Answer to Question #135917 in Statistics and Probability for Mbali

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.