Question

In: Statistics and Probability

A random survey of autos parked in the student lot and the staff lot at 3...

A random survey of autos parked in the student lot and the staff lot at 3 universities are provided in the table below. Are there differences between the universities in terms of cars driven by students and staff?

Student Staff
University 1 214 210
University 2 66 24
University 3 55 47

a) Write appropriate hypotheses.

b) How many degrees of freedom are there?

c) Find ?2 and the P-value.

d) State your conclusion (use α = 0.05).

All workings must be shown.

Solutions

Expert Solution

Observed Frequencies
Student Staff Total
University 1 214 210 424
University 2 66 24 90
University 3 55 47 102
Total 335 281 616
Expected Frequencies
Student Staff Total
University 1 335 * 424 / 616 = 230.5844 281 * 424 / 616 = 193.4156 424
University 2 335 * 90 / 616 = 48.9448 281 * 90 / 616 = 41.0552 90
University 3 335 * 102 / 616 = 55.4708 281 * 102 / 616 = 46.5292 102
Total 335 281 616
(fo-fe)²/fe
University 1 (214 - 230.5844)²/230.5844 = 1.1928 (210 - 193.4156)²/193.4156 = 1.422
University 2 (66 - 48.9448)²/48.9448 = 5.943 (24 - 41.0552)²/41.0552 = 7.0851
University 3 (55 - 55.4708)²/55.4708 = 0.004 (47 - 46.5292)²/46.5292 = 0.0048

a) Ho: There is no difference between the universities in terms of cars driven by students and staff.

Ha: There is a difference between the universities in terms of cars driven by students and staff.

b) Number of Rows = 2

Number of Columns = 3

df = (r-1)(c-1) = 2

c) Test statistic:  

χ² = ∑ ((fo-fe)²/fe) = 1.1928 + 1.4220 + 5.9430 + 7.0851 + 0.0040 + 0.0048 = 15.6517

p-value = CHISQ.DIST.RT(15.6517, 2) = 0.0004

d) Conclusion:

p-value < α, Reject the null hypothesis.

There is enough evidence to conclude that there is a difference between the universities in terms of cars driven by students and staff.


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