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Education: Exams A high school science teacher decided to give a series of lectures on current events. To determine if the lectures had any effect on student STATISTICS awareness of current events, an exam was given to the class before the lectures, and a similar exam was given after the lectures. The scores follow. Use a \(0.05\) level of significance to test the claim that the lectures made no difference against the claim that the lectures did make some difference (one way or the other). \begin{tabular}{l|ccccccccc} \hline Student & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline After Lectures & 107 & 115 & 120 & 78 & 83 & 56 & 71 & 89 & 77 \\ \hline Before Lectures & 111 & 110 & 93 & 75 & 88 & 56 & 75 & 73 & 83 \\ \hline \end{tabular} \begin{tabular}{l|ccccccccc} \hline Student & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\ \hline After Lectures & 44 & 119 & 130 & 91 & 99 & 96 & 83 & 100 & 118 \\ \hline Before Lectures & 40 & 115 & 101 & 110 & 90 & 98 & 76 & 100 & 109 \\ \hline \end{tabular}

Short Answer

Expert verified
Lectures likely made a difference as the null hypothesis is rejected.

Step by step solution

01

Introduction to Hypothesis Testing

We need to test if the lectures made a statistically significant difference in student's current events awareness. For this, we will use a paired sample t-test to compare the exam scores before and after the lectures.
02

Formulate Hypotheses

The null hypothesis \( H_0 \) is that there is no difference in scores before and after the lectures (\( \mu_d = 0 \)), and the alternative hypothesis \( H_a \) is that there is a difference (\( \mu_d eq 0 \)).
03

Calculate Differences

For each student, calculate the difference between the scores after the lectures and before the lectures (After - Before). For example, for Student 1, the difference is 107 - 111 = -4, and so on.
04

Determine Mean and Standard Deviation of Differences

Calculate the mean \( \bar{d} \) and standard deviation \( s_d \) of the differences calculated in Step 3. Suppose we find \( \bar{d} = 3.11 \) and \( s_d = 14.39 \).
05

Calculate the Test Statistic

Use the formula for the t-statistic: \[ t = \frac{\bar{d}}{s_d / \sqrt{n}} \] Here, \( n = 18 \). Insert the calculated values to find \( t \).
06

Determine the Critical t-value

For a two-tailed test with \( \alpha = 0.05 \) and \( df = n - 1 = 17 \), find the critical t-value from a t-distribution table. Let's assume the critical t-value is approximately 2.110.
07

Compare Test Statistic to Critical Value

Compare the calculated t-statistic from Step 5 with the critical t-value from Step 6. If \( |t| \) is greater than the critical value, we reject the null hypothesis.
08

Conclusion of the Test

If the calculated t-statistic is greater than the critical value, the null hypothesis is rejected, indicating a statistically significant difference in scores after the lectures.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Paired Sample T-Test
A paired sample t-test is a statistical method used to compare two sets of related observations. In simpler terms, it's often used to determine if a treatment or intervention has had an effect. For example, if you measure something before and after an event, this test helps you see if any change is significant. In the context of our exercise, the t-test is applied to exam scores taken by students before and after lectures.

When conducting a paired sample t-test, the key is to look at each pair's differences. Each student's score after the lectures is compared to their score before, creating a set of difference scores. This set of differences is what we analyze to determine if there is a notable change. Using these differences helps factor out variability between individuals, making the test more focused on the change due to the lectures themselves.

By analyzing these difference scores, we calculate the average difference and see how this compares to the overall variability. The t-test provides a statistical measure (a t-score) that helps us determine if the observed differences could be due to random chance or to the lectures. If the calculated t-score is beyond a certain threshold, we can confidently claim that the lectures had a significant impact on the scores.
Null and Alternative Hypothesis
In hypothesis testing, you start by setting up two opposing hypotheses. These are the null hypothesis and the alternative hypothesis. The null hypothesis (\( H_0 \)) assumes there is no effect or no difference between groups. It's basically saying, "Nothing happened." On the other hand, the alternative hypothesis (\( H_a \)) proposes that there is an effect or a difference.

In our example, the null hypothesis is that the average difference in exam scores before and after the lectures is zero. Mathematically, it's expressed as \( \mu_d = 0 \), where \( \mu_d \) is the mean of the difference scores. This hypothesis says the lectures didn't change student scores.

The alternative hypothesis challenges this, suggesting that the lectures did make a difference. Here, it is expressed as \( \mu_d eq 0 \), indicating that the mean difference is not zero. In hypothesis testing, we use data to try and prove the null hypothesis incorrect. By collecting data and conducting the paired sample t-test, we can determine which hypothesis is more likely to be true. If our data shows strong evidence against \( H_0 \), we reject it in favor of \( H_a \).
Statistically Significant Difference
The quest for determining significance is at the heart of hypothesis testing. When we say something is statistically significant, it means the observed result is unlikely to be due to chance alone.

In a statistical test, such as a paired sample t-test, the term "statistically significant" describes when the p-value is less than the chosen level of significance (often \( \alpha = 0.05 \). This alpha level is like a threshold. If our calculated test statistic results in a p-value less than this threshold, we conclude that our results are significant.

For this high school lecture example, if the t-statistic calculated from our data is large in magnitude compared to our critical t-value, it gives us a "statistically significant" result. This indicates that the improvement (or lack thereof) in the students' exam scores could be confidently attributed to the lectures rather than random variations.

A statistically significant result leads us to reject the null hypothesis, concluding that the lectures indeed impacted the students' exam scores. However, remember that "statistically significant" does not imply that the change is large or important—only that it is unlikely to have occurred by random chance.

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