Get the information you need quickly and easily with IDNLearn.com. Our platform offers reliable and comprehensive answers to help you make informed decisions quickly and easily.

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]


Sagot :

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]