Question

1. The University is interested in determining if the number of people coming to work bipedal bicycle has changed from 2016 to 2017. Over the course of a week (Mon-Sun) they count the number of cyclists who enter campus at the North campus cycle path entrance between 8am-10am each day. The University has data from the same week in 2016. The counts for each day of the week for 2016 and2017 are given in Table

	Mon	Tues	Wed	Thurs	Fri	Sat	Sun
2016	243	175	255	187	241	53	32
2017	254	242	250	263	235	51	35

Table 1: Counts of cyclists at the University North campus cycle path.

1. Describe one positive and one negative aspect of the University's sampling strategy.
2. Explain why a paired approach is appropriate for this test.
3. Write down appropriate null and alternative hypotheses for this test.
4. Calculate the mean and variance of the differences di=xi-yi.
5. What assumptions are required to conduct your hypothesis test from (c), does this data satisfy them?
6. Using your answer to part (d), or otherwise, calculate a 90% confidence interval for the difference in an average number of cyclists.
7. Using your answer to part (f), or otherwise, test your hypothesis from (c)  at the 10% level. You should clearly state the conclusion of your test.
8. From Table 1 it looks like Tuesday and Thursday have the largest difference from one year to the next.  Why is it wrong to now testing the hypothesis  “2017 has more cyclists on Tuesdays and Thursdays than in 2016”?

Answer 1

a. Positive aspect: the week is same in both the years, hence, seasonal effect will be as low as possible. We can say the data is collected almost under same environment.

Negative aspect: Students have changed over one year. Thus, population itself has changed. Hence, it is not a very good idea to draw interpretation based on this data.

b. To overcome the negative aspect, we should use paired approach. We can check those students who were there both the years and check how their way of coming has changed.

c. H0: di=0 against H1: di is not 0.

di=xi-yi

d. Mean: -20.6

variance: 1249.6 (using minitab)

e. Assumption of Normality.

f.

where, Z is the upper 5% point of a standard Normal ditribution