Please answer B, C, & D. The average number of accidents at controlled intersections per year is 5.6. Is this average a differentnumber for intersections with cameras installed? The 59 randomly observed intersections with camerasinstalled had an average of 5.4 accidents per year and the standard deviation was 1.21. What can beconcluded at the a = 0.01 level of significance?a. For this study, we should use t-test for a population meanb. The null and alternative hypotheses would be:Ho: ?v Select an answer vH1:?v Select an answer vc. The test statistic ?v =(please show your answer to 3 decimal places.)d. The p-value =(Please show your answer to 4 decimal places.)e. The p-value isf. Based on this, we should fail to rejectthe null hypothesis.g. Thus, the final conclusion is that ...The data suggest that the population mean is not significantly different from 5.6 at a = 0.01,so there is statistically insignificant evidence to conclude that the population mean number ofaccidents per year at intersections with cameras installed is different from 5.6 accidents.The data suggest that the populaton mean is significantly different from 5.6 at a = 0.01, sothere is statistically significant evidence to conclude that the population mean number ofaccidents per year at intersections with cameras installed is different from 5.6 accidents.O The data suggest that the sample mean is not significantly different from 5.6 at a = 0.01, sothere is statistically insignificant evidence to conclude that the sample mean number ofaccidents per year at intersections with cameras installed is different from 5.4 accidents.%3D

Answered: b. The null and alternative hypotheses…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

Please answer B, C, & D.

The average number of accidents at controlled intersections per year is 5.6. Is this average a different
number for intersections with cameras installed? The 59 randomly observed intersections with cameras
installed had an average of 5.4 accidents per year and the standard deviation was 1.21. What can be
concluded at the a = 0.01 level of significance?
a. For this study, we should use t-test for a population mean
b. The null and alternative hypotheses would be:
Ho: ?v Select an answer v
H1:
?v Select an answer v
c. The test statistic ?v =
(please show your answer to 3 decimal places.)
d. The p-value =
(Please show your answer to 4 decimal places.)
e. The p-value is
f. Based on this, we should fail to reject
the null hypothesis.
g. Thus, the final conclusion is that ...
The data suggest that the population mean is not significantly different from 5.6 at a = 0.01,
so there is statistically insignificant evidence to conclude that the population mean number of
accidents per year at intersections with cameras installed is different from 5.6 accidents.
The data suggest that the populaton mean is significantly different from 5.6 at a = 0.01, so
there is statistically significant evidence to conclude that the population mean number of
accidents per year at intersections with cameras installed is different from 5.6 accidents.
O The data suggest that the sample mean is not significantly different from 5.6 at a = 0.01, so
there is statistically insignificant evidence to conclude that the sample mean number of
accidents per year at intersections with cameras installed is different from 5.4 accidents.
%3D

Expert Solution

Step 1

Let $μ$ denotes the population mean number of accidents at controlled intersections per year. The average number of accidents at controlled intersections per year is 5.3. The claim of test is that there is average different number for intersection with cameras installed. The hypothesis is,

Null hypothesis:

$H_{0} : μ = 5.6$

Alternative hypothesis:

$H_{1} : μ \neq 5.6$

The sample size is 59, sample mean is 5.4, sample standard deviation is 1.21.

The test statistic is,

$\begin{array}{rcl} t & = & \frac{\bar{x} - μ}{(\frac{s}{\sqrt{n}})} \\ = & \frac{5.4 - 5.6}{(\frac{1.21}{\sqrt{59}})} \\ = & \frac{- 0.2}{0.1575} \\ = & - 1.270 \end{array}$

Thus, the test statistic is $t = - 1.270$ .

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