Questions: Students at a high school are asked to evaluate their experience in the class at the end of each school year. The courses are evaluated on a 1-4 scale - with 4 being the best experience possible. In the History Department, the courses typically are evaluated at 10% 1's, 15% 2's, 34% 3's, and 41% 4's. Mr. Goodman sets a goal to outscore these numbers. At the end of the year, he takes a random sample of his evaluations and finds 101 1's, 132 2's, 483 3's, and 524 4's. At the 0.05 level of significance, can Mr. Goodman claim that his evaluations are significantly different than the History Department's?
Transcript text: Students at a high school are asked to evaluate their experience in the class at the end of each school year. The courses are evaluated on a $1-4$ scale - with 4 being the best experience possible. In the History Department, the courses typically are evaluated at $10 \% 1$ 's, $15 \% 2$ 's, $34 \% 3$ 's, and $41 \% 4^{\prime} \mathrm{s}$. Mr. Goodman sets a goal to outscore these numbers. At the end of the year he takes a random sample of his evaluations and finds 101 s ' $132 \mathrm{2} \mathrm{s}, 483 \mathrm{~s}$ s, and 524 s s. At the 0.05 level of significance, can Mr. Goodman claim that his evaluations are significantly different than the History Department's?
Solution
Solution Steps
Step 1: Observed and Expected Frequencies
The observed frequencies of evaluations from Mr. Goodman's class are:
\( O_1 = 101 \) (1's)
\( O_2 = 132 \) (2's)
\( O_3 = 483 \) (3's)
\( O_4 = 524 \) (4's)
The expected frequencies based on the History Department's evaluations are calculated as follows:
\( E_1 = 0.10 \times 1240 = 124 \)
\( E_2 = 0.15 \times 1240 = 186 \)
\( E_3 = 0.34 \times 1240 = 422.4 \)
\( E_4 = 0.41 \times 1240 = 508.4 \)
Step 2: Chi-Square Test Statistic Calculation
The Chi-Square test statistic (\( \chi^2 \)) is calculated using the formula:
\[
\chi^2 = \sum_i \frac{(O_i - E_i)^2}{E_i}
\]
Substituting the observed and expected values:
\[
\chi^2 = \frac{(101 - 124)^2}{124} + \frac{(132 - 186)^2}{186} + \frac{(483 - 422.4)^2}{422.4} + \frac{(524 - 508.4)^2}{508.4}
\]
Calculating this gives:
\[
\chi^2 = 29.3643
\]
Step 3: Critical Value and P-Value
For a Chi-Square test with \( df = 3 \) (degrees of freedom) at \( \alpha = 0.05 \), the critical value is:
\[
\chi^2(0.95, 3) = 7.8147
\]
The p-value associated with the test statistic is:
\[
P = P(\chi^2 > 29.3643) = 0.0
\]
Step 4: Conclusion
Since the test statistic \( \chi^2 = 29.3643 \) is greater than the critical value \( 7.8147 \), we reject the null hypothesis. This indicates that Mr. Goodman's evaluations are significantly different from those of the History Department.