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. Ten college students were selected and asked how many soft drinks they drink over a twoweek period. These students were asked to replace some of the soft drinks with water in order to cut down on the amount of soft drinks that they consumed. At \(\alpha=0.10,\) was there a decrease in the amount of soft drinks consumed over a two-week period? The results are shown. $$ \begin{array}{l|cccccccccc} \text { Student } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } & \text { G } & \text { H } & \text { I } & \text { J } \\ \hline \text { Before } & 6 & 12 & 15 & 20 & 18 & 24 & 9 & 7 & 26 & 21 \\ \hline \text { After } & 8 & 10 & 12 & 17 & 14 & 21 & 11 & 8 & 23 & 17 \end{array} $$

Short answer

There was a significant decrease in soft drink consumption at \( \alpha = 0.10 \).

Step-by-step solution

State the Hypotheses and Identify the Claim

The null hypothesis () is that there is no decrease in the number of soft drinks consumed: \( H_0: \mu_d = 0 \). The alternative hypothesis () is that there is a decrease in consumption: \( H_1: \mu_d > 0 \). The claim is that there is a decrease in the number of soft drinks consumed.

Find the Critical Value

This is a right-tailed test at significance level \( \alpha = 0.10 \). For a paired sample t-test with 9 degrees of freedom, the critical value from the t-distribution table is \( t_{\text{critical}} = 1.383 \).

Compute the Test Value

First, calculate the differences ("After" - "Before") for each student, then find the mean difference \( \bar{d} \) and the standard deviation of the differences \( s_d \).\[ d_i = \text{After}_i - \text{Before}_i = \{-2, -2, -3, -3, -4, -3, 2, 1, -3, -4\} \]\[ \bar{d} = \frac{-19}{10} = -1.9 \]\[ s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}} = \sqrt{\frac{41.9}{9}} \approx 2.15 \]Compute the t-test statistic:\[ t = \frac{\bar{d} - 0}{s_d/\sqrt{n}} = \frac{-1.9}{2.15/\sqrt{10}} \approx -2.80 \]

Make the Decision

Since the calculated t-value (\( -2.80 \)) is less than the critical t-value (\( -1.383 \)) in the context of a right-tailed test (where we compare to \( |t| \)), we reject the null hypothesis.

Summarize the Results

The test results indicate that there is sufficient evidence at \( \alpha = 0.10 \) to support the claim that replacing soft drinks with water resulted in a significant decrease in the number of soft drinks consumed over a two-week period.

Key concepts

Paired Sample t-Test

A paired sample t-test is a statistical method used to determine if there is a significant difference between the means of two related groups. In this exercise, we're evaluating whether the amount of soft drinks consumed decreased when the participants replaced some of these drinks with water. The key principle here is that we have measurements from the same subjects before and after the intervention (in this case, replacing soft drinks with water).

The paired nature means that each participant's results are compared to themselves, i.e., they are their own control. This is an advantage because it reduces variability as each "pair" accounts for any individual differences, focusing solely on the effect of the intervention.

• We first calculate the difference between the 'after' and 'before' scenarios.
• Then, we analyze these difference scores using the t-statistic to check if these differences are statistically significant.

This approach is typically applied in studies where you want to observe the effect of a treatment or condition over time on the same subjects.

Null Hypothesis

The null hypothesis, often denoted as \(H_0\), is a statement used in hypothesis testing that proposes there is no effect or no difference. It serves as the default hypothesis that is subject to testing, and it assumes that any observed effect is due to random chance.

In our soft drinks exercise, our null hypothesis states: "There is no decrease in the number of soft drinks consumed". Mathematically, this is written as \(H_0: \mu_d = 0\), where \(\mu_d\) represents the mean difference between the paired samples.

Rejecting the null hypothesis would mean that we have sufficient statistical evidence to support our claim that there is indeed a significant decrease in consumption when soft drinks are partially replaced with water. However, if we fail to reject the null hypothesis, it would imply that the observed changes could simply be due to random variation and not necessarily due to our intervention.

Critical Value

The critical value is a threshold value that the test statistic must exceed in order to reject the null hypothesis. It is determined based on the significance level (\(\alpha\)) and the degrees of freedom in your data.

In the context of our exercise, the significance level is set at \(\alpha = 0.10\). This means we are accepting a 10% chance of rejecting the null hypothesis when it is actually true. We use a t-distribution because we're working with small sample sizes and unknown population standard deviations.

• With 9 degrees of freedom (one less than the number of paired observations), the critical value from a t-distribution table is found.
• For a right-tailed test, this value equates to \(t_{\text{critical}} = 1.383\).

If our calculated test statistic is more extreme than this critical value in the appropriate direction, we reject the null hypothesis. This step is crucial in deciding whether the observed difference in our study is statistically significant.

Significance Level

The significance level, denoted as \(\alpha\), is a critical concept in hypothesis testing that represents the probability of making a Type I error, which occurs when we incorrectly reject a true null hypothesis. This value is pre-determined before an analysis is conducted, setting the threshold for what is considered statistically significant.

In the exercise, the significance level is established at \(\alpha = 0.10\). This indicates that there is a 10% risk of concluding that a difference exists when, in fact, it does not. When choosing a significance level, researchers must balance the risk of making Type I errors (false positives) against Type II errors (false negatives).
Practically speaking:

• A lower \(\alpha\) reduces the chances of a false positive but requires a stronger evidence to claim a statistically significant result.
• A higher \(\alpha\) makes it easier to find a significant result but increases the chance of mistaking a random variation for a true effect.

This crucial setup helps determine the rigor of the statistical test and the reliability of the conclusions drawn from the analysis.