Question

the weights (in pounds) of eight vehicles and the variabilities of their braking distances (in feet) when stopping on a dry surface are shown in the table. at $alpha = 0.01$, is there enough evidence to conclude that there is a significant linear correlation between vehicle weight and variability in braking distance on a dry surface?
weight, x|5900|5400|6500|5100|5800|4800|5700|5820|
variability, y|1.69|2.06|1.87|1.62|1.55|1.50|1.68|1.70|
use technology to calculate the test statistic, rounding to two decimal places.
t = 0.97
the test statistic, t, follows a t - distribution with n - 2 degrees of freedom, where n is the number of pairs of data. identify the number of degrees of freedom for the given data.
d.f. = 6
use the same technology output that was used to find the test statistic to calculate the p - value, rounding to three decimal places.

Explanation:

Step1: Recall P - value concept for t - distribution

The P - value is calculated based on the t - distribution with degrees of freedom \(d.f.=n - 2\). Here \(n = 8\) (number of data pairs), so \(d.f.=6\) and \(t = 0.97\). For a two - tailed test of correlation, we find the P - value associated with this t - statistic in a t - distribution.

Step2: Use technology for P - value calculation

Using a t - distribution calculator or statistical software (e.g., TI - 84 Plus: 2 - SampTTest, R: 2pt(0.97,6,lower.tail = FALSE)), we calculate the P - value for a two - tailed test. The formula for a two - tailed P - value in R is \(2(1 - pt(abs(0.97),6))\).
The result is \(P=0.369\)

Answer:

\(0.369\)