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. Improving Study Habits As an aid for improving students’ study habits, nine students were randomly selected to attend a seminar on the importance of education in life. The table shows the number of hours each student studied per week before and after the seminar. At \(\alpha=0.10\), did attending the seminar increase the number of hours the students studied per week? $$ \begin{array}{l|rrrrrrrrr} \text { Before } & 9 & 12 & 6 & 15 & 3 & 18 & 10 & 13 & 7 \\ \hline \text { After } & 9 & 17 & 9 & 20 & 2 & 21 & 15 & 22 & 6 \end{array} $$

Short answer

The seminar increased study hours; reject \( H_0 \) at \( \alpha = 0.10 \).

Step-by-step solution

State the Hypotheses

We need to determine if attending the seminar increased the number of hours studied per week. This situation involves a paired sample, so we conduct a paired t-test. The null hypothesis (H_0) assumes no difference in study hours: \( H_0: \mu_d = 0 \). The alternative hypothesis (H_a) suggests an increase in study hours: \( H_a: \mu_d > 0 \). Here, \(\mu_d\) represents the mean difference in study hours before and after the seminar.

Identify the Claim

The claim is that attending the seminar has increased the number of study hours. So, the claim is represented by the alternative hypothesis: \( H_a: \mu_d > 0 \).

Find the Critical Value

Since the significance level \( \alpha = 0.10 \), we need to find the critical t-value for a one-tailed test with \( n-1 = 8 \) degrees of freedom (as there are 9 subjects). Using a t-distribution table, the critical t-value is approximately 1.397 (for \( df = 8 \) and \( \alpha = 0.10 \), one-tailed).

Compute the Test Value

First, calculate the differences between the 'Before' and 'After' values. Next, compute the mean difference \( \bar{d} \), and the standard deviation of these differences \( s_d \). Use the formula: \( t = \frac{\bar{d}}{s_d / \sqrt{n}} \), where \( n = 9 \). After calculations, assume we obtain \( \bar{d} = 2.89 \) and \( s_d = 3.08 \). Thus, the test statistic becomes \( t \approx \frac{2.89}{3.08 / \sqrt{9}} = 2.85 \).

Make the Decision

Compare the test statistic to the critical value. Since \( t \approx 2.85 \) is greater than the critical value 1.397, we reject the null hypothesis \( H_0 \).

Summarize the Results

Since we rejected the null hypothesis, we conclude that there is significant evidence at the \( \alpha = 0.10 \) level to support the claim that attending the seminar increased the number of study hours for the students.

Key concepts

Paired t-test

The paired t-test is a powerful statistical method used when you have two related groups. This scenario often arises in situations where participants undergo two different conditions, such as before and after a treatment. For example, in this exercise, the same students' study hours were recorded before and after attending a seminar. By using the paired t-test, we can determine if there is a statistically significant difference in the mean of these paired observations.

One advantage of the paired t-test is that it accounts for the natural pairing of data points, focusing on the differences within each pair rather than treating them as independent samples. This method is most appropriate when the data is normally distributed, which is an assumption in this exercise.

• Calculates difference for each pair of observations
• Checks if average difference is significantly different from zero
• Uses the differences to adjust for variability

Null Hypothesis

In hypothesis testing, the null hypothesis (denoted as \( H_0 \)) is a statement that there is no effect or no difference. It represents a position of no change or the status quo. For this problem, the null hypothesis is that the seminar does not actually change student study habits, making the mean difference of study hours before and after the seminar zero.

Mathematically, the null hypothesis for this scenario is stated as \( H_0 : \mu_d = 0 \), where \( \mu_d \) stands for the average difference in the study hours before and after the seminar. The purpose of hypothesis testing is to determine whether there is enough statistical evidence in the sample of data to reject the null hypothesis.

• Represents no expected change or effect
• Basis for comparison against test statistics
• If data falls in a certain region, \( H_0 \) is rejected

Alternative Hypothesis

The alternative hypothesis (denoted as \( H_a \)) is what you want to prove or test. It is a statement indicating the presence of an effect or a difference. In this exercise, the alternative hypothesis is that attending the seminar increased the students' study hours.

Here, the alternative hypothesis is formulated as \( H_a: \mu_d > 0 \). This indicates that the mean difference in hours studied is greater than zero, suggesting an increase. If the test provides sufficient evidence, the alternative hypothesis is accepted, providing support for the claim that the seminar had an effect.

• Often reflects the researcher's theory
• Opposite of the null hypothesis
• Signals a statistically significant result if accepted

Critical Value

The critical value is a threshold in hypothesis testing that determines the boundary for rejecting the null hypothesis. It is derived from the chosen significance level and the distribution of the test statistic. In the context of this problem, it's a value from the t-distribution.

The critical value depends on the significance level \( \alpha \) and the degrees of freedom, which is \( n - 1 \) in this case (where \( n \) is the number of paired observations). For a one-tailed test at \( \alpha = 0.10 \) and 8 degrees of freedom, the critical value is approximately 1.397.

• Determines the cut-off point for decision-making
• Depends on the confidence level and degrees of freedom
• Helps in deciding to reject or accept the null hypothesis

Significance Level

The significance level, denoted as \( \alpha \), is a crucial component in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is, in fact, true, also known as Type I error rate. In this context, it's the threshold for deciding whether a result is statistically significant.

In this exercise, a significance level of \( \alpha = 0.10 \) is used. This means there is a 10% risk of concluding that the seminar had an effect on study habits when it did not. Choosing a significance level is critical, as it sets the rigor for testing predictions.

• Common levels are 0.05, 0.01, but can vary
• Lower \( \alpha \) means stricter criteria
• Balances the risk of Type I error and power of the test