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 statistics professor wants to investigate the relationship between a student's midterm examination score and the score on the final. Eight students were randomly selected, and their scores on the two examinations are noted. At the 0.10 level of significance, is there sufficient evidence to conclude that there is a difference in scores? $$ \begin{array}{l|rrrrrrrr} \text { Student } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \text { Midterm } & 75 & 92 & 68 & 85 & 65 & 80 & 75 & 80 \\ \hline \text { Final } & 82 & 90 & 79 & 95 & 70 & 83 & 72 & 79 \end{array} $$

Short answer

There is a significant difference in scores at the 0.10 significance level; the scores differ.

Step-by-step solution

State the Hypotheses

Identify the null hypothesis (H_0) and the alternative hypothesis (H_a). In this case, we will test if there is no difference between the midterm and final scores.\[H_0: \mu_d = 0\]\[H_a: \mu_d eq 0\]where \( \mu_d \) is the mean difference in scores. The claim is that there is a difference in scores.

Find the Critical Value(s)

Since this is a two-tailed test with a significance level of 0.10, you need to find the critical value for tdistribution with (n-1)degrees of freedom, where \(n = 8\). So, degrees of freedom (d.f.) = 7.For \alpha = 0.10(two-tailed), the critical values using a t-distribution table are approximately \(\pm t_{\alpha/2, 7} = \pm 1.895\).

Compute the Test Value

Calculate the differences between the midterm and final scores and then find the test statistic:\[\text{Difference (D): } D_i = \text{Midterm}_i - \text{Final}_i\]Calculate:\[D = [75-82, 92-90, 68-79, 85-95, 65-70, 80-83, 75-72, 80-79] = [-7, 2, -11, -10, -5, -3, 3, 1]\]Calculate the mean of these differences (\overline{D}) and their standard deviation (S_D). Finally, compute the test statistic: \[t = \frac{\overline{D}}{S_D/\sqrt{n}}\].Mean difference: \[\overline{D} = \frac{-30}{8} = -3.75\]Standard deviation of differences:\[S_D = \sqrt{\frac{\sum(D_i - \overline{D})^2}{n-1}} = \sqrt{\frac{80}{7}} \approx 3.38\]Test statistic:\[t = \frac{-3.75}{3.38/\sqrt{8}} \approx -3.15\].

Make the Decision

Compare the calculated test statistic with the critical values. Since \(t = -3.15\) is beyond \(-1.895\), we reject the null hypothesis. The evidence suggests that there is a significant difference in scores at the 0.10 significance level.

Summarize the Results

There is sufficient evidence at the 0.10 level of significance to conclude that there is a difference between students' midterm and final examination scores. The claim that the scores differ is supported.

Key concepts

t-distribution

Imagine you're trying to figure out if there's a notable difference between two sets of scores. Maybe it's the results of midterm and final exams. For such small sample sizes, you can't rely on the normal distribution. Enter the **t-distribution**! The t-distribution is a variation of the normal distribution that's wider and has thicker ends, making it perfect for small samples.
A unique feature of the t-distribution is that it's dependent on degrees of freedom, which is basically how much room your data has to vary. It's calculated as the number of data points minus one. So, if you have scores from 8 students, your degrees of freedom would be 7.
This distribution helps you understand how your sample compares to the entire population, allowing you to determine if your findings are due to chance or something noteworthy.

Critical Values

When you're testing a hypothesis, you want to know the boundaries of the typical data range. These boundaries are called **Critical Values**. They are the cut-off points on the tail ends of your distribution curve that define when you'll reject the null hypothesis.
In a two-tailed test (like in our exam score example), you have two critical values: a positive one and a negative one. They mark the two ends where, if your test statistic lands beyond these points, you reject the null hypothesis.
With a significance level of 0.10, these critical values using the t-distribution and 7 degrees of freedom are approximately \(\pm 1.895\). These values help you decide if the observed test statistic shows a significant effect or just random chance.

Test Statistic

The **Test Statistic** is a calculated value that tells you if the observed effect is real or not. It's derived from the sample data and compared against the critical values to make a decision about the null hypothesis.
For differences in test scores, the **t-test statistic** is used. In our case, we first find the differences between each pair of scores. Then, we calculate the mean and standard deviation of these differences to find the t-statistic using the formula:
\[t = \frac{\overline{D}}{S_D/\sqrt{n}}\]where \(\overline{D}\) is the mean difference, \(S_D\) is the standard deviation, and \(n\) is the number of differences. In this scenario, the calculated test statistic was approximately \(-3.15\). When comparing this value to our critical values, we get an indication of whether the difference in scores is statistically significant.

Significance Level

When you conduct a hypothesis test, you need a benchmark to decide if your results are noteworthy. That's where the **Significance Level** comes in. It's denoted by \(\alpha\) and represents the likelihood you'll reject the null hypothesis incorrectly.
A common level used is 0.05, but our exercise opts for 0.10. This level indicates that we're willing to accept a 10% chance of mistakenly rejecting the null hypothesis, known as a Type I error.
A lower significance level means you're stricter, accepting less risk of being wrong. With our \(\alpha = 0.10\), it means our results suggest there's enough evidence to note a difference between midterm and final scores, beyond what might occur by chance.