Question

In: Statistics and Probability

When designing quizzes for students, I always try to assure students will have ample time to...

When designing quizzes for students, I always try to assure students will have ample time to complete all of the questions. Suppose to assure my quizzes are “doable” I design a hypothesis test to determine if a quiz has an acceptable time limit or not. I consider a time limit to be incorrect if more than 10% of students do not complete the quiz on time. After my first year of using the quiz, I take a random sample of 130 students and see if it can be shown that the proportion of students that ran out of time on the quiz exceeds 10%.

a.) State this study as a formal hypothesis test.

??0:

????:

b.) How would I make a Type II Error in this study?

Suppose for the Module 8 Quiz, I collect my sample of 130 students, and I find that 22 did not complete the quiz in time. Use this information and the ?? = 0.05 level to help me make my decision about whether the Module 8 quiz has an appropriate time length.

c.) Calculate the sample proportion that ran out of time from my sample of 130 students.

d.) Calculate the test statistic for this study.

e.) Using either a p-value or a critical value. Make your decision. Should you Reject ??0, or Fail to Reject ??0? Be sure to justify your decision using either the p-value or the critical value/test statistic.

f.) Interpret your decision in the context of the problem. Is the quiz time limit appropriate, or does it look like the Module 8 Quiz time limit needs to be adjusted?

Solutions

Expert Solution

a) The null and the alternate hypothesis will be:

H0: 10% or less will not complete the test.

HA: More than 10% will not complete the test.

b) A Type II Error is made when we incorrectly fail to reject a false null hypothesis.

c) As 22 out of 130 sample students did not finish on time, the sample proportion is x = 22/130 = 0.1692.

d) From the hypothesis test, we have:

As the sample size is greater than 30, we use the z-table. So, the test statistic of the study

e) From the z-table, we get the value of the probability for the z-score of 2.63 as 0.9957.

A significance level of 0.05 corresponds to a probability level of 1-0.05 = 0.95.

As the probability is greater than 0.95, we reject the null hypothesis H0.

f) This means that the alternate hypothesis is accepted. The quiz time limit is inadequate and it should be increased.


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