AnswersProbability and Statistics with Applications Honors Sem-2-FL-1210300 2024-2025Cumulative Exam — Cumulative examQuestion 48
48
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Cumulative Exam — Cumulative exam

Question 48 • Probability and Statistics with Applications Honors Sem-2-FL-1210300 (2024-2025)

A school administrator claims that 85% of the students at his large school plan to attend college after graduation. The statistics teacher at this school selects a random sample of 50 students from this school and finds that 76% of them plan to attend college after graduation. The administrator would like to know if the data provide convincing evidence that the true proportion of all students from this school who plan to attend college after graduation is less than 85%. What are the values of the test statistic and P-value for this test?Find the z-table here.

Answer
A
z = –1.78, P-value = 0. 0375
B
z = –1.78, P-value = 0. 0750
C
z = 1.78, P-value = 0.0375
D
z = 1.78, P-value = 0. 0750

Explanation

For a one-proportion z-test, use the claimed proportion in the standard error: z = (0.76 − 0.85) / √[(0.85)(0.15)/50] ≈ −1.78. Because the question asks whether the true proportion is less than 85%, this is a left-tailed test. The area to the left of z = −1.78 is about 0.0375, so the test statistic is −1.78 and the P-value is 0.0375.

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