## Related questions with answers

The following observations are on stopping distance (ft) of a particular truck at 20 mph under specified experimental conditions (“Experimental Measurement of the Stopping Performance of a Tractor-Semitrailer from Multiple Speeds,” NHTSA, DOT HS 811 488, June 2011):

$\begin{matrix} \text{32.1} & \text{30.6} & \text{31.4} & \text{30.4} & \text{31.0} & \text{31.9}\\ \end{matrix}$

The cited report states that under these conditions, the maximum allowable stopping distance is 30. A normal probability plot validates the assumption that stopping distance is normally distributed. Does the data suggest that true average stopping distance exceeds this maximum value? Test the appropriate hypotheses using $\alpha=.01$.

Solution

VerifiedGiven:

$n=6$

$\alpha=0.01$

32.1, 30.6, 31.4, 30.4, 31, 31.9

The mean is the sum of all values divided by the number of values:

$\overline{x}=\dfrac{32.1+ 30.6+31.4+30.4+31+31.9}{6}=\dfrac{187.4}{6}\approx 31.2333$

The variance is the sum of squared deviations from the mean divided by $n-1$. The standard deviation is the square root of the variance:

$s=\sqrt{\dfrac{(32.1-31.2333)^2+....+(31.9-31.2333)^2}{6-1}}\approx 0.6890$

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