Question

. A textbook company states that the average time a student needs to take a quiz from its book is 30 minutes with a standard deviation of 3 minutes. A teacher using the book is not sure that this is correct for her classes and wants to check. She collects data on 10 random students and finds that the average time to take the quiz was only 25 minutes. As a result, the teacher performs a two-tailed hypothesis test with a significance level of 5%. Which conclusion is valid based on the results of the test? A. Her students, on average, do not take 30 minutes on the quiz, contrary to what the textbook company stated.
B. The teacher should pick up all unfinished quizzes at 25 minutes because her students are so much faster than average.

C. The teacher does not have enough information to make a conclusion about the average time on the test.

D. Her students, on average, do take the 30 minutes as the textbook company stated.

Answer 1

Answer:

A. Her students, on average, do not take 30 minutes on the quiz, contrary to what the textbook company stated.

Step-by-step explanation:

This is a normal distribution problem with a two-tailed hypothesis test. This characteristics helps to determine the hypothesis, which is the first step in a hypothesis test.

So, the null hypothesis would be: $H_{o} : u=30 min.$.

The alternative hypothesis would be: $H_{a} : u\neq 30min$.

In this problem, we don't use inequality relations because the teacher performed a two-tailed test, which only have to options, equal or not equal.

In second place, we have to determine the significance level, which is 5%, this means that the two tales rejecting zones must be equal to 5%, that is, 2.5% for each tale. This significance level is a percentage that refers to the probability of the null hypothesis to reach that zone, as you can see, it's a low probability, so if the null hypothesis reaches that zone, it means that the data has significance, it's reliable.

Now, we have to calculate the z-score, which is gonna tell us if the null hypothesis can reach the rejecting zone or not. The equation to find z-value is: $Z = \frac{x - u}{\frac{o}{\sqrt{n} } }$; where x is the sample mean, u is the population mean, o is the standard deviation and n is the sample.

Replacing all values:

$Z = \frac{25-30}{\frac{3}{\sqrt{10} } } = \frac{-5}{0.95} =-5.26$.

Hence, the z-value is -5.26, which is located in the left side of the population mean. Now, we have to use the z-table to know if the null hypothesis is rejected or accepted.

So, according to the table the probability value (p-value), for a 5% of significance level, is 0.0001. But, this is a two-tailed test, so the p-value is actually 0.0001(2) = 0.0002.

Therefore, the p-value is way lower than the level of significance 0.05, we must reject the null hypothesis, which means that there's no enough evidence to accept it. So, according with the test, the students don't take 30 minutes on the quiz, the correct answer is A.

Answer 2

The conclusion that is valid based on the result of the test would be : A. Her students on average, do not take 30 minutes on the quiz, contrary to what the textbook company stated

we can see in the statement that the average student only need about 25 minutes to take the quiz

hope this helps