The weights (in pounds) of elght vehicles and the variabilities of their braking distances (in feet) when stopping on a wet surface are shown in the table. At $\alpha=0.01$, is there enough evidence to conclude that there is a significant linear cortclation batween vehicle weight and variability in braking distance on a wet surface?
\begin{tabular}{|l|c|c|c|c|c|c|c|c|}
\hline Weight, $\mathrm{x}$ & 5970 & 5360 & 6500 & 5100 & 5870 & 4800 & 5700 & 5880 \\
\hline Variability, $\mathrm{y}$ & 293 & 240 & 4.07 & 178 & 294 & 249 & 2.25 & 270 \\
\hline
\end{tabular}
Setup the hypothesis for the test
\[
\begin{array}{l}
H_{0} p=0 \\
H_{a} p=0
\end{array}
\]
Calculate the test statistic
$t=33$ (Round to two decimal places as needed)
Cilculate the Pivalue
\[
\text { P.value }=\square \text { (Round to three decirtial places as needect) }
\]
Answer
Final Answer: The p-value is \(\boxed{5.15 \times 10^{-8}}\).
Step 1 ：Setup the hypothesis for the test: \(H_{0}: p=0\) (There is no correlation) and \(H_{a}: p \neq 0\) (There is a correlation).
Step 2 ：Calculate the test statistic: \(t=33\).
Step 3 ：Calculate the degrees of freedom for this test: \(df = n-2 = 8-2 = 6\).
Step 4 ：Calculate the p-value using the t-distribution with the given test statistic and degrees of freedom: \(p = 5.151762394461912 \times 10^{-8}\).
Step 5 ：Compare the p-value with the significance level \(\alpha = 0.01\). Since the p-value is much less than the significance level, we reject the null hypothesis.
Step 6 ：Conclude that there is a significant linear correlation between vehicle weight and variability in braking distance on a wet surface.
Step 7 ：Final Answer: The p-value is \(\boxed{5.15 \times 10^{-8}}\).
