Question

are annual data for various years. the data are the numbers of cars sold (thousands) and the numbers ed 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 es? would it be reasonable to expect a correlation?
ales 8173 8214 8515 8995 8635 8535 8273 8145
bowl points 60 70 44 76 43 55 54 53
000 9000 0 8000 9000 0 8000 9000 0 8000 9000
car sales car sales car sales car sales
ear correlation coefficient is r = 0.194
to three decimal places as needed.)
st statistic is t = 0.48
d to two decimal places as needed.)
-value is
nd to three decimal places as needed.)

Explanation:

Step1: Recall t - test for correlation

The formula for the t - statistic in a correlation test is \(t=\frac{r\sqrt{n - 2}}{\sqrt{1-r^{2}}}\), and we can also use the t - distribution to find the p - value. The degrees of freedom \(df=n - 2\), where \(n = 8\) (number of data points), so \(df=8 - 2=6\).

We know that \(r = 0.194\) and \(t = 0.48\). To find the p - value, we consider a two - tailed test (since we are testing for a linear correlation, positive or negative). We use the t - distribution with \(df = 6\) and \(t=0.48\).

Step2: Use t - distribution table or calculator

Using a t - distribution calculator or software (such as a TI - 84 Plus, or statistical software like R or Python), for a two - tailed test with \(df = 6\) and \(t = 0.48\), we calculate the p - value.

The cumulative distribution function for the t - distribution \(P(T\leq t)\) for \(t = 0.48\) and \(df = 6\) can be found. The p - value for a two - tailed test is \(2\times(1 - P(T\leq|t|))\).

Using a calculator: In R, we can use the command 2*(1 - pt(0.48,6)). Calculating \(pt(0.48,6)\) (the cumulative probability for a t - distribution with \(df = 6\) at \(t = 0.48\)):

\(pt(0.48,6)\approx0.673\)

Then \(1-0.673 = 0.327\)

And \(2\times0.327=0.654\)

Answer:

The p - value is \(\boxed{0.654}\)