Question

You are tasked with investigating whether fake news media platform and personality are independent. Consider the table below to help you with your task.
Celebrity. C/ Politician .P/ Sport Star .SS/ Total
Social Media .SM/ 1800 700 515 3015
Traditional Media. TM/ 485 350 150 985
Total 2285 1050 775 4000

Consider the statements below. Assume the level of significance is 1%
A. We reject the null hypothesis
B. We do not reject the null hypothesis
C. Media platform and personality are independent

Which statements are correct?
1. Only A
2. Only B
3. Only C
4. A and C Only
5. B and C only

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} & C & P & SS & Total \\ \hline SM & 1800 & 700 & 515 & 3015 \\ \hdashline TM & 485 & 350 & 150 & 985 \\ \hdashline Total & 2285 & 1050 & 665 & 4000 \end{array}$$

The expected values are computed in terms of row and column totals. In fact, the formula is

$$E_{ij}=\dfrac{R_i\times C_j}{T},$$

where $R_i$ corresponds to the total sum of elements in row $i,$ $C_j$ corresponds to the total sum of elements in column $j,$ and $T$ is the grand total.

$$E_{11}=\dfrac{3015\times 2285}{4000}=1722.31875$$ $$E_{12}=\dfrac{3015\times 1050}{4000}=791.4375$$

$$E_{13}=\dfrac{3015\times 665}{4000}=501.24375$$ $$E_{21}=\dfrac{985\times 2285}{4000}=562.68125$$ $$E_{22}=\dfrac{985\times 1050}{4000}=258.5625$$ $$E_{23}=\dfrac{985\times 665}{4000}=163.75625$$

The table below shows the calculations to obtain the table with expected values:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} Expected\ Values & C & P & SS & Total \\ \hline SM & 1722.32 & 791.44& 501.24 & 3015 \\ \hdashline TM & 562.68 & 258.56 & 163.76 & 985 \\ \hdashline Total & 2285 & 1050 & 665 & 4000 \end{array}$$

Based on the observed and expected values, the squared distances can be computed according to the following formula: $\dfrac{(E-O)^2}{E}.$ The table with squared distances is shown below:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} Squared\ Distances & C & P & SS \\ \hline SM & 3.504 & 10.564 & 0.378 \\ \hdashline TM & 10.724 & 32.336 & 1.156 \end{array}$$

The following null and alternative hypotheses need to be tested:

$H_0:$ The two variables are independent.

$H_1:$ The two variables are dependent.

This corresponds to a Chi-Square test of independence.

Based on the information provided, the significance level is $\alpha=0.05,$ the number of degrees of freedom is $df=(2-1)\times(3-1)=2,$ so then the rejection region for this test is $R=\{\chi^2:\chi^2>5.991\}.$

The Chi-Squared statistic is computed as follows:

$$\chi^2=3.504+10.724+10.564+32.336+$$ $$+0.378+1.156=58.661$$

Since it is observed that $\chi^2=58.661>5.991=\chi_c^2,$ 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 0.05 significance level.

1. Only A statement is correct.