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. Click the icon to view the data table. (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 reaction times are different. C. This is a good idea in designing the experiment because the sample size is not large enough. (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 insufficient evidence to conclude there is a difference in braking time with impaired vision and normal vision. There is sufficient evidence to conclude there is a difference in braking time with impaired vision and normal vision.

Answer 1

(a) If the type of vision was not randomly selected, then would not be able to determine if the differences are due to the type of vision or due to experience.

(b)

Sample 1	Sample 2	Difference = Sample 1 - Sample 2
	4.47	5.77	-1.3
	4.24	5.85	-1.61
	4.58	5.51	-0.93
	4.65	5.32	-0.67
	4.31	5.9	-1.59
	4.83	5.49	-0.66
	4.59	5.23	-0.64
	5	5.63	-0.63
	4.79	5.63	-0.84
Average	4.607	5.592	-0.986
St. Dev.	0.246	0.229	0.408
n	9	9	9

From the sample data, it is found that the corresponding sample means are: X = 4.607 X2 = 5.592 Also, the provided sample standard deviations are: $i = 0.246 S2 = 0.229 and the sample size is n = 9. For the score differences we have D = -0.986 SD = 0.408

CONFIDENCE INTERVAL

Determine the critical value using the Students T distribution table in the appendix in the row with df = n - 1 = 9 - 1 = 8 and in the column with alpha = (1 - c )/2 =0.025:

tc​=2.306

Margin of error is

$E=tc * \frac{s_{d}}{\sqrt{n}}$

$E=2.306* \frac{0.408}{\sqrt{9}}$

=0.3136

The endpoints of the confidence interval

dbar - E= −0.986 - 0.3136 =-1.2996

dbar + E=  −0.986 + 0.3136= -0.6724

95% confidence interval is −1.299<μD​<−0.672.