Question

Many freeways have service (or logo) signs that give information on attractions, camping, lodging, food, and gas services prior to off-ramps. These signs typically do not provide information on distances. The article "Evaluation of Adding Distance Information to Freeway-Specific Service (Logo) Signs" \((J\). of Transp. Engr., 2011: 782-788) reported that in one investigation, six sites along Virginia interstate highways where service signs are posted were selected. For each site, crash data was obtained for a three-year period before distance information was added to the service signs and for a one-year period afterward. The number of crashes per year before and after the sign changes were as follows:

\(\begin{array}{*{20}{l}}{Before:\;\;15\;26\;66\;115\;62\;64}\\{After:\;\;\;\;\;16\;24\;42\;80\;\;78\;73}\end{array}\)

a. The cited article included the statement "A paired\(t\)test was performed to determine whether there was any change in the mean number of crashes before and after the addition of distance information on the signs." Carry out such a test. (Note: The relevant normal probability plot shows a substantial linear pattern.)

b. If a seventh site were to be randomly selected among locations bearing service signs, between what values would you predict the difference in number of crashes to lie?

Short answer

(a) There is not sufficient evidence to support the claim that there was any change in the mean number of crashes before and after the addition of distance information on the signs.

(b) \(\left( { - 48.8508,60.5174} \right)\)

Step-by-step solution

A)Step 1: Find the difference

Let us assume:

\(\alpha = 0.05\)

Determine the difference in value of each pair.

Sample 1	Sample 2	Difference D
15	16	-1
26	24	2
66	42	24
115	80	35
62	78	-16
64	73	-9

Find the value of test statistics and P

Given claim: a change

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain the value mentioned in the claim.

\(\begin{array}{l}{H_0}:{\mu _d} = 0\\{H_a}:{\mu _d} \ne 0\end{array}\)

Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.

\(\bar d = \frac{{ - 1 + 2 + 24 + 35 - 16 - 9}}{6} \approx 5.8333\)

Determine the sample standard deviation of the differences:

\({s_d} = \sqrt {\frac{{{{( - 1 - 5.8333)}^2} + \ldots + {{( - 9 - 5.8333)}^2}}}{{6 - 1}}} \approx 19.6918\)

Determine the value of the test statistic:

\(t = \frac{{\bar d - d}}{{{s_d}/\sqrt n }} = \frac{{5.8333 - 0}}{{19.6918/\sqrt 6 }} \approx 0.726\)

The\({\rm{P}}\)-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. The Pvalue is the number (or interval) in the column title of the Student's T distribution in the appendix containing the t-value in the row \(df = n - 1 = 6 - 1 = 5\) :

\(P > 2 \times 0.10 = 0.20\)If the P-value is less than the significance level, reject the null hypothesis.

\(P > 0.05 \Rightarrow {\rm{ Fail to reject }}{H_0}\)

There is not sufficient evidence to support the claim that there was any change in the mean number of crashes before and after the addition of distance information on the signs.

B)Step 3: Find the value of difference

We will determine a\(95\% \) prediction interval. Other confidence levels can be determined similarly.

\(c = 95\% = 0.95\)

Determine the difference in value of each pair.

Sample 1	Sample 2	Difference D
15	16	-1
26	24	2
66	42	24
115	80	35
62	78	-16
64	73	-9

Find the sample standard deviation

Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.

\(\bar d = \frac{{ - 1 + 2 + 24 + 35 - 16 - 9}}{6} \approx 5.8333\)

Determine the sample standard deviation of the differences: \({s_d} = \sqrt {\frac{{{{( - 1 - 5.8333)}^2} + \ldots + {{( - 9 - 5.8333)}^2}}}{{6 - 1}}} \approx 19.6918\)

Find the endpoint of the prediction interval

Determine the t-value by looking in the row starting with degrees of freedom\(df = n - 1 = 6 - 1 = 5\)and in the column with\(\alpha = 1 - c/2 = 0.025\)in the Student's t distribution table in the appendix:

\({t_{\alpha /2}} = 2.571\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot {s_d}\sqrt {1 + \frac{1}{n}} = 2.571 \cdot 19.6918\sqrt {1 + \frac{1}{6}} \approx 54.6841\)

The endpoints of the prediction interval for\({\mu _d}\)are:

\(\begin{array}{l}\bar d - E = 5.8333 - 54.6841 = - 48.8508\\\bar d + E = 5.8333 + 54.6841 = 60.5174\end{array}\)