Continuing with the previous question, here is the contingency table. Using the expected frequencies found in the previous part, determine the test statistic, [tex]$\chi_0^2$[/tex], for this test of independence.

\begin{tabular}{|c|c|c|c|}
\hline & CDs & No CDs & Row Total \\
\hline Smartphone & 5 & 25 & 30 \\
\hline No Smartphone & 17 & 20 & 37 \\
\hline Column Total & 22 & 45 & 67 \\
\hline
\end{tabular}

Select the correct answer below:
A. [tex]$\chi_0^2=1.3$[/tex]
B. [tex][tex]$\chi_0^2=3.3$[/tex][/tex]
C. [tex]$\chi_0^2=6.6$[/tex]
D. [tex]$\chi_0^2=9.2$[/tex]
E. [tex][tex]$\chi_0^2=13.2$[/tex][/tex]



Answer :

Let's solve the problem step-by-step.

First, let's denote the observed frequencies from the contingency table:

Observed frequencies:
[tex]\[ \begin{array}{|c|c|c|c|} \hline & \text{CDs} & \text{No CDs} & \text{Row Total} \\ \hline \text{Smartphone} & 5 & 25 & 30 \\ \hline \text{No Smartphone} & 17 & 20 & 37 \\ \hline \text{Column Total} & 22 & 45 & 67 \\ \hline \end{array} \][/tex]

### Step 1: Calculate the Expected Frequencies
To calculate expected frequencies for each cell, we use the formula:
[tex]\[ E_{ij} = \frac{( \text{Row Total}_i \times \text{Column Total}_j )}{\text{Grand Total}} \][/tex]
where [tex]\( E_{ij} \)[/tex] is the expected frequency for cell [tex]\( (i, j) \)[/tex].

For the cell (Smartphone, CDs):
[tex]\[ E_{11} = \frac{(30 \times 22)}{67} \approx 9.85 \][/tex]

For the cell (Smartphone, No CDs):
[tex]\[ E_{12} = \frac{(30 \times 45)}{67} \approx 20.15 \][/tex]

For the cell (No Smartphone, CDs):
[tex]\[ E_{21} = \frac{(37 \times 22)}{67} \approx 12.15 \][/tex]

For the cell (No Smartphone, No CDs):
[tex]\[ E_{22} = \frac{(37 \times 45)}{67} \approx 24.85 \][/tex]

### Step 2: Calculate the Chi-Squared Test Statistic
To determine the test statistic [tex]\(\chi_0^2\)[/tex], we use the formula:
[tex]\[ \chi_0^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \][/tex]
where [tex]\( O_{ij} \)[/tex] are the observed frequencies and [tex]\( E_{ij} \)[/tex] are the expected frequencies.

For cell (1,1):
[tex]\[ \frac{(5 - 9.85)^2}{9.85} \approx 2.38 \][/tex]

For cell (1,2):
[tex]\[ \frac{(25 - 20.15)^2}{20.15} \approx 1.18 \][/tex]

For cell (2,1):
[tex]\[ \frac{(17 - 12.15)^2}{12.15} \approx 1.97 \][/tex]

For cell (2,2):
[tex]\[ \frac{(20 - 24.85)^2}{24.85} \approx 0.94 \][/tex]

Now, we add these values together to get the test statistic:
[tex]\[ \chi_0^2 = 2.38 + 1.18 + 1.97 + 0.94 \approx 6.47 \][/tex]

### Final Step: Conclusion & Selection
Therefore, the calculated value for the chi-squared test statistic [tex]\(\chi_0^2\)[/tex] is approximately 6.44.

Given the provided answer choices, the closest one is:
[tex]\[ \chi_0^2 = 6.6 \][/tex]

Thus, we select:
[tex]\(\chi_0^2 = 6.6\)[/tex]