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A survey asks a group of students if they buy CDs or not. It also asks if the students own a smartphone or not. These values are recorded in the contingency table below. Which of the following tables correctly shows the expected values for the chi-square independence test?

\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:

\begin{tabular}{|c|c|c|c|c|c|}
\hline & & CDs & No CDs & Row Total & \\
\hline & Smartphone & \begin{tabular}{c}
5 \\
10.9
\end{tabular} & \begin{tabular}{c}
25 \\
18.1
\end{tabular} & 30 & \\
\hline & No Smartphone & \begin{tabular}{c}
17 \\
13.1
\end{tabular} & \begin{tabular}{c}
20 \\
26.9
\end{tabular} & 37 & \\
\hline & Column Total & 22 & 45 & 67 & \\
\hline & & CDs & No CDs & Row Total & \\
\hline & Smartphone & \begin{tabular}{c}
5 \\
10.9
\end{tabular} & \begin{tabular}{c}
25 \\
22.1
\end{tabular} & 30 & \\
\hline & No Smartphone & \begin{tabular}{c}
17 \\
11.1
\end{tabular} & \begin{tabular}{c}
20 \\
25.9
\end{tabular} & 37 & \\
\hline
\end{tabular}

Sagot :

GinnyAnswer
Let me walk you through the process of finding the expected values in a Chi-Square test for independence based on the provided contingency table.

First, we need to collect the totals from our contingency table:
[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]

Next, we use the formula for expected values in each cell:
[tex]\[ E = \frac{{(\text{Row Total}) \times (\text{Column Total})}}{{\text{Grand Total}}} \][/tex]

Now, let’s calculate each expected value step by step:

1. Expected value for Smartphones & CDs:
[tex]\[ E(\text{Smartphone, CDs}) = \frac{{\text{Row Total (Smartphones)} \times \text{Column Total (CDs)}}}{{\text{Grand Total}}} = \frac{{30 \times 22}}{67} \approx 9.85 \][/tex]

2. Expected value for Smartphones & No CDs:
[tex]\[ E(\text{Smartphone, No CDs}) = \frac{{\text{Row Total (Smartphones)} \times \text{Column Total (No CDs)}}}{{\text{Grand Total}}} = \frac{{30 \times 45}}{67} \approx 20.15 \][/tex]

3. Expected value for No Smartphones & CDs:
[tex]\[ E(\text{No Smartphone, CDs}) = \frac{{\text{Row Total (No Smartphones)} \times \text{Column Total (CDs)}}}{{\text{Grand Total}}} = \frac{{37 \times 22}}{67} \approx 12.15 \][/tex]

4. Expected value for No Smartphones & No CDs:
[tex]\[ E(\text{No Smartphone, No CDs}) = \frac{{\text{Row Total (No Smartphones)} \times \text{Column Total (No CDs)}}}{{\text{Grand Total}}} = \frac{{37 \times 45}}{67} \approx 24.85 \][/tex]

Placing these expected values in our table, we get:
[tex]\[ \begin{array}{|c|c|c|c|} \hline & \text{CDs} & \text{No CDs} & \text{Row Total} \\ \hline \text{Smartphone} & 5\ (9.85) & 25\ (20.15) & 30 \\ \hline \text{No Smartphone} & 17\ (12.15) & 20\ (24.85) & 37 \\ \hline \text{Column Total} & 22 & 45 & 67 \\ \hline \end{array} \][/tex]

Using the above table, we identify the correct answer from the given options is:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \multirow{4}{*}{ O } & & \text{CDs} & \text{No CDs} & \text{Row Total} & \\ \hline & \text{Smartphone} & 5\ (9.85) & 25\ (20.15) & 30 & \\ \hline & \text{No Smartphone} & 17\ (12.15) & 20\ (24.85) & 37 & \\ \hline & \text{Column Total} & 22 & 45 & 67 & \\ \hline \end{array} \][/tex]
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