Question

Transit Railroads is interested in the relationship between travel distance and ticket class
purchased. A random sample of 200 passengers is taken. Table 3 shows the results. The railroad wants to know if a passenger’s choice in ticket class is independent of the distance they must
travel.

Traveling
Distance
Third class Second class First class Total
1-100 miles 21 14 6 41
101-200 miles 18 16 8 42
201-300 miles 16 17 15 48
301-400 miles 12 14 21 47
401-500 miles 6 6 10 22
Total 73 67 60 200

Table 3

a. State the hypotheses.
b. State the degree of freedom
c. How many passengers are expected to travel between 201 and 300 miles and
purchase second-class tickets?
d. How many passengers are expected to travel between 401 and 500 miles and
purchase first-class tickets?
e. What is the test statistic?
f. What is the p-value?
g. What can you conclude at the 5% level of significance?

Answer 1

a. $H_0:$ The ticket class chosen is independent on the traveling distance.

$H_a:$ The ticket class chosen is dependent on the traveling distance.

b. df=(number of raws – 1)(numbers of columns – 1)

$df=(5-1)(3-1)= 4\cdot2 = 8$

c. How many passengers are expected to travel between 201 and 300 miles and

purchase second-class tickets?

$$expected\ frequency=\frac{raw\ total\ \cdot\ column\ total}{grand\ total}$$

$$\frac{67\cdot48}{200}=16.08$$

d. How many passengers are expected to travel between 401 and 500 miles and

purchase first-class tickets?

$$\frac{60\cdot22}{200}=6.6$$

e. test statistic:

$$\chi^2=\sum\frac{(observed -expected)^2}{expected}$$

for each category.

Expected 1-100miles 3rd class = $\frac{73\cdot41}{200}=14.965$

Expected 1-100miles 2nd class = $\frac{67\cdot41}{200}=13.735$

Expected 1-100miles 1st class = $\frac{60\cdot41}{200}=12.3$

Expected 101-200miles 3rd class = $\frac{73\cdot42}{200}=15.33$

Expected 101-200miles 2nd class = $\frac{67\cdot42}{200}=14.07$

Expected 101-200miles 1st class = $\frac{60\cdot42}{200}=12.6$

Expected 201-300miles 3rd class = $\frac{73\cdot48}{200}=17.52$

Expected 201-300miles 2nd class = $\frac{67\cdot48}{200}=16.08$

Expected 201-300miles 1st class = $\frac{60\cdot48}{200}=14.4$

Expected 301-400miles 3rd class = $\frac{73\cdot47}{200}=17.155$

Expected 301-400miles 2nd class = $\frac{67\cdot47}{200}=15.745$

Expected 301-400miles 1st class = $\frac{60\cdot47}{200}=14.1$

Expected 401-500miles 3rd class = $\frac{73\cdot22}{200}=8.03$

Expected 401-500miles 2nd class = $\frac{67\cdot22}{200}=7.37$

Expected 401-500miles 1st class = $\frac{60\cdot22}{200}=6.6$

$\chi^2=\frac{(21-14.965)^2}{14.965}+\frac{(14-13.735)^2}{13.735}+\frac{(6-12.3)^2}{12.3}+\frac{(18-15.33)^2}{15.33}+\frac{(16-14.07)^2}{14.07}+\frac{(8-12.6)^2}{12.6}+\frac{(16-17.52)^2}{17.52}+\frac{(17-16.08)^2}{16.08}+\frac{(15-14.4)^2}{14.4}+\frac{(12-17.155)^2}{17.155}+\frac{(14-15.745)^2}{15.745}+\frac{(21-14.1)^2}{14.1}+\frac{(6-8.03)^2}{8.03}+\frac{(6-7.37)^2}{7.37}+\frac{(10-6.6)^2}{6.6}=15.92$

f. For chi-square 15.92 and 8 df the p-value is 0.0435.

g. Since the p-value=0.0435 is less than the significance level 0.05 we cannot accept the null hypothesis. Or according to the critical values of chi square table with 5% significance level and 8 df the critical value is 15.51. Since 15.92 > 15.51 we can reject the null hypothesis. At the 5% significance level the data do provide sufficient evidence to conclude that the selection of ticket class is dependent on travel distance.