Mathematics, 06.05.2020 05:29 only1cache

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: 14 22 66 119 68 65 After: 15 21 43 83 79 74 (a) The 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.] State and test the appropriate hypotheses. (Use α = 0.05.)

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Answer from: moisealafleur

As we are testing for any difference, this test is two-tailed and the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

As the P-value (0.749) 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 was a change in the mean number of crashes before and after the addition of distance information on the signs.

Step-by-step explanation:

We start by calculating the mean and standard deviation for the number of crashes BEFORE the installation of the signs:

$M=\dfrac{1}{6}\sum_{i=1}^6(14+22+66+119+68+65)\\\\\\ M=\dfrac{354}{6}=59$

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^6(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{5}\cdot [(14-59)^2+(22-59)^2+(66-59)^2+(119-59)^2+(68-59)^2+(65-59)^2]}\\\\\\$

$s=\sqrt{\dfrac{1}{5}\cdot [(2025)+(1369)+(49)+(3600)+(81)+(36)]}=\sqrt{\dfrac{7160}{5}}=\sqrt{1432}\\\\\\s=37.842$

The mean and standard deviation of the number of crashes AFTER the installation of the signs is:

$M=\dfrac{1}{6}\sum_{i=1}^6(15+21+43+83+79+74)\\\\\\ M=\dfrac{315}{6}=52.5$

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^6(x_i-M)^2}\\\\\\$

$s= \sqrt{ \dfrac{1}{5} [(15-52.5)^2+(21-52.5)^2+(43-52.5)^2+(83-52.5)^2+(79-52.5)^2+(74-52.5)^2]}$

$s=\sqrt{\dfrac{1}{5}\cdot [(1406.25)+(992.25)+(90.25)+(930.25)+(702.25)+(462.25)]}\\\\\\s=\sqrt{\dfrac{4583.5}{5}}=\sqrt{916.7}\\\\\\s=30.277$

This is a hypothesis test for the difference between populations means.

The claim is that there was a change in the mean number of crashes before and after the addition of distance information on the signs.

As we are testing for any difference, this test is two-tailed and the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

The significance level is 0.05.

The sample 1, of size n1=6 has a mean of 59 and a standard deviation of 37.842.

The sample 1, of size n1=6 has a mean of 52.5 and a standard deviation of 30.277.

The difference between sample means is Md=6.5.

M_d=M_1-M_2=59-52.5=6.5

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{37.842^2}{6}+\dfrac{30.277^2}{6}}\\\\\\s_{M_d}=\sqrt{238.669+152.783}=\sqrt{391.452}=19.785$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{6.5-0}{19.785}=\dfrac{6.5}{19.785}=0.329$

The degrees of freedom for this test are:

df=n_1+n_2-1=6+6-2=10

This test is a two-tailed test, with 10 degrees of freedom and t=0.329, so the P-value for this test is calculated as (using a t-table):

$P-value=2\cdot P(t0.329)=0.749$

As the P-value (0.749) 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 was a change in the mean number of crashes before and after the addition of distance information on the signs.

Answer from: Quest

it would rotate about 180 degrees.

Answer from: Quest

where is the graph? need more info