Question

QUESTION 3 (13 MARKS)

A faculty is interested in whether there is a relationship between gender and favourite subject at his college students. He tabulated some men and women on campus and asked them if their favourite subject was Mathematics (M), Geography (G), and Science (S).

M

G

S

Total

Women

10

14

10

34

Men

11

22

14

23

Total

21

36

24

57



Use the data provided above to test if there is any relationship between gender and favourite subject. By using α = 0.01, Answer the following questions

3.1) Formulate the null and alternative hypotheses (2)

3.2) Select and compute the test statistic(5)

3.3) Obtain the critical value and formulate the decision.(3)

3.4) Based on the information obtained in Questions 3.1 ; 3.2 & 3.3, what can the faculty conclude? (3)

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} & M & G & S & Total \\ \hline Women & 10 & 14 & 10 & 34 \\ Men & 11 & 22 & 14 & 47 \\ \hdashline Total & 21 & 36 & 24 & 81 \end{array}$$

3.1) The following null and alternative hypotheses need to be tested:

$H_0:$ The two variables are independent

$H_a:$ The two variables are dependent

3.2) This corresponds to a Chi-Square test of independence.

3.3) Based on the information provided, the significance level is $\alpha=0.01,$ the number of degrees of freedom is $df=(2-1)\times(3-1)=2,$ so then the rejection region for this test is $R=\{\chi^2:\chi^2>9.21\}$  

3.4)

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} Expected\ values& M & G & S & Total \\ \hline Women & \dfrac{21(34)}{81} & \dfrac{36(34)}{81} & \dfrac{24(34)}{81} & 34 \\ Men & \dfrac{21(47)}{81} & \dfrac{36(47)}{81} & \dfrac{24(47)}{81} & 47 \\ \hdashline Total & 21 & 36 & 24 & 81 \end{array}$$

Based on the observed and expected values, the squared distances can be computed according to the following formula: $(E-O)^2/E$

$$\dfrac{(10-8.8148)^2}{8.8148}=0.1594$$

$$\dfrac{(14-8.8148)^2}{8.8148}=0.0817$$

$$\dfrac{(10-10.0741)^2}{10.0741}=0.0005$$

$$\dfrac{(11-12.1852)^2}{12.1852}=0.1153$$

$$\dfrac{(22-20.8889)^2}{20.8889}=0.0591$$

$$\dfrac{(14-13.9259)^2}{13.9259}=0.0004$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} Squared\ distances& M & G & S \\ \hline Women & 0.1594 & 0.0817 & 0.0005 \\ Men & 0.1153 & 0.0591 & 0.0004 \\ \end{array}$$

The Chi-Squared statistic is computed as follows:

$$\chi^2=\sum_{i, j}\dfrac{(O_{ij}-E_{ij})^2}{E_{ij}}=$$

$$=0.1594+0.0817+0.0005+0.1153+0.0591+0.0004=$$

$$=0.4164$$

Since it is observed that $\chi^2=0.4164<9.21=\chi^2_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is NOT enough evidence to claim that the two variables are dependent, at the 0.01 significance level.

The corresponding p-value for the test is $p=P(\chi^2>0.4164)=0.812045.$