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. An article reported that in one investigation, six sites along 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.
Before: 15 26 66 115 62 64
After: 16 24 42 80 78 73
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.
B. If a seventh site were to be randomly selected among bearing service signs, between what values predict the difference in number of crashes to lie?

Answer 1

Answer:

a) The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that there are less accidents after the sign changes (P-value=0.25).

b) The expected value for the difference for a seventh site is within the interval (-26.498, 14.832) with a 95% of confidence.

Step-by-step explanation:

We have the data:

Before: 15 26 66 115 62 64

After: 16 24 42 80 78 73

As this is a paired t-test, we will calculate the difference d=(after-before) and test the claim that there are less accidents after the sign changes.

The sample difference d is [1, -2, -24, -35, 16, 9].

It has a sample mean of -5.833 and a standard deviation of 19.692.

This is a hypothesis test for the population mean.

The claim is that there are less accidents after the sign changes.

Then, the null and alternative hypothesis are:

$H_0: \mu=0\\\\H_a:\mu< 0$

The significance level is 0.05.

The sample has a size n=6.

The sample mean is M=-5.833.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=19.692.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{19.692}{\sqrt{6}}=8.039$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{-5.833-0}{8.039}=\dfrac{-5.833}{8.039}=-0.726$

The degrees of freedom for this sample size are:

$df=n-1=6-1=5$

This test is a left-tailed test, with 5 degrees of freedom and t=-0.726, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t$

As the P-value (0.25) is bigger than the significance level (0.05), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that there are less accidents after the sign changes.

b) To answer what is the expected value for the difference for a seventh hypothetical site, we have to calculate a 95% confidence interval for the mean.

The sample standard error is s_M=8.039

The t-value for a 95% confidence interval is t=2.571.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.571 \cdot 8.039=20.665$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = -5.833-20.665=-26.498\\\\UL=M+t \cdot s_M = -5.833+20.665=14.832$

The 95% confidence interval for the mean is (-26.498, 14.832).