Question

are annual data for various years. the data are the numbers of cars sold (thousands) and the numbers of d in the super bowl. construct a scatterplot, find the value of the linear correlation coefficient r, and find using α=0.05. is there sufficient evidence to conclude that there is a linear correlation between those s? would it be reasonable to expect a correlation? les 8173 8214 8515 8995 8635 8535 8273 8145 bowl points 60 70 44 76 43 55 54 53 ear correlation coefficient is r= to three decimal places as needed.)

Explanation:

Step1: Identify Variables

Let \( x \) be Car Sales (thousands) and \( y \) be Super Bowl Points. The data points are:
\( x: [8173, 8214, 8515, 8995, 8635, 8535, 8273, 8145] \)
\( y: [60, 70, 44, 76, 43, 55, 54, 53] \)

Step2: Calculate Necessary Sums

• \( \sum x = 8173 + 8214 + 8515 + 8995 + 8635 + 8535 + 8273 + 8145 = 67485 \)
• \( \sum y = 60 + 70 + 44 + 76 + 43 + 55 + 54 + 53 = 455 \)
• \( \sum xy = (8173×60)+(8214×70)+(8515×44)+(8995×76)+(8635×43)+(8535×55)+(8273×54)+(8145×53) \)

\( = 490380 + 574980 + 374660 + 683620 + 371305 + 469425 + 446742 + 431685 = 3842797 \)

• \( \sum x^2 = 8173^2 + 8214^2 + 8515^2 + 8995^2 + 8635^2 + 8535^2 + 8273^2 + 8145^2 \)

\( = 66807929 + 67469796 + 72505225 + 80910025 + 74563225 + 72846225 + 68442529 + 66341025 = 571886989 \)

• \( \sum y^2 = 60^2 + 70^2 + 44^2 + 76^2 + 43^2 + 55^2 + 54^2 + 53^2 \)

\( = 3600 + 4900 + 1936 + 5776 + 1849 + 3025 + 2916 + 2809 = 26811 \)

Step3: Use Correlation Coefficient Formula

The formula for \( r \) is:

$$ r = \frac{n\sum xy - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}} $$

where \( n = 8 \).

• Numerator: \( 8×3842797 - 67485×455 = 30742376 - 30705675 = 36701 \)
• Denominator part 1: \( 8×571886989 - 67485^2 = 4575095912 - 4554225225 = 20870687 \)
• Denominator part 2: \( 8×26811 - 455^2 = 214488 - 207025 = 7463 \)
• Denominator: \( \sqrt{20870687×7463} \approx \sqrt{155757000000} \approx 124802.64 \)
• \( r = \frac{36701}{124802.64} \approx 0.294 \)

Answer:

\( r \approx 0.294 \)