Question

Open your data in Excel and answer the following in complete sentences. In class this week we've learned about chi-squared tests, which allow us to answer more complicated problems with categorical data. 1. Create a contingency table using Carrier and Origin as your columns and rows, respectively. 2. What is the null hypothesis for a test of independence based on this table? For our purposes, we'll pretend this data set is a random sample of all recent flights out of Oregon. 3. Calculate this hypothesis test, and report your p-value. You may calculate this "by hand," but it will be faster and easier in XLSTAT. Remember: These are five steps to a hypothesis test. I would like to see all five steps.. 4) What does this p-value tell you about flight carrier and origin airport, in everyday language? In other words, what is your conclusion in context? 5) Give an example of a different test of independence we could perform using this data. Be sure your data is qualitative. Qualitative. Are you trying to choose delay time as one of your variables??? Stop that. It's not qualitative.

How to do that? Please give me a example.

Answer 1

Here, we have taken a data to create a contingency table in execl.

part 1) We use excel sheet to create the following contingency table-

Flight Carrier
Origin Airport	AA	AS	BS	TOTAL
EUG(obs)	5	3	2	10
MFR(obs)	3	3	1	7
TOTAL	8	6	3	17

part 2)

To perform the test for independence we take the following hypothesis-

H0: Flight carrier and origin airport are independent vs H1: They are dependent

part 3) Formulas used-

expected frequency=(row total*coloumn total)/total frequency

$\chi ^{2}=\frac{\sum (O-E)^{2}}{E}$

where, O=observed frequency and E=expected frequency

df=(row-1)*(column-1)

Here, we assume .

= 0,05

	AA	AS	BS
EUG(exp)	4.705882	3.529412	1.764706
O-E	0.294118	-0.52941	0.235294
(O-E)^2	0.086505	0.280277	0.055363
(O-E)^2/E	0.018382	0.079412	0.031373
	AA	AS	BS
MFR(exp)	3.294118	2.470588	1.235294
O-E	-0.29412	0.529412	-0.23529
(O-E)^2	0.086505	0.280277	0.055363
(O-E)^2/E	0.026261	0.113445	0.044818
Chi sq	0.31369
df	2
Chi sq crit	5.991

Comment: Here, the calculated chi-square value is less than the chi-square value obtained from the chi-square table. Hence, we accept the null hypothesis at level $\alpha$ . So, we conclude that the flight carriers and origin air[port are independent.

part 4)

Here, we have to calculate the p-value-

Step 1: copy the observed frequencies and expected frequencies in 2 different columns.

Step 2: select formula bar and then select more functions.

Step 3: from the drop down menu select statistical and then choose CHISQ.TEST

Step 4: select the observed frequency column in actual range nd expected frequency colum in expected range and press ok.

	Observed frequency	Expected frequency
	5	4.705882
	3	3.529412
	2	1.764706
	3	3.294118
	3	2.470588
	1	1.235294
p-value	0.997377642

Here, we assume .

= 0,05

Commet : Here, we see the p- value is more than $\alpha$ so we accept the null hypothesis i.e, the flight carrier and origin airport are independent.