a textbook company states that the average time a student needs to take a quiz from its book is 30 minutes…

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? her students, on average, do not take 30 minutes on the quiz, contrary to what the textbook company stated. the teacher should pick up all unfinished quizzes at 25 minutes because her students are so much faster than average. the teacher does not have enough information to make a conclusion about the average time on the test. her students, on average, do take the 30 minutes as the textbook company stated.

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? her students, on average, do not take 30 minutes on the quiz, contrary to what the textbook company stated. the teacher should pick up all unfinished quizzes at 25 minutes because her students are so much faster than average. the teacher does not have enough information to make a conclusion about the average time on the test. her students, on average, do take the 30 minutes as the textbook company stated.

Answer

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $SE=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 3$ (standard - deviation) and $n = 10$ (sample size). So, $SE=\frac{3}{\sqrt{10}}\approx\frac{3}{3.162}= 0.95$.

Step2: Calculate the z - score

The formula for the z - score is $z=\frac{\bar{x}-\mu}{SE}$, where $\bar{x}=25$ (sample mean), $\mu = 30$ (population mean) and $SE\approx0.95$. So, $z=\frac{25 - 30}{0.95}=\frac{- 5}{0.95}\approx - 5.26$.

Step3: Determine the critical values

For a two - tailed test with a significance level of 5% ($\alpha=0.05$), the critical z - values are $z_{\alpha/2}=\pm1.96$.

Step4: Make a conclusion

Since the calculated z - score ($z\approx - 5.26$) is less than the lower critical value ($z=-1.96$), we reject the null hypothesis. The null hypothesis is that the average time is 30 minutes.

Answer:

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