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. 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? $$ \begin{array}{l|ccccccccc}{\text { Before }} & {9} & {12} & {6} & {15} & {3} & {18} & {10} & {13} & {7} \\ \hline \text { After } & {9} & {17} & {9} & {20} & {2} & {21} & {15} & {22} & {6}\end{array} $$

Short answer

The seminar increased the students' study hours at the 0.10 significance level.

Step-by-step solution

State the Hypotheses

We are testing if attending the seminar increased the number of hours the students studied per week. The null hypothesis (H_0) states that there is no increase in the average number of hours studied, while the alternative hypothesis (H_a) claims that the seminar increased the study time.\[H_0: \mu_d \leq 0\]\[H_a: \mu_d > 0\]The claim is in the alternative hypothesis.

Find the Critical Value(s)

Since we are conducting a one-tailed t-test at a significance level of \(\alpha=0.10\) with \(n-1=8\) degrees of freedom, we look up the critical value in the t-distribution table. This gives us a critical t-value of approximately 1.397.

Compute the Test Value

Calculate the differences between the before and after study hours for each student. Compute the mean and standard deviation of these differences.\[\text{Differences: } 0, 5, 3, 5, -1, 3, 5, 9, -1\]Mean difference, \(\overline{d} = 3.22\), and standard deviation, \(s_d = 3.23\). Use the formula for the t-test:\[t = \frac{\overline{d} - 0}{s_d/\sqrt{n}}\]Compute the test statistic:\[t = \frac{3.22 - 0}{3.23/\sqrt{9}} \approx 2.98\]

Make the Decision

Compare the calculated t-test statistic (2.98) to the critical t-value (1.397). Since 2.98 is greater than 1.397, we reject the null hypothesis.

Summarize the Results

By rejecting the null hypothesis, we conclude that there is sufficient evidence to support the claim that attending the seminar increased the number of hours students studied per week at the \(\alpha=0.10\) level of significance.

Key concepts

t-test

Imagine you want to understand the impact of a seminar on students’ study hours. To analyze this, the t-test is your go-to tool. It's a statistical test that helps you compare the means of two groups to see if there's a significant difference between them. In this case, we compare study hours before and after the seminar.
You use a t-test because:

• It determines if the means are statistically different from each other.
• It's ideal for small sample sizes, like our nine students.
• Assumes the differences in study hours follow a normal distribution.

The t-test formula, \( t = \frac{\overline{d}}{\frac{s_d}{\sqrt{n}}} \), calculates the test statistic by considering the mean difference \(\overline{d}\), standard deviation \(s_d\), and sample size \(n\).
The result tells you how many standard deviations the sample mean is from the null hypothesis mean.

critical value

The critical value is like a benchmark for your hypothesis test—it's the threshold you compare your test statistic against.
In hypothesis testing, you set a significance level \(\alpha\), which in this exercise is 0.10. The critical value is derived from a statistical distribution table based on this \(\alpha\) and degrees of freedom (\(n-1\)).
In our example, with 8 degrees of freedom, the critical value is approximately 1.397 for a one-tailed test.
Why critical values?

• They help determine the rejection region for the null hypothesis.
• Provide a clear comparison point for the calculated test statistic.

If the test statistic exceeds this value, like ours which is 2.98, it leads to rejecting the null hypothesis.

null hypothesis

The null hypothesis, symbolized as \(H_0\), is a statement suggesting no effect or no difference. It's the hypothesis that the experiment seeks to contradict.
In this exercise, \(H_0 : \mu_d \leq 0\) assumes that attending the seminar did not increase the study hours.
Key points about the null hypothesis:

• It's often a statement of "no effect" or "no change."
• Serves as a baseline for testing your research hypothesis.
• Must be testable.

The goal of hypothesis testing is to determine whether the observed data provides enough evidence to reject \(H_0\) in favor of an alternative hypothesis.

alternative hypothesis

The alternative hypothesis, \(H_a\), presents a statement you aim to prove with the data. It often suggests a new effect or a difference.
Here, \(H_a : \mu_d > 0\) claims that the seminar increased study hours. This is the hypothesis we're testing against the null.
Think of \(H_a\) as:

• The hypothesis you're interested in proving.
• Supports the research question or claim.
• Indicates a potential effect or relationship.

When the test statistic crosses the critical value threshold, we have sufficient evidence to reject the null in favor of the alternative hypothesis. This implies the data supports \(H_a\).

test statistic

The test statistic is a value calculated from the sample data that you compare against your critical value. It's a key component of the hypothesis testing process.
In this exercise, the test statistic \(t = 2.98\) was computed using the sample mean difference \(\overline{d}\), the standard deviation \(s_d\), and the sample size \(n\).

• Acts as a bridge between observed data and probability distributions.
• Helps quantify how far the observed data is from the null hypothesis.
• Is specific to the type of test you're performing (e.g., t-test in this case).

The goal is to see if this value lies within the critical region. If it does, like in our case, we conclude that the evidence is strong enough to reject the null hypothesis.