Question

please

A sample of 123 students is classified with respect to appearance and with respect to academic performance. The following table shows the results of the classification. Academic Performance Appearance High Fair Low Poor Total Attractive 5 10 11 14 40 Ordinary 14 16 16 12 58 Unattractive 10 7 4 4 25 Total 29 33 31 30 123 (a) To determine whether there is any significant relationship between the appearance and academic performance of the students, we can perform a chi-square test. Use EXCEL or otherwise, to calculate the chi-square test statistic for this 3 x 4 table. (b) Determine the p-value of the statistic calculated from part (a), using the EXCEL function QCHIDIST(observed value of the Chi-square statistic, 6). What can be concluded? (c) The number, 6, used in the EXCEL function in part (b) is called the degrees of freedom of the Chi-square test statistic. Can you speculate how to determine this degrees of freedom from the dimension of the table? (Hint: the degrees of freedom of the Chi-square test statistic for a 2 x 2 table is equal to 1.)

Answer 1

a)

The null and alternative hypotheses are     
Ho :   Appearance and performance are independent     
Ha :   Appearance and performance are not independent     
     
α = 0.05     
We use Chi Square test of independence     
     
Grand Total of frequencies = 123     
To find Expected Frequencies     
Expected Frequencies = (Row Total * Column Total)/Grand Total     

Expected Frequencies (E)	High	Fair	Low	Poor
Attractive	9.4309	10.7317	10.0813	9.7561
Ordinary	13.6748	15.561	14.6179	14.1463
Unattractive	5.8943	6.7073	6.3008	6.0976

Following table gives the value of (Observed - Expected)² / Expected

	High	Fair	Low	Poor
Attractive	2.0818	0.0499	0.0837	1.8461
Ordinary	0.0077	0.0124	0.1307	0.3257
Unattractive	2.8598	0.0128	0.8402	0.7216

Chi Square Test Statistic

Chi Square Value = ∑[(Observed - Expected)² / Expected]     
χ² = 8.9722     

b)

From chi square table or CHISQ.DIST.RT(X, df) function of Excel     
we find the p-value     

We calculated χ² = 8.9722     
p-value = CHISQ.DIST.RT( 8.9722, 6) = 0.1751     
p-value = 0.1751     

c)

The dimension of the table is 3 x 4

that is there are 3 rows and 4 columns

degrees of freedom = (number of rows - 1) x (number of columns - 1)

                                 = (3 - 1) x ( 4 - 1)

                                 = 2 x 3

                                 = 6

Degrees of freedom = 6

Decision :

Considering level of significance = 0.05
0.1751 > 0.05     
that is p-value > α     
Hence we DO NOT REJECT Ho     
         
Conclusion :
There does not exist enough statistical evidence at α = 0.05 to show that    
appearance and performance are dependent

That is there is no relationship between appearances and performance