Answer to Question #143283 in Statistics and Probability for Margaret

Question

Let X=level of income and Y=political preference. Use the results shown in the table below abd test on a1% level of significance whether the political preference and the level of income are independent
Political party: Level of income A - 23, 40, 16, 2
B - 11, 75, 107, 14
C - 1, 31, 60, 10
Suppose that the test statistic calculated is 69, 3875, which statement is incorrect?
1. H0 The variable are independent
H1 The variables are dependent
2. The degrees of freedom df = 12
3. The critical value =16, 812
4. There are 390 people
5 H0 is rejected The two variable are dependent

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & Y_1 & Y_2 & Y_3 & Y_4 & Total \\ \hline A & 23 & 40 & 16 & 2 & 81 \\ B & 11 & 75 & 107 & 14 &207 \\ C & 1 & 31 & 60 & 10 & 102 \\ \hdashline Total & 35 & 146 & 183 & 26 & 390 \end{array}$$

 The following null and alternative hypotheses need to be tested:

$H_0:$ The two variables are independent

$H_a:$ The two variables are dependent

This corresponds to a Chi-Square test of independence.

Based on the information provided, the significance level is $\alpha=0.01,$ the number of degrees of freedom is $df=(3-1)\times(4-1)=6,$ so then the rejection region for this test is $R=\{\chi^2:\chi^2>16.8119\}$ 

$$\dfrac{35(81)}{390}=7.2692$$

$$\dfrac{146(81)}{390}=30.3231$$

$$\dfrac{183(81)}{390}=38.0077$$

$$\dfrac{26(81)}{390}=5.4$$

$$\dfrac{35(207)}{390}=18.5769$$

$$\dfrac{146(207)}{390}=77.4923$$

$$\dfrac{183(207)}{390}=97.1308$$

$$\dfrac{26(207)}{390}=13.8$$

$$\dfrac{35(102)}{390}=9.1538$$

$$\dfrac{146(102)}{390}=38.1846$$

$$\dfrac{183(102)}{390}=47.8615$$

$$\dfrac{26(102)}{390}=6.8$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} E_{ij} & Y_1 & Y_2 & Y_3 & Y_4 & Total \\ \hline A & 7.2692 & 30.3231 & 38.0077 & 5.4 & 81 \\ B & 18.5769 & 77.4923 & 97.1308 & 13.8 &207 \\ C & 9.1538 & 38.1846 & 47.8615 & 6.8 & 102 \\ \hdashline Total & 35 & 146 & 183 & 26 & 390 \end{array}$$

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

$$\dfrac{(23-7.2692)^2}{7.2692}=34.0416$$

$$\dfrac{(40-30.3231)^2}{30.3231}=3.0882$$

$$\dfrac{(16-38.0077)^2}{38.0077}=12.7432$$

$$\dfrac{(2-5.4)^2}{5.4}=2.1407$$

$$\dfrac{(11-18.5769)^2}{18.5769}=3.0904$$

$$\dfrac{(75-77.4923)^2}{77.4923}=0.0802$$

$$\dfrac{(107-97.1308)^2}{97.1308}=1.0028$$

$$\dfrac{(14-13.8)^2}{13.8}=0.0029$$

$$\dfrac{(1-9.1538)^2}{9.1538}=7.2630$$

$$\dfrac{(31-38.1846)^2}{38.1846}=1.3518$$

$$\dfrac{(60-47.8615)^2}{47.8615}=3.0785$$

$$\dfrac{(10-6.8)^2}{6.8}=1.5059$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} Squared\ distances& Y_1 & Y_2 & Y_3 & Y_4 \\ \hline A & 34.0416 & 3.0882 & 12.7432 & 2.1407 \\ B & 3.0904 & 0.0802 & 1.0028 & 0.0029 \\ C & 7.2630 & 1.3518 & 3.0785 & 1.5059 \\ \end{array}$$

The Chi-Squared statistic is computed as follows:

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

$$=34.046+3.0882+12.7432+2.1407+$$

$$+3.0904+0.0802+1.0028+0.0029+$$

$$+7.2630+1.3518+3.0785+1.5059=$$

$$=69.3936$$

Since it is observed that $\chi^2=69.3936>16.8119=\chi^2_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the two variables are dependent, at the 0.01 significance level.

The corresponding p-value for the test is $p=P(\chi^2>69.3936)<0.0001.$ 

The statement

2. The degrees of freedom df = 12

is incorrect.