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.)

Answers: 3

Show answers

Answers

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

henry ordered 5 large pizzas

step-by-step explanation:

sorry no step by step

Answer from: Quest

P(chicago, illinois)

Another question on Mathematics

Mathematics, 21.06.2019 14:00

Pokój na planie w skali 1: 100 jest prostokątem i ma wymiary 5,2cmx4,4cm. jaką powierzchnię w m2 ma pokój? ile m bieżących wykładziny dywanowej o szerokości 5m trzeba kupić by wyłożyć cały pokój jednym kawałkiem przy minimalnych stratach ? ile zapłacimy za wykładzinę jeśli 1m2 kosztuje 30 zł ?

Answers: 1

Answer

Mathematics, 21.06.2019 17:30

Find the difference (8ab+a+2) - (3ab+6)

Answers: 1

Answer

Mathematics, 21.06.2019 17:30

The marriott family bought a new apartment three years ago for $65,000. the apartment is now worth $86,515. assuming a steady rate of growth, what was the yearly rate of appreciation? what is the percent of the yearly rate of appreciation?

Answers: 1

Answer

Mathematics, 22.06.2019 04:30

Me. brady is putting trim around the edge of circular merry-go-round that has diameter of 15 feet. how much trim does he need to buy to the nearest tenth

Answers: 1

Answer

You know the right answer?

Many freeways have service (or logo) signs that give information on attractions, camping, lodging, f...

Questions

Mathematics, 29.08.2019 10:30

Slope -6/5,passes through (0,-13/2)...

Mathematics, 29.08.2019 10:30

What is the value of log2 18 rounded to the nearest hundredth...