Question

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Obstacle Course Times An obstacle course was set up on a campus, and 8 randomly selected volunteers were given a chance to complete it while they were being timed. They then sampled a new energy drink and were given the opportunity to run the course again. The "before" and "after" times in seconds are shown. Is there sufficient evidence at \(\alpha=0.05\) to conclude that the students did better the second time? Discuss possible reasons for your results. $$ \begin{array}{l|ccccccc}{\text { Student }} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} \\ \hline \text { Before } & {67} & {72} & {80} & {70} & {78} & {82} & {69} & {75} \\ \hline \text { After } & {68} & {70} & {76} & {65} & {75} & {78} & {65} & {68}\end{array} $$

Short answer

There is enough evidence to conclude that performance improved after consuming the energy drink.

Step-by-step solution

State the Hypotheses

We need to determine if the energy drink has led to better performance (i.e., lower times) on the obstacle course. - Null Hypothesis \( H_0 \): The mean time "before" is equal to the mean time "after". In mathematical terms, \( H_0: \mu_\text{before} = \mu_\text{after} \).- Alternative Hypothesis \( H_1 \): The mean time "before" is greater than the mean time "after", \( H_1: \mu_\text{before} > \mu_\text{after} \), indicating better performance in the second trial. This is a claim test as we want to see if the performance improves (i.e., the time decreases) after the energy drink.

Find the Critical Value

Since \( \alpha = 0.05 \) and we are conducting a one-tailed test for a difference in means (improvement implies a reduction in time), we use the t-distribution table to find the critical value. With 8 students, and thus 7 degrees of freedom ( = 8), the critical value for a one-tailed test at \( \alpha = 0.05 \) is approximately \( t_{critical} = 1.895 \).

Compute the Test Value

First, we calculate the differences between the "before" and "after" times for each student:\[ \text{Differences} = [67-68, 72-70, 80-76, 70-65, 78-75, 82-78, 69-65, 75-68] = [-1, 2, 4, 5, 3, 4, 4, 7] \]Find the mean and standard deviation of these differences.The mean difference \( \bar{d} \) is:\[ \bar{d} = \frac{-1 + 2 + 4 + 5 + 3 + 4 + 4 + 7}{8} = 3.5 \]Calculate the standard deviation \( s_d \) of the differences:\[ s_d = \sqrt{\frac{1}{8-1} \left( (-1 - 3.5)^2 + (2 - 3.5)^2 + (4 - 3.5)^2 + (5 - 3.5)^2 + (3 - 3.5)^2 + (4 - 3.5)^2 + (4 - 3.5)^2 + (7 - 3.5)^2 \right)} approx 2.41 \]The test statistic \( t \) is given by:\[ t = \frac{\bar{d}}{\frac{s_d}{\sqrt{n}}} = \frac{3.5}{\frac{2.41}{\sqrt{8}}} \approx 4.11 \]

Make the Decision

Compare the test statistic with the critical value. Since \( t = 4.11 \) is greater than \( t_{critical} = 1.895 \), we reject the null hypothesis at \( \alpha = 0.05 \). This suggests there is sufficient evidence to conclude that the students performed better (i.e., took less time) after consuming the energy drink.

Summarize the Results

The hypothesis test indicates that there is sufficient evidence at a 5% level of significance to conclude that the energy drink improved students' performance on the obstacle course. Possible reasons could include increased energy or focus, but other factors like learning effect (practice) could also contribute to improved times.

Key concepts

Null Hypothesis

The null hypothesis is an initial claim that states there is no effect or no difference in a given scenario. In hypothesis testing, it's the hypothesis that researchers try to disprove. In our exercise involving the energy drink and obstacle course, the null hypothesis (\( H_0 \)) states that the mean course completion time before drinking the energy supplement is the same as after drinking it. Mathematically, this is written as:
\( H_0: \mu_{\text{before}} = \mu_{\text{after}} \).
This hypothesis assumes that the energy drink has no impact on the students' performance. Testing this hypothesis allows us to objectively determine if any observed differences in performance are significant or could have occurred by chance.

Alternative Hypothesis

The alternative hypothesis offers a different perspective and usually contradicts the null hypothesis. It represents what the researcher aims to prove. In this exercise, we are looking to see if the mean time of running the obstacle course before consuming the energy drink is greater than the time afterward. This can indicate improved performance, meaning the time decreased after drinking the supplement. Formally, the alternative hypothesis (\( H_1 \)) is expressed as:
\( H_1: \mu_{\text{before}} > \mu_{\text{after}} \).
If evidence from the data supports the alternative hypothesis strongly enough, then we can reject the null hypothesis. This indicates there's statistical proof showing the energy drink's impact on the students' times.

t-distribution

The t-distribution is a probability distribution that is used in hypothesis testing, particularly when the sample size is small and the population standard deviation is unknown. In our situation with the 8 students, we're operating with small sample sizes, making the t-distribution suitable for calculating probabilities and critical values.
The shape of the t-distribution resembles the standard normal distribution (bell-shaped) but with heavier tails, meaning it accounts better for data variability in small samples.

• Degrees of Freedom: For our test, the degrees of freedom are calculated as the sample size minus one (\( df = n - 1 \)), so with 8 students, we have 7 degrees of freedom.
• This distribution helps determine the critical value or threshold which our test statistic is compared against.

Using a t-table, given our sample and a significance level (\( \alpha \)) of 0.05, we find the necessary critical value for deciding whether to reject the null hypothesis.

Critical Value

In hypothesis testing, the critical value is the threshold beyond which we reject the null hypothesis. It is derived from the chosen significance level (\( \alpha \)), which is the probability of making a Type I error – rejecting a true null hypothesis.
For this test, we set our significance level at \( \alpha = 0.05 \), which represents a 5% risk of incorrectly rejecting the null hypothesis.

• Using a t-distribution table and 7 degrees of freedom (since we have 8 students), the critical value is approximately 1.895 for a one-tailed test.
• This benchmark means that if our calculated test statistic exceeds the critical value, we have sufficient evidence to reject \( H_0 \).

In this case, our test statistic turned out to be 4.11, which is greater than 1.895. Thus, we reject the null hypothesis, concluding that the energy drink indeed had a significant effect on improving the students' course completion times.