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Several retired bicycle racers are coaching a large group of young prospects. They randomly select seven of their riders to take part in a test of the effectiveness of a new dietary supplement that is supposed to increase strength and stamina. Each of the seven riders does a time trial on the same course. Then they all take the dietary supplement for 4 weeks. All other aspects of their training program remain as they were prior to the time trial. At the end of the 4 weeks, these riders do another time trial on the same course. The times (in minutes) recorded by each rider for these trials before and after the 4-week period are shown in the following table. $$ \begin{array}{l|lllrrll} \hline \text { Before } & 103 & 97 & 111 & 95 & 102 & 96 & 108 \\ \hline \text { After } & 100 & 95 & 104 & 101 & 96 & 91 & 101 \\ \hline \end{array} $$ a. Construct a \(99 \%\) confidence interval for the mean \(\mu_{d}\) of the population paired differences, where a paired difference is equal to the time taken before the dietary supplement minus the time taken after the dietary supplement. b. Test at the \(2.5 \%\) significance level whether taking this dietary supplement results in faster times in the time trials.

Step by step solution

01

Calculation of Paired Differences

First, we calculate the paired differences \(d_i = t_{before_i} - t_{after_i}\), where \(t_{before_i}\) and \(t_{after_i}\) are the times before and after the dietary supplement for the \(i-th\) rider respectively.

02

Mean and Standard Deviation of Differences

Now, calculate the mean \(\bar{d} = \frac{1}{n} \sum_{i=1}^{n} d_i\) and standard deviation \(s_d = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (d_i - \bar{d})^2}\) of these differences.

03

Calculation of Confidence Interval

The 99% confidence interval for the mean of the differences is given by \(\bar{d} \pm z \cdot \frac{s_d}{\sqrt{n}}\), where \(z\) is the z-value for a 99% confidence level.

04

Null and Alternative Hypothesis Setting

Set the null hypothesis \(H_0: µ_d = 0\) and the alternative hypothesis \(H_1: µ_d > 0\). Our null hypothesis is that the mean difference is zero, which implies that the supplement brings no change, while the alternative hypothesis is that the mean difference is greater than zero, implying that the supplement leads to faster times.

05

Test Statistic

Calculate the test statistic \(T = \frac{\bar{d} - μ_0}{s_d / \sqrt{n}}\).

06

Conclusion

Compare the test statistic with the critical value for a 2.5% significance level. If the test statistic is greater than the critical value, we reject the null hypothesis.

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Key Concepts

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

Paired Differences

When conducting studies such as the one with the dietary supplement for cyclists, we often deal with paired differences. In this context, a paired difference represents the change in performance for each individual before and after undergoing treatment. Specifically, if a rider clocked 103 minutes before and 100 minutes after taking the supplement, the paired difference would be 3 minutes.
This calculation is crucial because it isolates the effect of the treatment, removing individual variability and focusing purely on the treatment's impact.
In this exercise, the differences are calculated by subtracting the time after the treatment from the time before, which helps in evaluating whether the dietary supplement contributes to reduced time in trials. It simplifies the data into a single series, allowing for straightforward analysis and comparison.

Significance Level

The significance level is a key component when conducting hypothesis tests in statistics. In this exercise, a 2.5% significance level is used. This level indicates the probability of rejecting the null hypothesis when it is actually true.
A lower significance level, like 2.5%, implies a stricter test criterion and reduces the risk of a Type I error—incorrectly concluding that the supplement has an effect. It ensures that if the null hypothesis is rejected, it is done with a high level of confidence in the decision.
Researchers select significance levels depending on the stakes of the investigation. More stringent levels are chosen when the consequences of an incorrect decision are significant.

Null and Alternative Hypothesis

For this study, formulating null and alternative hypotheses is essential. They set up a framework for the statistical test.
- The **null hypothesis** ( H_0) states that the mean of paired differences, μ_d = 0 , meaning that the dietary supplement has no real effect on increasing speed; any observed change is due to random chance.
- The **alternative hypothesis** ( H_1) suggests that μ_d > 0 , meaning that the supplement is effective and results in faster times in trials.
Establishing these hypotheses helps analysts decide whether to accept the current state (null) or look for evidence of change (alternative). This decision is based on statistical tests that compare the collected data against probabilistic expectations.

Mean and Standard Deviation

Calculating the mean and standard deviation of the paired differences is vital for analysis. The mean, ar{d} , provides a central value that represents the average improvement or change due to the dietary supplement.

• The **mean** is calculated as the sum of all paired differences divided by the total number of differences. It gives a basic idea of the average effect of the treatment.
• The **standard deviation** measures the spread or variability of the differences. A smaller standard deviation implies that the results are more consistent among the sample riders.

Understanding both the mean and standard deviation helps in drawing conclusions about the treatment's effectiveness and reliability. High variability may suggest differing responses among subjects, while a small mean can indicate a minimal overall effect.