Questions: Question 5 (1 point) A college professor is curious if the location of a seat in class affects grades in the class. They are teaching in a lecture hall with 240 students. The lecture hall has 10 rows, so they split the rows into 5 sections - Rows 1-2, Rows 3-4, Rows 5-6, Rows 7-8, and Rows 9-10. At the end of the course, they determine the top 25% of grades in the class, and if the location of the seat makes no difference, they would expect that these top 25% of students would be equally dispersed throughout the classroom. Their observations are recorded below.
Run a Goodness of Fit test to determine whether or not location has an impact on the grade. Let α=0.05. After running a Goodness of Fit test, does the professor have evidence to conclude that location in the classroom has an impact on final grade and what is the p -value?
Transcript text: Question 5 (1 point)
A college professor is curious if the location of a seat in class affects grades in the class. They are teaching in a lecture hall with 240 students. The lecture hall has 10 rows, so they split the rows into 5 sections - Rows 1-2, Rows 3-4, Rows 5-6, Rows $7-8$, and Rows $9-10$. At the end of the course, they determine the top $25 \%$ of grades in the class, and if the location of the seat makes no difference, they would expect that these top $25 \%$ of students would be equally dispersed throughout the classroom. Their observations are recorded below.
Run a Goodness of Fit test to determine whether or not location has an impact on the grade. Let $\alpha=0.05$. After running a Goodness of Fit test, does the professor have evidence to conclude that location in the classroom has an impact on final grade and what is the p -value?
Solution
Solution Steps
To determine if the location of a seat affects grades, we can use a Chi-Square Goodness of Fit test. The null hypothesis is that the distribution of top 25% students is uniform across all sections. We compare the observed frequencies of top students in each section with the expected frequencies (which are equal if the distribution is uniform). We calculate the Chi-Square statistic and compare it to the critical value for the given alpha level (0.05) or use the p-value to make a decision.
Step 1: Define the Hypotheses
The null hypothesis (\(H_0\)) is that the distribution of top 25% students is uniform across all sections. The alternative hypothesis (\(H_a\)) is that the distribution is not uniform.
Step 2: Calculate the Chi-Square Statistic
The Chi-Square statistic is calculated using the formula:
\[
\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}
\]
where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency. For this problem:
Step 3: Compute the Chi-Square Statistic and p-value
Using the observed and expected frequencies, the Chi-Square statistic is calculated as:
\[
\chi^2 \approx 1.8333
\]
The p-value associated with this Chi-Square statistic is approximately:
\[
p \approx 0.7669
\]
Step 4: Decision Rule
With \(\alpha = 0.05\), we compare the p-value to \(\alpha\):
If \(p \leq \alpha\), reject \(H_0\).
If \(p > \alpha\), do not reject \(H_0\).
Since \(p \approx 0.7669 > 0.05\), we do not reject the null hypothesis.
Final Answer
The professor does not have sufficient evidence to conclude that the location in the classroom has an impact on the final grade. The p-value is \(\boxed{0.7669}\).