Question

Suppose that during an unexpected snowstorm, Mr. Wong decided to obtain a random sample of students in his AP Statistics class to examine their arrival times. He compared the difference between the students' arrival time with the time the class was supposed to begin.

Mr. Wong asks you (his assistant) to use the information below to answer the following questions (negative value means that the student arrived BEFORE class began).

Number of students	30
Mean	-1.067
Q1	-24
Q3	18
Q2	-10.5
Min	-41
Max	53
Variance	765.78850575713
Standard deviation	27.67288394344

* Please do not copy other experts’ solution.

Question A: Mr. Wong would like to determine if his students arrive to class late on average. He asks you to perform a hypothesis test @ 10% significance level. Clearly state the conclusion you would tell Mr. Wong using a critical value.

Question B:

i) Mr. Wong asks you to calculate a 90% confidence interval for the average difference in time.

ii) Interpret the interval you calculated.

iii) Based on the interval, what can you say about Mr. Wong's AP Statistics class's arrival time?

Question C: Did you expect the conclusion in Question B to be the same as the conclusion in Question A? Explain why or why not.

Answer 1

  -1.067 Sx =27.673

Question A: Mr. Wong would like to determine if his students arrive to class late on average. He asks you to perform a hypothesis test @ 10% significance level. Clearly state the conclusion you would tell Mr. Wong using a critical value.

We want to test if the students are late for the arrival. So if we are checking based on difference (class start time - arrival). To be late means arrival time is later than former and so the differece would be negative. or less than 0 numerically. This is one tailed test and we do not have the population SD so we will use t-dist.

Test

(The students are not late)

(The students are late)

Test Stat =

Where the null mean = 1750

Test stat = -0.211

Critical value with = 10%. This is a 1 tailed test therefore

C.V. = 1.311 .............using t-dist tables

Since |Test stat| < C.V.

We do not reject the null hypothesis at 10%. We conclude that the students are not significantly late.

Question B:

i) Mr. Wong asks you to calculate a 90% confidence interval for the average difference in time.

(1- )% is the confidence interval for population mean

Where -1.067

Alpha =1 - 0.90 = 0.10

Therefore the C.V. =

=

= 1.699 .............found using t-dist tables

Substituting the value

ii) Interpret the interval you calculated.

The true average difference in time lies between (-9.65, 7.518) with 90% confidence (certainty.)

iii) Based on the interval, what can you say about Mr. Wong's AP Statistics class's arrival time?

Since it has negative difference as well as positive, as well '0', we can say that students can be late, early or on time.

Question C: Did you expect the conclusion in Question B to be the same as the conclusion in Question A? Explain why or why not.

Since the interval includes null value = 0, it means we are not rejecting the null hypothesis. So we still hold the same decision as the hypothesis testing. This is because we used the same statistics and the same confidence level. So the same conclusion is expected.