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A student group maintains that the average student must travel for at least 25 minutes in order to reach college each day. The college admissions office thinks the average travel time is actually lower than 25 minutes and obtained one-way travel times to college from a random sample of 36 students. The sample had a mean of 19. 4 minutes. Assume that the population standard deviation is 9. 6 minutes. Does the admissions office have sufficient evidence to reject the students’ claim and conclude that the true mean travel time for all students is less than 25 minutes? The level of significance is α = 0. 5.

Answer the following three questions:

1) What is the alternative hypothesis for this test?

Group of answer choices

a) The mean one-way travel time for students is not equal to 25 minutes

b) The mean one-way travel time for students is 19. 4 minutes

c) The mean one-way travel time for students is less than 25 minutes

d) The mean one-way travel time for students is 25 minutes (or more)

2. What is the p-value for this test?

Group of answer choices

a) 0. 025

b) 0. 05

c) approximately zero

d) 1. 96

3. What is the conclusion for this test?

Group of answer choices

a) Data obtained by College Admissions Office provide sufficient evidence to say that the mean one-way travel time for students is less than 25 minutes

b) Data obtained by College Admissions Office provide sufficient evidence to say that the mean one-way travel time for students is 25 minutes (or more)

c) Data obtained by College Admissions Office provide sufficient evidence to say that the mean one-way travel time for students is not equal to 25 minutes

d) Data obtained by College Admissions Office provide sufficient evidence to say that the mean one-way travel time for students is 19. 4 minutes

1 Answer

4 votes

Let's break down the problem step by step.

1) The first thing we need to identify is the alternative hypothesis for this test. The student group claims that the average travel time is at least 25 minutes. The college admissions office thinks it's less. The alternative hypothesis is what the admissions office is trying to prove, which is that the average travel time is less than 25 minutes.

Answer to Question 1:

c) The mean one-way travel time for students is less than 25 minutes.

2) Next, we'll calculate the p-value. The p-value tells us how likely it is to get a sample like the one the admissions office got if the student group’s claim (that the average travel time is at least 25 minutes) is true. The smaller the p-value, the stronger the evidence against the student group’s claim.

To find the p-value, we can use the formula for the z-score:

Z = (sample mean - population mean under null hypothesis) / (population standard deviation / sqrt(sample size))

= (19.4 - 25) / (9.6 / sqrt(36))

= (19.4 - 25) / (9.6 / 6)

= -5.6 / 1.6

≈ -3.5

Now, we look up the z-score in a Z-table or use a calculator to find the p-value. For a z-score of -3.5, the p-value is very close to 0.

Answer to Question 2:

c) approximately zero

3) Finally, we have to decide whether this p-value is small enough to reject the student group’s claim. We compare it to the level of significance, α = 0.05. If the p-value is smaller than α, that means that the evidence is strong enough to reject the student group’s claim. Since the p-value is almost 0, which is much smaller than 0.05, the admissions office has enough evidence to say that the average travel time is less than 25 minutes.

Answer to Question 3:

a) Data obtained by College Admissions Office provide sufficient evidence to say that the mean one-way travel time for students is less than 25 minutes.

In simple terms, think of the p-value like a measuring tape. The admissions office is trying to show that the student group's claim doesn't hold up, and the p-value tells us how much the data supports the admissions office. Since the p-value is super tiny, it's like the measuring tape showing that the student group's claim is way off.

User ErasmoOliveira
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