Answer:
Step-by-step explanation:
Part A:
To calculate the proportion of STEM students who participate in at least one AP course, we need to add up the number of STEM students who indicated they are enrolled in at least one AP course and divide by the total number of STEM students in the sample. From the table, we see that 55 STEM students are enrolled in at least one AP course. The total number of STEM students in the sample is 80. Therefore, the proportion of STEM students who participate in at least one AP course is:
55/80 = 0.688
To calculate the proportion of regular students who participate in at least one AP course, we follow the same procedure. We add up the number of regular students who indicated they are enrolled in at least one AP course and divide by the total number of regular students in the sample. From the table, we see that 30 regular students are enrolled in at least one AP course. The total number of regular students in the sample is 120. Therefore, the proportion of regular students who participate in at least one AP course is:
30/120 = 0.25
Part B:
To determine whether participating in two or more AP courses is independent of student status, we need to compare the proportion of STEM students and regular students who are enrolled in two or more AP courses. From the table, we see that 35 STEM students are enrolled in two or more AP courses, and 20 regular students are enrolled in two or more AP courses.
To test for independence, we need to use a chi-square test. The null hypothesis is that the two variables, student status and enrollment in two or more AP courses, are independent. The alternative hypothesis is that they are dependent.
The expected frequencies for each cell under the null hypothesis can be calculated as follows:
The expected frequency for STEM students enrolled in two or more AP courses is (55/200) * (35/200) * 200 = 4.8125.
The expected frequency for regular students enrolled in two or more AP courses is (30/200) * (20/200) * 200 = 0.75.
The observed and expected frequencies for each cell are summarized in the following table:
Enrolled in ≥2 AP Courses Not Enrolled in ≥2 AP Courses
STEM Students Observed: 35 Observed: 45
Expected: 4.8125 Expected: 75.1875
Regular Observed: 20 Observed: 100
Expected: 0.75 Expected: 29.25
The chi-square test statistic can be calculated using the formula:
χ² = ∑(observed - expected)² / expected
Plugging in the observed and expected frequencies from the table, we get:
χ² = [(35-4.8125)² / 4.8125] + [(45-75.1875)² / 75.1875] + [(20-0.75)² / 0.75] + [(100-29.25)² / 29.25]
χ² = 115.48
Using a chi-square distribution table with 1 degree of freedom (df = (2-1) * (2-1)), we find that the critical value of chi-square at the 0.05 level of significance is 3.84.
Since the calculated chi-square value of 115.48 is greater than the critical value of 3.84, we reject the null hypothesis and conclude that participating in two or more AP courses is not independent of student status. In other words, the participation in two or more AP courses is associated with student status. Specifically, STEM students are more likely to be enrolled in two or more AP courses than regular students.