Question

For Exercises 5 through \(20,\) perform these steps. 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. A researcher wishes to test the effects of a pill on a person's appetite. Twelve randomly selected subjects are allowed to eat a meal of their choice, and their caloric intake is measured. The next day, the same subjects take the pill and eat a meal of their choice. The caloric intake of the second meal is measured. The data are shown here. At \(\alpha=0.02,\) can the researcher conclude that the pill had an effect on a person's appetite? $$ \begin{array}{l|ccccccc} \text { Subject } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \text { Meal 1 } & 856 & 732 & 900 & 1321 & 843 & 642 & 738 \\ \hline \text { Meal 2 } & 843 & 721 & 872 & 1341 & 805 & 531 & 740 \end{array} $$ $$ \begin{array}{l|rrrrr} \text { Subject } & 8 & 9 & 10 & 11 & 12 \\ \hline \text { Meal 1 } & 1005 & 888 & 756 & 911 & 998 \\ \hline \text { Meal 2 } & 900 & 805 & 695 & 878 & 914 \end{array} $$

Short answer

The pill has a significant effect on appetite if the test statistic exceeds ±2.718, leading to a rejection of the null hypothesis.

Step-by-step solution

State the Hypotheses

We will begin by stating the null and alternative hypotheses. The null hypothesis states that there is no difference in caloric intake, thus the pill has no effect: \[ H_0: \mu_d = 0 \]The alternative hypothesis states that there is a difference in caloric intake after taking the pill: \[ H_1: \mu_d eq 0 \]The claim is that the pill has an effect on appetite.

Find the Critical Value(s)

Since we are using a significance level of \( \alpha = 0.02 \) and a two-tailed test, we will find the critical value for a t-distribution with \( n-1 = 11 \) degrees of freedom. Using a t-table or calculator, the critical t-values for \( \alpha = 0.02 \) and 11 degrees of freedom are approximately \( \pm 2.718 \).

Compute the Test Value

First, calculate the differences between each pair (Meal 1 - Meal 2):- Subject 1: \( 856 - 843 = 13 \)- Subject 2: \( 732 - 721 = 11 \)- Subject 3: \( 900 - 872 = 28 \)- Subject 4: \( 1321 - 1341 = -20 \)- Subject 5: \( 843 - 805 = 38 \)- Subject 6: \( 642 - 531 = 111 \)- Subject 7: \( 738 - 740 = -2 \)- Subject 8: \( 1005 - 900 = 105 \)- Subject 9: \( 888 - 805 = 83 \)- Subject 10: \( 756 - 695 = 61 \)- Subject 11: \( 911 - 878 = 33 \)- Subject 12: \( 998 - 914 = 84 \)Calculate the mean and standard deviation of these differences.\( \bar{d} = \frac{556}{12} = 46.33 \)\( s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}} \)Now calculate the t-test value:\[ t = \frac{\bar{d}}{s_d / \sqrt{n}} \]

Make the Decision

Since the calculated t-value is compared to the critical t-value \( \pm 2.718 \). If the test value falls within the critical region, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Summarize the Results

By analyzing the decision from the previous step, we summarize whether the researcher has enough evidence to claim that the pill had an effect on a person's appetite. This conclusion should relate back to the claim about the pill being effective or not in suppressing appetite.

Key concepts

t-distribution

The t-distribution is a probability distribution used when the sample size is small, typically less than 30, and the population standard deviation is unknown. It helps us determine where the sample mean falls in relation to a population mean. The t-distribution is similar to the normal distribution but has thicker tails, indicating more variability.
In hypothesis testing, the t-distribution is used to calculate the critical value and the test statistic. This distribution accounts for additional uncertainty because of the limited sample data, making it appropriate for smaller datasets like the one in the exercise, where we examined 12 subjects.

• The t-distribution becomes nearly identical to the normal distribution as the sample size increases.
• It is defined by its degrees of freedom, calculated as the sample size minus one (n-1).
• In our case, with 11 degrees of freedom, we utilized the t-distribution to find our critical values for decision-making.

null hypothesis

The null hypothesis, denoted as \( H_0 \), is a statement assumed to be true at the start of the hypothesis test. It usually suggests no effect or no difference between the groups or treatments being compared. The null hypothesis serves as a baseline for statistical testing.
For the pill effect study, the null hypothesis was that there is no difference in caloric intake after taking the pill. Mathematically, we expressed it as \( H_0: \mu_d = 0 \). This implies that the pill does not change the appetite of individuals.

• The null hypothesis provides a starting point and is tested for rejection in favor of the alternative hypothesis.
• If our calculated test statistic falls within the critical region, the null hypothesis is rejected.
• In many scientific studies, the null hypothesis reflects the status quo or an assumption of no effect.

alternative hypothesis

The alternative hypothesis, symbolized as \( H_1 \), proposes that there is an effect or a difference. It is what researchers seek to prove throughout the hypothesis testing process. Unlike the null hypothesis, the alternative hypothesis indicates the expected change subject to empirical validation.
In this exercise, the alternative hypothesis was \( H_1: \mu_d eq 0 \), suggesting that taking the pill does affect caloric intake. This reflects the researcher's claim that the pill has some impact on appetite.

• The alternative hypothesis is accepted when there is sufficient evidence to reject the null hypothesis.
• It is crucial because it defines the direction of the test (e.g., one-tailed or two-tailed tests).
• Our test was two-tailed, indicating that any change (either increase or decrease in caloric intake) supports the alternative hypothesis.

significance level

The significance level, denoted as \( \alpha \), represents the probability of rejecting the null hypothesis when it is true. This measure helps control the risk of making a Type I error in hypothesis testing. The significance level defines the threshold for identifying statistical significance in our test results.
In the context of the exercise, a significance level of \( \alpha = 0.02 \) was set. This implies a 2% risk of concluding that the pill affects caloric intake when it actually doesn't.

• Common significance levels include 0.05, 0.01, and 0.10, but the choice depends on how stringent the researcher wants to be.
• The lower the \( \alpha \), the stronger the evidence required to reject the null hypothesis.
• A significance level helps determine the critical region against which the test statistic is compared.

critical value

The critical value forms part of the decision rule in hypothesis testing. It is the threshold that the calculated test statistic must exceed to reject the null hypothesis. Critical values are determined based on the significance level and the t-distribution.
For our study with 12 subjects, we found critical t-values of approximately \( \pm 2.718 \) for a two-tailed test at \( \alpha = 0.02 \) with 11 degrees of freedom. Comparing the test statistic against these critical values helps decide whether to accept or reject the null hypothesis.

• If the test statistic lies beyond the critical value in the extreme tails, the null hypothesis is rejected.
• Critical values also determine the boundaries of the acceptance region within a confidence interval.
• Proper selection of critical values controls both Type I and Type II errors in hypothesis testing.