Question

The following table summarises the data from a survey on the ownership of iPods among families with different levels of income. Determine whether the two nominal variables are related. Ownership Level of income No 40 32 48 Yes 30 48 ! ................ 52

Answer 1

The two nominal variables are related.

The following table summarises the data from a survey on the ownership of iPods among families with different levels of income.  

Ownership           C1                  C2                  C3

No                         40                 32                   48

Yes                        30                 48                   52

The first thing to do in order to determine if they are related or not is to state our null and alternative hypothesis

Null hypothesis

$\mathbf{H_o: Two \nomial \ variables \ are \ related}$

Alternative hypothesis

$\mathbf{H_a: Two \nomial \ variables \ are \ not \ related}$

Using the Chi-square test statistics which can be expressed by using the formula  

$X ^2 = \sum \dfrac{(O-E)^2}{E}$

Ownership           C1                  C2                  C3                  Total

No                         40                 32                   48                 120

Yes                        30                 48                   52                 130

Total                      70                 80                   100                250

The expected values are calculated as:

$\mathsf{E_{a,b} = \dfrac{(row \ total \times column \ total )}{grand \ total }}$

$\mathsf{E_{1,1} = \dfrac{(70 \times120 )}{250 }}$

$\mathsf{E_{1,1} = 33.6}$

$\mathsf{E_{1,2} = \dfrac{(70 \times130 )}{250 }}$

$\mathsf{E_{1,2} = 36.4}$

$\mathsf{E_{2,1} = \dfrac{(80 \times120 )}{250 }}$

$\mathsf{E_{2,1} = 38.4}$

$\mathsf{E_{2,2} = \dfrac{(80 \times 130 )}{250 }}$

$\mathsf{E_{2,2} = 41.6}$

$\mathsf{E_{3,1} = \dfrac{(100 \times120 )}{250 }}$

$\mathsf{E_{3,1} = 48}$

$\mathsf{E_{3,2} = \dfrac{(70 \times130 )}{250 }}$

$\mathsf{E_{3,2} = 52}$

∴ Using the Chi-square test statistics, we have:

$X ^2 = \sum \dfrac{(O-E)^2}{E}$

$X ^2 = \Bigg( \dfrac{(40-33.6)^2}{33.6}+ \dfrac{(30-36.4)^2}{36.4}+ \dfrac{(32-38.4)^2}{38.4}+ \dfrac{(48-41.6)^2}{41.6} + \dfrac{(48-48)^2}{48}+ \dfrac{(52-52)^2}{52} \Bigg)$

$X ^2 = \Bigg( \dfrac{40.96}{33.6}+ \dfrac{40.96}{36.4}+ \dfrac{40.96}{38.4}+ \dfrac{40.96}{41.6} + \dfrac{0}{48}+ \dfrac{0}{52} \Bigg)$

$X ^2 = \Bigg( 1.2190+ 1.1253+ 1.0667+ 0.9846+0+ 0 \Bigg)$

$\mathbf{X ^2 =4.3956}$

The degree of freedom df = ((r - 1) × (c - 1))

= (3 - 1) (2 -1 )

= 2 × 1

= 2

∴

Assuming the level of significance = 5%

The p-value of the Chi-square test statistics at df of 2 is:

= $\mathbf{P(X^2 > 4.3956) \implies 0.111}$

Therefore, we can conclude that since the p-value (0.111) is greater than the level of significance (0.05), we fail to reject the null hypothesis.

Hence, the two nominal variables are related.

Learn more about Chi-square test statistics here:

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