/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Some sports that involve a signi... [FREE SOLUTION] | 91Ó°ÊÓ

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Some sports that involve a significant amount of running, jumping, or hopping put participants at risk for Achilles tendon injuries. A study in The American Journal of Sports Medicine looked at the diameter (in \(\mathrm{mm}\) ) of the injured tendons for patients who participated in these types of sports activities. \({ }^{6}\) Suppose that the Achilles tendon diameters in the general population have a mean of 5.97 millimeters \((\mathrm{mm})\). When the diameters of the injured tendon were measured for a random sample of 31 patients, the average diameter was \(9.80 \mathrm{~mm}\) 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 Achilles tendon injuries is greater than \(5.97 \mathrm{~mm}\) ? Test at the \(5 \%\) level of significance.

Short Answer

Expert verified
Answer: Yes, based on the hypothesis test at the 5% level of significance, we have sufficient evidence to conclude that the average diameter of the Achilles tendon in patients with injuries is greater than 5.97 mm.

Step by step solution

01

State the null and alternative hypothesis

The null hypothesis, \(H_0\), is that the mean diameter of the Achilles tendon in the general population is equal to \(5.97\mathrm{~mm}\). The alternative hypothesis, \(H_a\), is that the mean diameter of the Achilles tendon in patients with injuries is greater than \(5.97\mathrm{~mm}\). Thus, we have: $$H_0: \mu = 5.97$$ $$H_a: \mu > 5.97$$
02

Calculate the test statistic

To calculate the test statistic, we will use the following formula for one-sample t-test: $$t = \frac{\bar{X} - \mu_0}{\frac{s}{\sqrt{n}}}$$ where \(t\) is the test statistic, \(\bar{X}\) is the sample mean, \(\mu_0\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size. In this case, we have: $$t = \frac{9.80 - 5.97}{\frac{1.95}{\sqrt{31}}}$$ Now, substituting the values, we have: $$t \approx 10.77$$
03

Determine the critical value and the rejection region

To determine the critical value for a one-tailed test at the \(5\%\) level of significance, we need to find the \(t\) value for \(95\%\) of the distribution with \(31 - 1 = 30\) degrees of freedom. Using a t-table or calculator, we have: $$t_{0.95, 30} \approx 1.697$$ The rejection region is defined as the area where the test statistic is greater than the critical value. In this case: $$t > 1.697$$
04

Compare the test statistic to the critical value and make a decision

We found that the test statistic, \(t \approx 10.77\), is greater than the critical value, \(t_{0.95, 30} \approx 1.697\). This means that the test statistic is in the rejection region. Therefore, we reject the null hypothesis in favor of the alternative hypothesis.
05

Conclusion

Based on the hypothesis test at the \(5\%\) level of significance, we have sufficient evidence to conclude that the average diameter of the Achilles tendon in patients with injuries is greater than \(5.97 \mathrm{~mm}\).

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Key Concepts

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

Null and Alternative Hypothesis
In hypothesis testing for statistics, the null and alternative hypotheses are two mutually exclusive statements about a population parameter. The null hypothesis (\(H_0\)) is typically a statement of no effect or no difference; it is the hypothesis that researchers strive to nullify or disprove. For example, if we are investigating the average diameter of tendons, the null hypothesis may claim that the average diameter is a certain value, which in this case is 5.97 mm.The alternative hypothesis (\(H_a\text{ or }H_1\functions)), on the other hand, is a statement that indicates the presence of an effect or a difference. In our tendon diameter example, the alternative hypothesis suggests that the average diameter of injured tendons is greater than 5.97 mm. The formulation of these hypotheses is critical as it sets the stage for the entire hypothesis test and determines the direction and the scope of the study. In the case of a one-tailed test, like the one in this example, the alternative hypothesis is directional, pointing specifically to a 'greater than' scenario.
One-Sample t-Test
The one-sample t-test is a statistical procedure used to determine whether the mean of a single sample is significantly different from a known or hypothesized population mean. This method is particularly useful when the sample size is small or when the population standard deviation is unknown. In the case of our exercise, a one-sample t-test is appropriate since we have a sample (injured tendon diameters) we want to compare against a known population mean.

How It's Calculated

The formula for the one-sample t-test is: \[\begin{equation} t = \frac{\bar{X} - \frac{\text{population mean}}{\frac{s}{\text{sample standard deviation}}}}{\text{sample size}^{1/2}}\end{equation}\]The resulting value, known as the test statistic, tells us how many standard deviations the sample mean is from the population mean. If the test statistic falls beyond a critical value, the likelihood that the sample came from the same population with the hypothesized mean decreases, indicating a significant difference.
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to make a decision regarding the null hypothesis. Test statistics can follow different distributions; in the case of the one-sample t-test, it follows a t-distribution, which adjusts for the sample size with degrees of freedom.In the tendon diameter example, with a calculated test statistic of approximately 10.77, the test statistic is very high, indicating that the sample mean is far away from the hypothesized population mean. Such an extreme test statistic result leads us to suspect that the null hypothesis may not be true, as it implies a low probability of observing such a sample mean given that the null hypothesis is true.
Level of Significance
The level of significance, commonly denoted by alpha (\[\begin{equation} \text{sample standard deviation})\end{equation}\], is a threshold used to judge whether a test statistic provides enough evidence to reject the null hypothesis. The level of significance is set by the researcher before conducting the hypothesis test and often takes a value of 5% (\[\begin{equation} \text{sample standard deviation})\end{equation}\] or 1% (\[\begin{equation} \text{sample standard deviation})\end{equation}\]).This value corresponds to the probability of rejecting the null hypothesis when it is actually true (Type I error). In our exercise, a 5% level of significance is used, which implies that there is a 5% chance of concluding that the average diameter of injured Achilles tendons is greater than 5.97 mm when, in fact, it is not. Because the calculated test statistic is greater than the critical value from a t-distribution table, we would reject the null hypothesis at the 5% significance level.

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Most popular questions from this chapter

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