Question

1.   A Bloomberg Businessweek subscriber study asked, “in the past 12 months, when traveling for business, what type of airline ticket did you purchase most often?” A second question asked if the type of airline ticket purchased most often was for domestic or international travel. Sample data obtained are shown in the following table.

Type of Ticket	Domestic Flight	International Flight
First class	29	22
Business class	95	121
Economy class	518	135

a)   The study wants to test whether the type of ticket is independent of the type of flight. Clearly state the null and alternative hypotheses.

b)   Compute the expected frequencies by completing the table below.

Type of Ticket	Domestic Flight	International Flight	Total
First Class
Business Class
Economy Class
Total

c)   Compute the test statistic.
Please copy your R code and the result and paste them here.

d)   At 5% significance level, compute the critical value for the test statistic and the p value for the test. Draw your conclusion.
Please copy your R code and the result and paste them here.

e)   Use the function chisq.test() in R to run the test directly to confirm your results above are correct.

Please copy your R code and the result and paste them here.

Answer 1

The obtained sample data is

	Domestic flight	International flight	Total
First Class	29	22	51
Business Class	95	121	216
Economy Class	518	135	653
Total	642	278	920

a) Now we have to test whether the type of ticket is independent of the type of flight. Therefore the hypothesis is

Null hypothesis, H₀ : Type of ticket is independent of the type of flight

Alternate hypothesis, H₁ : Type of ticket is not independent of the type of flight.

b)

Now the expected frequencies are obtained by using following formula

Hence

	Domestic flight	International flight	Total
First Class	35.5891	15.4109	51
Business Class	150.7304	65.2696	216
Economy Class	455.6804	197.3196	653
Total	642	278	920

c)

The test statistic for testing above hypothesis is

Oij (Observed frequency)	Eij (Expected frequency)	(Oij-Eij)2/Eij
29	35.5891	1.2199
22	15.4109	2.8172
95	150.7304	20.6055
121	65.2696	47.5854
518	455.6804	8.5229
135	197.3196	19.6824
Total	920

Therefore the value of test statistic is 100.4334.

R-Code:

> Oij=c(29,22,95,121,518,135) ## Oij = Observed Frequency
> Eij=c(35.5891,15.4109,150.7304,65.2696,455.6804,197.3196) ## Eij = Expected Frequency
> chi_test=sum(((Oij-Eij)^2)/Eij) ## chisquare test statistic
> chi_test
[1] 100.4334

Therefore the value of test statistic is 100.4334.

d)

Let the level of significance = = 0.05 and (r-1)*(c-1) = 2

Hence the degrees of freedom = 2

Therefore the critical value for the test statistic is

Now value of A is computed by using following formula

Hence A = 5.991 -------------------(By using chisquare table)

Therefore, the critical value for the test statistic is 5.991.

P-value is

Therefore p- value is 0.

R code:

> Oij=c(29,22,95,121,518,135) ## Oij = Observed Frequency
> Eij=c(35.5891,15.4109,150.7304,65.2696,455.6804,197.3196) ## Eij = Expected Frequency
> chi_test=sum(((Oij-Eij)^2)/Eij) ## chisquare test statistic
> critical_value = qchisq(p=0.05, df=2, lower.tail=FALSE)
> critical_value
[1] 5.991465
> pvalue=pchisq(chi_test,df=2,lower.tail=FALSE) ## pvalue for test
> pvalue
[1] 1.552954e-22

e)

> data=matrix(c(29,95,518,22,121,135),nrow=3,ncol=2)
> data
[,1] [,2]
[1,] 29 22
[2,] 95 121
[3,] 518 135
> chisq.test(data)

Pearson's Chi-squared test

data: data
X-squared = 100.43, df = 2, p-value < 2.2e-16