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according to a u.s. news poll, 38% of students in the class of 2013 had…

Question

according to a u.s. news poll, 38% of students in the class of 2013 had done an internship during their time as an undergraduate student. dana is interested in finding out whether students at her university had an internship rate that was higher than the national average. she obtained a list of 25 randomly selected students from the population of all students at her university by requesting this information from the universitys institutional research office. she collected the responses and calculated that the proportion for her university was 43%. which one of the following statements about the z - test is correct? a. it is not safe to use the z - test for p, since n p₀ is not large enough. b. it is not safe to use the z - test for p, since n (1 - p₀) is not large enough. c. it is not safe to use the z - test for p, since the sample is not a random sample from the entire population (or cannot be considered as one). d. it is safe to use the z - test for p.

Explanation:

Step1: Identify sample size and proportion

The sample size $n = 25$, and the national - proportion $p_0=0.38$.

Step2: Check the condition $np_0$

Calculate $np_0=25\times0.38 = 9.5$.

Step3: Check the condition $n(1 - p_0)$

Calculate $n(1 - p_0)=25\times(1 - 0.38)=25\times0.62 = 15.5$.
For a one - sample z - test for a proportion, we need $np_0\geq10$ and $n(1 - p_0)\geq10$. Since $np_0 = 9.5<10$, it is not safe to use the z - test for $p$.

Answer:

A. It is not safe to use the z - test for p, since n * p<sub>0</sub> is not large enough.