Question

An educational researcher wishes to know if there is a difference in academic performance for college freshmen that live on campus and those that commute. Data was collected from 214214 students. Can we conclude that freshman housing location and academic performance are related?

Location	Average	Below Average	Above Average	Total
On campus	7777	4242	3939	158158
Off campus	2828	1414	1414	5656
Total	105105	5656	5353	214214

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Step 1 of 8 :  

State the null and alternative hypothesis.

Step 2 of 8 : Find the expected value for the number of students that live on campus and have academic performance that is average. Round your answer to one decimal place.

Step 3 of 8 : Find the expected value for the number of students that live on campus and have academic performance that is above average. Round your answer to one decimal place.

Step 4 of 8 : Find the value of the test statistic. Round your answer to three decimal places.

Step 5 of 8 : Find the degrees of freedom associated with the test statistic for this problem.

Step 6 of 8 : Find the critical value of the test at the

0.01

0.01

level of significance. Round your answer to three decimal places.

Step 7 of 8 : Make the decision to reject or fail to reject the null hypothesis at the

0.01

0.01

level of significance.

Step 8 of 8 : State the conclusion of the hypothesis test at the

0.01

0.01

level of significance.

Answer 1

step 1

null hypothesis: Ho: freshman housing location and academic performance are independent

Ha: freshman housing location and academic performance are related

step 2:

expected value =row total*column total/grand total =158*105/214 =77.50

step 3: expected value =158*53/214=39.10

step 4:

applying chi square test:

Expected	Ei=Σrow*Σcolumn/Σtotal	average	below average	above average	Total
	on campus	77.500	41.300	39.100	157.9
	Off campus	27.500	14.700	13.900	56.1
	Total	105	56	53	214
chi square	=(Oi-Ei)2/Ei	average	below average	above average	Total
	on campus	0.003	0.012	0.000	0.015
	Off campus	0.009	0.033	0.001	0.043
	Total	0.012	0.045	0.001	0.058

from above test statistic =0.058 ( please try 0.055 if this comes wrong)

Step 5 of 8 : degree of freedom =(row-1)*(column-1)=(2-1)*(3-1)=2

step 6: crtiical value =9.210

Step 7 of 8 :as test statistic is less than crtiical value ; fail to reject null hypothesis

Step 8 of 8 : we do not have sufficient evidence to conclude that freshman housing location and academic performance are related