Questions: An automobile company is running a new television commercial in five cities with approximately the same population. The following table shows the number of times the commercial is run on TV in each city and the number of car sales (in hundreds). Find the Pearson correlation coefficient r for the data given in the table. Round any intermediate calculations to no less than six decimal places, and round your final answer to three decimal places.
Number of TV commercials, x 4 5 13 14 18
Car sales, y (in hundreds) 2 4 9 6 7
Transcript text: An automobile company is running a new television commercial in five cities with approximately the same population. The following table shows the number of times the commercial is run on TV in each city and the number of car sales (in hundreds). Find the Pearson correlation coefficient $r$ for the data given in the table. Round any intermediate calculations to no less than six decimal places, and round your final answer to three decimal places.
\begin{tabular}{|l|c|c|c|c|c|}
\hline Number of TV commercials, $x$ & 4 & 5 & 13 & 14 & 18 \\
\hline Car sales, $y$ (in hundreds) & 2 & 4 & 9 & 6 & 7 \\
\hline
\end{tabular}
Solution
Solution Steps
Step 1: Calculate Covariance
The covariance between the number of TV commercials \( X \) and car sales \( Y \) is calculated as follows:
\[
\text{Cov}(X,Y) = 13.15
\]
Step 2: Calculate Standard Deviations
Next, we calculate the standard deviations of \( X \) and \( Y \):
\[
\sigma_X = 6.058
\]
\[
\sigma_Y = 2.702
\]
Step 3: Calculate Pearson Correlation Coefficient
The Pearson correlation coefficient \( r \) is computed using the formula:
\[
r = \frac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y}
\]
Substituting the values we have:
\[
r = \frac{13.15}{6.058 \times 2.702} \approx 0.803
\]
Final Answer
The Pearson correlation coefficient, rounded to three decimal places, is \\(\boxed{0.803}\\).