Question

To illustrate the effects of driving under the influence​ (DUI) of​ alcohol, a police officer brought a DUI simulator to a local high school. Student reaction time in an emergency was measured with unimpaired vision and also while wearing a pair of special goggles to simulate the effects of alcohol on vision. For a random sample of nine​ teenagers, the time​ (in seconds) required to bring the vehicle to a stop from a speed of 60 miles per hour was recorded. Complete parts​ (a) and​ (b).

​Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.

TABLE IS BELOW.

​(a) Whether the student had unimpaired vision or wore goggles first was randomly selected. Why is this a good idea in designing the​ experiment?

A.

This is a good idea in designing the experiment because it controls for any​ "learning" that may occur in using the simulator.

B.

This is a good idea in designing the experiment because the sample size is not large enough.

C.

This is a good idea in designing the experiment because reaction times are different.

​(b) Use a​ 95% confidence interval to test if there is a difference in braking time with impaired vision and normal vision where the differences are computed as​ "impaired minus​ normal."

The lower bound is

?

The upper bound is

?

​(Round to the nearest thousandth as​ needed.)

State the appropriate conclusion. Choose the correct answer below.

There is sufficient evidence to conclude there is a difference in braking time with impaired vision and normal vision.

There is insufficient evidence to conclude there is a difference in braking time with impaired vision and normal vision.

Click to select your answer(s).

Data Table

Subject	1	2	3	4	5	6	7	8	9
​Normal, Upper X Subscript iXi	4.47	4.24	4.58	4.56	4.31	4.80	4.55	5.00	4.79
​Impaired, Upper Y Subscript iYi	5.86	5.85	5.45	5.32	5.83	5.49	5.23	5.61	5.63

PrintDone

Answer 1

a.
Whether the student had unimpaired vision or wore goggles first was randomly selected.
option :C
because
This is a good idea in designing the experiment because reaction times are different.
Given that,
mean(x)=5.5856
standard deviation , s.d1=0.2324
number(n1)=9
y(mean)=4.5889
standard deviation, s.d2 =0.2424
number(n2)=9
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.306
since our test is two-tailed
reject Ho, if to < -2.306 OR if to > 2.306
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =5.5856-4.5889/sqrt((0.05401/9)+(0.05876/9))
to =8.9042
| to | =8.9042
critical value
the value of |t α| with min (n1-1, n2-1) i.e 8 d.f is 2.306
we got |to| = 8.90418 & | t α | = 2.306
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 8.9042 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 8.9042
critical value: -2.306 , 2.306
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that there is a difference in braking time with impaired vision and normal vision.
b.
TRADITIONAL METHOD
given that,
mean(x)=5.5856
standard deviation , s.d1=0.2324
number(n1)=9
y(mean)=4.5889
standard deviation, s.d2 =0.2424
number(n2)=9
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((0.054/9)+(0.059/9))
= 0.112
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 8 d.f is 2.306
margin of error = 2.306 * 0.112
= 0.258
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (5.5856-4.5889) ± 0.258 ]
= [0.739 , 1.255]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=5.5856
standard deviation , s.d1=0.2324
sample size, n1=9
y(mean)=4.5889
standard deviation, s.d2 =0.2424
sample size,n2 =9
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 5.5856-4.5889) ± t a/2 * sqrt((0.054/9)+(0.059/9)]
= [ (0.997) ± t a/2 * 0.112]
= [0.739 , 1.255]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [0.739 , 1.255] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion