Question

G. W. Marino investigated the variables related to a hockey player's ability to make a fast start from a stopped position. \({ }^{11}\) In the experiment, each skater started from a stopped position and attempted to move as rapidly as possible over a 6-meter distance. The correlation coefficient \(r\) between a skater's stride rate (number of strides per second) and the length of time to cover the 6 -meter distance for the sample of 69 skaters was -.37 . a. Do the data provide sufficient evidence to indicate a correlation between stride rate and time to cover the distance? Test using \(\alpha=.05 .\) b. Find the approximate \(p\) -value for the test. c. What are the practical implications of the test in part a?

Short answer

Yes, there is sufficient evidence to indicate a correlation between a hockey player's stride rate and the time it takes them to cover a 6-meter distance. The test statistic is more extreme than the critical value, and the p-value is less than the significance level, leading to the rejection of the null hypothesis and the conclusion that there is a significant correlation between the two variables.

Step-by-step solution

Hypotheses

The null hypothesis \(H_0\) states that there is no correlation between stride rate and time to cover the distance (i.e., \(ρ = 0\)). The alternative hypothesis \(H_a\) states that there is a correlation between stride rate and time to cover the distance (i.e., \(ρ ≠ 0\)). $$H_0: ρ = 0$$ $$H_a: ρ ≠ 0$$ **Step 2: Calculate the test statistic**

Test Statistic

The test statistic for the correlation coefficient is given by the formula: $$t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$$ where \(r\) is the correlation coefficient, and \(n\) is the sample size. Plug in the given values, \(r=-0.37\) and \(n=69\): $$t = \frac{-0.37\sqrt{69-2}}{\sqrt{1-(-0.37)^2}}$$ Calculate the value of the test statistic: $$t ≈ -3.278$$ **Step 3: Determine the critical value**

Critical Value

For a two-tailed test with a significance level of \(α = 0.05\) and degrees of freedom \((n-2) = 67\), we can use a t-distribution table or a calculator to find the critical value. The critical value is: $$t_{\alpha/2,DF} ≈ ± 1.997$$ **Step 4: Compare the test statistic to the critical value and draw conclusions**

Conclusion

Our test statistic \(t ≈ -3.278\) is more extreme than the critical value \(± 1.997\). Therefore, we reject the null hypothesis \(H_0\) and conclude that there is a significant correlation between stride rate and the time to cover the 6-meter distance. **Step 5: Find the approximate p-value for the test**

p-value

Using a calculator or software, we can find the p-value associated with our test statistic \(t = -3.278\) with degrees of freedom \((n-2) = 67\). The approximate p-value is: $$p ≈ 0.0017$$ **Step 6: Discuss the practical implications of the test in part a**

Practical Implications

The result of the hypothesis test indicates that there is a significant correlation between a hockey player's stride rate and the time it takes them to cover a 6-meter distance. The practical implication is that improving a skater's stride rate could potentially lead to a faster start from a stopped position and thus better performance in a game. However, it's important to note that the correlation coefficient of \(-0.37\) is only moderate, and other factors may also influence a player's performance.

Key concepts

Probability and Statistics

At the core of understanding and analyzing data lies the field of probability and statistics. This branch of mathematics allows us to collect, analyze, interpret, present, and organize data.

Probability measures the likelihood that an event will occur, which we often express as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. Statistics, on the other hand, deals with the data itself — how we gather it, how we analyze it, and how we use it to make informed decisions.

In the context of the exercise regarding hockey players, statistics is used to determine if there's a meaningful relationship between two variables: stride rate and time to cover a distance. By calculating the correlation coefficient and performing hypothesis testing, we can infer if changes in one variable are associated with changes in the other — a fundamental task in statistical analysis.

T-test for Correlation

When it comes to identifying whether a correlation between two variables exists, the t-test for correlation is a pivotal tool. This statistical test determines if the observed correlation coefficient, denoted as 'r', is significantly different from zero — which implies no correlation.

To conduct a t-test, we calculate a test statistic that, under the null hypothesis (assuming no correlation), follows a t-distribution. The formula incorporates the correlation coefficient, the sample size, and adjusts for the number of degrees of freedom.

In the case of the hockey player study, the t-test helps to conclude whether the negative correlation observed (stride rate and time to cover distance) is statistically significant or merely a result of random sampling.

Significance Level

The significance level, often denoted by \( \alpha \), is a threshold we set to determine whether we can reject the null hypothesis. It's the probability of rejecting the null hypothesis when it is, in fact, true — known as a Type I error. A common choice is \( \alpha = 0.05 \) or 5%, which means we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis.

The significance level defines the critical value(s) in our hypothesis testing. When the test statistic falls beyond this threshold, it implies that the observed data is unlikely to have occurred by random chance, thus leading us to reject the null hypothesis. The exercise with the skaters uses a significance level of 5% to test the correlation, showing that the researchers are accepting a 5% risk of concluding there is a substantial relationship when there might not be one.

P-value

The p-value is a pivotal concept in hypothesis testing. It represents the probability of observing a test statistic as extreme as, or more extreme than, the one calculated from the sample data, assuming the null hypothesis is true.

A small p-value, usually less than the chosen significance level \( \alpha \), suggests that such an extreme observed result would be very unlikely under the null hypothesis. Thus, we reject the null hypothesis in favor of the alternative. In the exercise provided, the p-value (approximately 0.0017) is significantly less than the significance level of 0.05, buttressing the conclusion that there is a significant correlation between stride rate and time to cover a 6-meter distance among the skaters.