

Step by step solution

01

Set up the null and alternative hypotheses

The null hypothesis (\(H_0\)) represents the current belief, which is that the average diameter of the affected tendon for patients with AT is equal to 5.97 mm. The alternative hypothesis (\(H_1\)) represents the claim we want to test; in this case, that the average diameter of the affected tendon for patients with AT is greater than 5.97mm. \(H_0: \mu = 5.97 \mathrm{~mm}\) \(H_1: \mu > 5.97 \mathrm{~mm}\)
02

Calculate the test statistic

The test statistic, or t-score, is calculated using the following formula: \( t = \dfrac{(\bar{x} - \mu)}{(s/\sqrt{n})}\) Plugging in the values for the sample mean (\(\bar{x} = 9.80 \text{ mm}\)), population mean (\(\mu = 5.97 \text{ mm}\)), sample standard deviation (\(s = 1.95 \text{ mm}\)), and sample size (\(n = 31\)): \(t = \dfrac{(9.80 - 5.97)}{(1.95/\sqrt{31})} = 9.32\)
03

Determine the critical value and test region

Since we're conducting a one-tailed t-test at a 5% level of significance, we need to find the critical value \(t_{\alpha}\) associated with 30 degrees of freedom (DF = \(n - 1 = 31 - 1 = 30\)). You can use the t-distribution table or statistical software to find the critical value. The critical value for a t-distribution at 5% significance level and 30 degrees of freedom is 1.697.
04

Compare test statistic to the critical value and make a decision

We have the test statistic \(t = 9.32\), and the critical value \(t_{\alpha} = 1.697\). Since our test statistic is in the test region (\(t > t_{\alpha}\)), we can reject the null hypothesis (\(H_0\)) and conclude that there is sufficient evidence to suggest that the average diameter of the affected tendon for patients with AT is greater than 5.97 mm at a 5% level of significance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

t-distribution
The t-distribution is a key concept in statistics, especially used when dealing with small sample sizes. This distribution helps us make inferences about a population mean when the sample size is small, typically less than 30, or when the population standard deviation is unknown.

Some important characteristics of the t-distribution include:
  • Symmetry around zero, similar to a standard normal distribution.
  • It has heavier tails than the normal distribution, meaning there is a greater probability of values far from the mean, particularly useful when sample size is small or variance is high.
  • The shape of the distribution becomes more like a normal distribution as the sample size increases (central limit theorem).

In the problem, the t-distribution helps in calculating the critical value for the given degrees of freedom, determining whether we should reject the null hypothesis.
Null Hypothesis
The null hypothesis (\(H_0\)) is a fundamental concept in hypothesis testing. It represents the current assumption or a default statement about a population parameter.

In our exercise, the null hypothesis is that the average diameter of the Achilles tendon for patients with AT is equal to the known population mean of 5.97 mm.
The mathematical representation is:
\[H_0: \mu = 5.97 \text{ mm} \]

Essentially, the null hypothesis is the hypothesis that there is no effect or change. Our goal in hypothesis testing is to determine whether there is enough evidence to reject this null hypothesis in favor of an alternative hypothesis. If our test statistic falls into a region of rejection, we can reject the null hypothesis at the given level of significance.
Alternative Hypothesis
The alternative hypothesis (\(H_1\)) is what researchers seek to prove with their data. It's the statement that there is an effect or a change as opposed to the null hypothesis.

In this exercise, the alternative hypothesis posits that the average diameter of the Achilles tendon in patients with AT is greater than 5.97 mm. The mathematical notation for this is:
\[H_1: \mu > 5.97 \text{ mm} \]

Testing this hypothesis helps determine if the average diameter is statistically significantly greater than the general population mean. If the data provides sufficient evidence, the alternative hypothesis is accepted, suggesting a significant effect or difference from the null hypothesis assumption.
Test Statistic
The test statistic is a crucial part of hypothesis testing. It is a standardized value that helps compare the estimated population parameter against a hypothesized value.

For our exercise, the test statistic is calculated using the formula:
\[ t = \dfrac{(\bar{x} - \mu)}{(s/\sqrt{n})} \]
  • \(\bar{x}\) is the sample mean, which is 9.80 mm in this case.
  • \(\mu\) is the mean of the population under the null hypothesis (5.97 mm).
  • \(s\) is the standard deviation of the sample, given as 1.95 mm.
  • \(n\) is the sample size, 31 patients here.

By plugging in these values, we find that the t-statistic is 9.32. This value is compared against the critical value from the t-distribution table to decide if the null hypothesis should be rejected.

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