Question

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. $$ \begin{array}{l|rrrrrrrr} \text { Student } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \text { Before } & 67 & 72 & 80 & 70 & 78 & 82 & 69 & 75 \\ \hline \text { After } & 68 & 70 & 76 & 65 & 75 & 78 & 65 & 68 \end{array} $$

Short answer

Reject \(H_0\); sufficient evidence to conclude students improved their times.

Step-by-step solution

State the Hypotheses

We need to determine if there is a significant improvement in times after consuming the energy drink. The null hypothesis \(H_0\) assumes no improvement, meaning the mean difference in times (Before - After) is zero. The alternative hypothesis \(H_1\) assumes an improvement, meaning the mean difference is greater than zero. Mathematically, let \(d_i = \text{Before}_i - \text{After}_i\).- Null Hypothesis \(H_0: \mu_d = 0\) - Alternative Hypothesis \(H_1: \mu_d > 0\) The claim is that students did better the second time, which corresponds to \(H_1\).

Find the Critical Value

Since the alternative hypothesis is \(H_1: \mu_d > 0\), this is a right-tailed test. Given \(\alpha = 0.05\) and \(n = 8\), we use a t-distribution with \(n-1 = 7\) degrees of freedom. From the t-distribution table, the critical value \(t_c\) corresponding to \(\alpha = 0.05\) for 7 degrees of freedom is approximately 1.895.

Compute the Test Value

First, we calculate the differences for each student \(d_i = \text{Before}_i - \text{After}_i\). Then find the mean \(\bar{d}\) and standard deviation \(s_d\) of these differences.- Differences: \([67-68, 72-70, 80-76, 70-65, 78-75, 82-78, 69-65, 75-68] = [-1, 2, 4, 5, 3, 4, 4, 7]\)- Mean difference: \(\bar{d} = \frac{1}{8}(-1 + 2 + 4 + 5 + 3 + 4 + 4 + 7) = 3.5\) - Standard deviation: \(s_d \approx 2.44\)Compute the test statistic using: \[ t = \frac{\bar{d} - \mu_0}{s_d/\sqrt{n}} = \frac{3.5 - 0}{2.44/\sqrt{8}} \approx 4.07 \]

Make the Decision

Compare the computed test statistic \(t = 4.07\) with the critical value \(t_c = 1.895\). Since \(4.07 > 1.895\), we reject the null hypothesis \(H_0\).

Summarize the Results

We have sufficient evidence at \(\alpha = 0.05\) to conclude that the students improved their times after consuming the energy drink. This suggests that the energy drink may have a positive effect, though other factors might also contribute to the observed improvement, such as familiarity with the course.

Key concepts

t-distribution

The t-distribution is a probability distribution that is used frequently in statistics, especially when dealing with small sample sizes, typically below 30, or when the population standard deviation is unknown. It resembles the normal distribution but has heavier tails, which means it is more prone to producing values far from its mean. This property allows for greater variability in results when handling smaller datasets.
The t-distribution is characterized by its degrees of freedom (df), which are determined by the sample size. In our obstacle course exercise, the sample size is 8, leading to 7 degrees of freedom (since degrees of freedom = sample size - 1).
We use the t-distribution in hypothesis testing when calculating the critical value, as it helps determine the point at which we reject or fail to reject a null hypothesis. The versatility of the t-distribution makes it essential in situations where precise parameters of the population are unknown and must be estimated from sample data.

null hypothesis

In hypothesis testing, the null hypothesis denotes the assumption that there is no effect or no difference. It's the default position and forms the basis for statistical analysis. We denote the null hypothesis by \(H_0\). In the context of our exercise, the null hypothesis states that there is no improvement in the students' obstacle course times after consuming the energy drink. Mathematically, it is represented as \(H_0: \mu_d = 0\), where \(\mu_d\) is the mean of the differences between the before and after times.
Accepting the null hypothesis implies that any observed difference in times is purely due to random variation and is not statistically significant. The objective of hypothesis testing is to assess whether sufficient evidence exists to reject \(H_0\) in favor of the alternative hypothesis. This forms the backbone of statistical inquiry by providing a methodological way to test assumptions quantitatively.

alternative hypothesis

The alternative hypothesis is the statement we aim to support through evidence. Denoted as \(H_1\) or \(H_a\), it represents the outcome that contradicts the null hypothesis. It suggests that there is a real effect or difference. In our exercise, the alternative hypothesis asserts that the students’ performance improved after drinking the energy beverage. This can be written mathematically as \(H_1: \mu_d > 0\).
By supporting \(H_1\), we claim that the observed improvements in obstacle course times are not due to random variation alone. Instead, some factors, possibly the energy drink, contributed to the enhanced performance. The alternative hypothesis is critical because it is usually aligned with the goal of the research or experiment. In academia and applied fields, designing a study often begins with a clear alternative hypothesis.

test statistic

The test statistic is a standardized value calculated from sample data during a hypothesis test. It determines how far our sample statistic is from the null hypothesis parameter, relative to the expected standard error. In our obstacle course example, the test statistic is computed using the formula:

• \( t = \frac{\bar{d} - \mu_0}{s_d/\sqrt{n}} \)

where \(\bar{d}\) is the mean of the differences, \(\mu_0\) is the mean difference under the null hypothesis (which is 0), \(s_d\) is the standard deviation of the differences, and \(n\) is the sample size.
The computed test value in this case, \(t \approx 4.07\), informs us about the relative size and direction of our sample statistic compared to what is expected under the null hypothesis. A large test statistic value suggests that the sample data is very different from the null hypothesis assumption, hence providing grounds for potentially rejecting \(H_0\).

critical value

The critical value is a key threshold in hypothesis testing that determines the boundary of acceptance or rejection of the null hypothesis. It is derived from the probability distribution based on the sample size and the significance level \(\alpha\). For a right-tailed test like the one in our exercise, at \(\alpha=0.05\) and with 7 degrees of freedom, the critical value is approximately 1.895.
In testing, the computed test statistic is compared against this critical value. If the test statistic exceeds the critical value, the null hypothesis \(H_0\) is rejected in favor of the alternative hypothesis. For this exercise, our test statistic \(t \approx 4.07\) is larger than the critical value of 1.895, thus leading to a rejection of \(H_0\), suggesting significant improvement in student performance after consuming the energy drink. Understanding critical values is essential as it aids in making informed decisions based on statistical evidence.