Question

the weights (in pounds) of eight vehicles and the variabilities of their braking distances (in feet) when stopping on a wet surface are shown in the table. at α = 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 wet surface? weight, x: 5600 5400 6000 5100 5900 4800 5700 5800 variability, y: 2.19 2.46 4.04 1.77 2.86 2.51 2.31 2.79 use technology to calculate the test statistic, rounding to two decimal places. t = 3.42 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. p - value = 0.014

Explanation:

Step1: Identify the test - statistic formula

The test - statistic \(t\) for testing the significance of the linear correlation coefficient \(r\) is given by \(t = r\sqrt{\frac{n - 2}{1 - r^{2}}}\), where \(n\) is the number of data pairs. Here, \(t = 3.42\) is already given. The degrees of freedom \(df=n - 2\). Since \(n = 8\) (number of pairs of data for weight \(x\) and variability \(y\)), \(df=8 - 2=6\).

Step2: Understand the P - value concept

The P - value is the probability of obtaining a result as extreme or more extreme than the observed result, assuming the null hypothesis (\(H_0:
ho = 0\), no linear correlation) is true. Given \(t = 3.42\) and \(df = 6\), using a \(t\) - distribution table or statistical software, the P - value is calculated as \(0.014\).

Answer:

The test - statistic \(t = 3.42\), degrees of freedom \(df = 6\), and P - value \(=0.014\)