Question

The authors of the paper "Ultrasound Techniques Applied to Body Fat Measurement in Male and Female Athletes" (Journal of Athletic Training [2009]: \(142-147\) ) compared two different methods for measuring body fat percentage. One method uses ultrasound and the other method uses X-ray technology. The accompanying table gives body fat percentages for 16 athletes using each of these methods (a subset of the data given in a graph that appeared in the paper). For purposes of this exercise, you can assume that the 16 athletes who participated in this study are representative of the population of athletes. Do these data provide convincing evidence that the mean body fat percentage measurement differs for the two methods? Test the appropriate hypotheses using \(\alpha=0.05\). $$ \begin{array}{crr} \text { Athlete } & \text { X-ray } & \text { Ultrasound } \\ \hline 1 & 5.00 & 4.75 \\ 2 & 7.00 & 3.75 \\ 3 & 9.25 & 9.00 \\ 4 & 12.00 & 11.75 \\ 5 & 17.25 & 17.00 \\ 6 & 29.50 & 27.50 \\ 7 & 5.50 & 6.50 \\ 8 & 6.00 & 6.75 \\ 9 & 8.00 & 8.75 \\ 10 & 8.50 & 9.50 \\ 11 & 9.25 & 9.50 \\ 12 & 11.00 & 12.00 \\ 13 & 12.00 & 12.25 \\ 14 & 14.00 & 15.50 \\ 15 & 17.00 & 18.00 \\ 16 & 18.00 & 18.25 \end{array} $$

Short answer

The final decision to reject or accept the null hypothesis depends on the computed t-statistic and P-value. If the absolute t-statistic is greater than the critical t-value and the P-value is less than \(\alpha=0.05\), there is convincing evidence that the mean body fat percentage measurements from the two methods significantly differ. Conversely, if these conditions are not met, there isn't sufficient evidence to assert that the mean measurements from the two methods significantly differ.

Step-by-step solution

Formulate the Null and Alternative Hypotheses

The null hypothesis \(H_0\) assumes that there is no significant difference between the means of the two observation sets. Therefore, it assumes that the difference between the X-ray and Ultrasound measurements would be 0 on average. The alternative hypothesis \(H_A\) posits that there is a significant difference, hence, the mean difference is not 0. Formally, \(H_0: \mu_{diff} = 0\) and \(H_A: \mu_{diff} \neq 0\).

Calculate the Paired Differences and their Mean

Subtract the Ultrasound measurements from the X-ray measurements for each athlete. After obtaining the differences, calculate their mean, \(\bar{d}\). The mean of the differences represents the average discrepancy between the two methods.

Calculate the Standard Deviation of Differences

Calculate the standard deviation, s, of the previously calculated differences. This can be done using the standard deviation formula for a sample.

Conduct the T-Test

Calculate the t-statistic using the formula: \(t = \frac{\bar{d}}{s/\sqrt{n}}\), where \(\bar{d}\) is the mean difference, s is the standard deviation of the differences, and n is the total number of pairs. This t-statistic will help measure the size of the difference relative to the variability in the data.

Compare t-statistic with critical t-value

The t-distribution table is used to find the critical t-value for a two-tailed test with (n-1) degrees of freedom and a significance level of \(\alpha=0.05\). If the absolute t-statistic is greater than the critical t-value, we reject the null hypothesis.

Calculate the P-Value

The P-value is the smallest level of significance at which the null hypothesis would be rejected. It is obtained utilizing the t-distribution. If the P-value is less than \(\alpha=0.05\), we reject the null hypothesis.

Key concepts

Paired Sample T-Test

The paired sample t-test is a statistical procedure used to determine whether there is a significant difference between two sets of paired data. Its usage in sports statistics can be particularly valuable when assessing the effectiveness of different measurement techniques or treatments on the same subjects. For example, if researchers wish to compare two methods for measuring body fat percentage in athletes—such as ultrasound and X-ray—this test is an appropriate choice.

In our exercise, we started by formulating hypotheses. The null hypothesis posited no significant difference between the two techniques, while the alternative hypothesis suggested a difference existed. We then computed the mean of the differences between paired measurements to assess the overall discrepancy in the techniques. The calculation of the standard deviation followed, shedding light on the variability of the differences.

To conclude the test, we compared a calculated t-statistic, which measures the mean difference relative to the variability in the sample, against a critical value from the t-distribution table. If our computed t-statistic had been larger in magnitude than this critical value, or if the P-value had been less than 0.05, we would have evidence to reject the null hypothesis, suggesting a significant difference between the two body fat measurement methods.

Body Fat Percentage Measurement

Measuring body fat percentage accurately is crucial for athletes as it can influence their training and nutrition plans. In the context of our exercise, two different methods were compared: X-ray and ultrasound. Each technique comes with its advantages and limitations regarding accuracy, cost, and ease of use.

The X-ray method, often referred to as DXA (Dual-Energy X-ray Absorptiometry), is considered highly accurate but also more expensive and less accessible due to the equipment required. On the other hand, the ultrasound method is more portable and can be less costly, though it may sometimes be less precise due to operator dependency and variability in measurement technique.

To ascertain which method gives a more consistent and accurate representation of an athlete's body fat percentage, a paired sample t-test provides an excellent statistical approach. By measuring the same athletes with both methods and analyzing the data, researchers can understand whether the difference in measurements is due to chance or significant discrepancies between the two techniques.

Statistical Significance

Statistical significance plays a pivotal role in hypothesis testing, indicating whether the difference observed in a study is likely due to something other than mere random chance. Essentially, it helps researchers to make inferences about the population based on sample data.

In the context of our exercise, we used a significance level of \(\alpha=0.05\), which means we're willing to accept a 5% probability of wrongly rejecting the null hypothesis (Type I error). If the P-value calculated from the t-test is less than this \(\alpha\) level, we have sufficient evidence to conclude that the difference between body fat percentage measurements by X-ray and ultrasound is statistically significant.

Understanding statistical significance is vital for interpreting sports statistics and, indeed, any scientific data. It informs whether an observed effect is strong enough to warrant a deeper look or whether it could just be due to random variation within the sample. The careful application of these principles helps to ensure that conclusions drawn from studies are sound and reliable.