Problem

Visa Card USA studied bow frequently consumers of…

10:07

Problem 21 Hard Difficulty

One of the questions on the Business Week Subscriber Study was, "In the past 12 months,
when traveling for business, what type of airline ticket did you purchase most often? "The
data obtained are shown in the following contingency table.
Use $\alpha=.05$ and test for the independence of type of type of flight and type of ticket. What is your
conclusion?

Answer

There is sufficient evidence to support the claim that the variables are not
independent.

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Intro Stats / AP Statistics

Essentials of Modern Business Statistics

Chapter 11

Comparisons Involving Proportions and a Test of Independence

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Video Transcript

Washington use us a table here and it wants us to use our particular awful level to be 0.5 And it wants us to test for the independence of type, fight and the type of ticket. So essentially, what we have to first do is to complete that observed frequency here, which is going to be essentially just the aggregate of you Step a flight. So it's just gonna be so for observed, so observed here, it's gonna be 29.5518 20 to 1 to one and 135 Um, to make it more simple, weaken re calculate this or redistribute to be 29 22 95 1 21 518 and 1 35 This makes it such that the first two for each set of two are gonna be on course want into the type of ticket. The second thing that we have to calculate is going to be expected frequency, which is essentially the product of the column and row total divided by the sample size of end. So, essentially, if we take all of these observed frequent sees an accurate Adam up and then divided by our total sample size. This is what we're gonna get for are expected frequency. So in this part tickler case here, we're going to get on 35.6 15.4 1 50.7, uh, 65.3 455.7 and 197.3. So after here, we can find the value of chi squared, which is gonna be the aggregate for the some of O minus E squared over E, which is going to give us approximately the value of 100.435 So using rt table values here, we can basically see that First of all, we have a degrees of freedom of to, as they're essentially is only two or n minus one sets through which we get a P value. That's gonna be less than is your five. So essentially, if r p values less than this particular significant cellphone here, then we can reject are no hypothesis. So, essentially, as our Valley is actually less than 0.5 here, we can reject the no hypothesis here

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