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. Retention Test Scores A random sample of nonEnglish majors at a selected college was used in a study to see if the student retained more from reading a 19 th-century novel or by watching it in DVD form. Each student was assigned one novel to read and a different one to watch, and then they were given a 100 -point written quiz on each novel. The test results are shown. At \(\alpha=0.05,\) can it be concluded that the book scores are higher than the DVD scores? $$ \begin{array}{l|lllllll} \text { Book } & 90 & 80 & 90 & 75 & 80 & 90 & 84 \\ \hline \text { DVD } & 85 & 72 & 80 & 80 & 70 & 75 & 80 \end{array} $$

Short answer

Reject the null hypothesis; book scores are higher than DVD scores.

Step-by-step solution

State the Hypotheses

We need to determine what our null and alternative hypotheses are. The null hypothesis ( \( H_0 \) ) usually represents no effect or no difference, while the alternative hypothesis ( \( H_a \) ) represents a significant effect or difference. In this case, we want to test if the book scores are higher than the DVD scores. Thus, we have: \[ H_0: \mu_{\text{Book}} \leq \mu_{\text{DVD}} \] \[ H_a: \mu_{\text{Book}} > \mu_{\text{DVD}} \] Here, \( \mu_{\text{Book}} \) and \( \mu_{\text{DVD}} \) are the population means for book and DVD scores. The claim is represented by \( H_a \), stating that the book scores are higher than DVD scores.

Find the Critical Value

To find the critical value, we need to determine the distribution to use. Since the sample is small (less than 30), we will use the t-distribution. We are conducting a one-tailed test at a significance level of \( \alpha = 0.05 \) with \( n-1 \) degrees of freedom, where \( n \) is the number of pairs. With 7 pairs, df = 6.Using a t-table or calculator, the critical value \( t_c \) for a one-tailed test at \( \alpha = 0.05 \) and df = 6 is approximately 1.943.

Compute the Test Value

First, calculate the mean and standard deviation of the differences between book and DVD scores. The differences \( d_i \) are 5, 8, 10, -5, 10, 15, and 4.Mean difference \( \bar{d} \):\[ \bar{d} = \frac{5 + 8 + 10 - 5 + 10 + 15 + 4}{7} = 6.71 \]Standard deviation of the differences \( s_d \):\[ s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}} = \sqrt{\frac{(5-6.71)^2 + (8-6.71)^2 + ... + (4-6.71)^2}{6}} \approx 6.18 \]Now, compute the t-test value:\[ t = \frac{\bar{d} - 0}{s_d / \sqrt{n}} = \frac{6.71}{6.18 / \sqrt{7}} \approx 3.02 \]

Make the Decision

We compare the computed test value (t = 3.02) with the critical value (t_c = 1.943).Since \( t = 3.02 \) is greater than \( t_c = 1.943 \), we reject the null hypothesis \( H_0 \). This indicates that there is a significant difference and the book scores are indeed higher than the DVD scores.

Summarize the Results

After completing the hypothesis test, we conclude that at the significance level of 0.05, there is sufficient evidence to support the claim that students retained more from reading a 19th-century novel than from watching it on DVD. Therefore, the book scores are significantly higher than the DVD scores.

Key concepts

Null Hypothesis

In hypothesis testing, the null hypothesis (denoted as \( H_0 \)) is a statement that suggests there is no effect or difference in the population. It serves as a starting point.
We test the null hypothesis to provide a basis upon which we can determine if the effect we observe through data is genuinely present or if it might have occurred by random chance. In our context, \( H_0 \) proposes that the average book scores are less than or equal to the DVD scores, implying no superior retention from books.
This foundational step is where we lay out what we aim to disprove or reject through our test's findings. It is crucial to clearly define \( H_0 \) because the entire test relies on comparing this statement against the real sample data.

Alternative Hypothesis

The alternative hypothesis (\( H_a \)) is what researchers often anticipate or hope to prove. It's the statement indicating there is an effect or a significant difference.
Contrary to the null hypothesis, the alternative hypothesis suggests that the event or association we're investigating does in fact affect the results.
In this example, \( H_a \) claims that book scores are indeed higher than the DVD scores. Simply put, \( H_a \) puts forth the scenario that students retain more information after reading than watching.

• \( H_0: \mu_{\text{Book}} \leq \mu_{\text{DVD}} \)
• \( H_a: \mu_{\text{Book}} > \mu_{\text{DVD}} \)

Identifying \( H_a \) allows us to focus on proving this claim via our analysis. The test supports \( H_a \) by providing evidence to reject \( H_0 \), thus showing a significant difference exists.

Critical Value

The critical value is a key element in our decision-making process in hypothesis testing. It's the threshold that the test statistic must exceed to reject \( H_0 \).
Determining this value requires us to know the test's degree of freedom, significance level (\( \alpha \)), and whether the test is one-tailed or two-tailed.
In this study, since our sample size is small (7 paired scores), we employ the t-distribution to find our critical value.

• Given that \( \alpha = 0.05 \) and the test is one-tailed (checking if book scores are higher), we calculate the critical value for \( df = 6 \) (since \( n-1 = 6 \)).
• We use a t-table or a calculator to find that the critical value is approximately 1.943.

Our computed test statistic must be greater than this critical value for us to reject the null hypothesis. So, it's this pivotal value that guides the 'reject or not' decision.

t-distribution

The t-distribution is a type of probability distribution that is symmetric and bell-shaped, similar to the normal distribution but with heavier tails. It's especially useful when dealing with small sample sizes (typically less than 30).
This study employs the t-distribution because we are working with only 7 sets of data. The broader tails of the t-distribution account for the additional variability inherent in estimating the population standard deviation from a small sample.
Key aspects of the t-distribution include:

• It becomes closely aligned with the normal distribution as sample size increases.
• Each variation of the t-distribution is defined by its degrees of freedom (\( df = n-1 \)), which adjusts for the sample size's effects.

The t-distribution provides a robust framework for testing means when variance is unknown, allowing for conclusions even in small samples when paired with a critical value.