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.
Subject 1 2 3 4 5 6 7 8 9
Normal, Xi 4.47 4.24 4.58 4.65 4.31 4.80 4.55 5.00 4.79
Impaired, Yi 5.77 5.67 5.51 5.32 5.83 5.49 5.23 5.61 5.6
(a) Whether the student had unimpaired vision or wore goggles first was randomly selected. Why is this a good idea in designing the experiment?
(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. "Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.

Answer 1

Answer:

a) If we design the experiment on this way we can check if we have an improvement with the method used.

We assume that we have the same individual and we take a value before with the normal impaired condition and the final condition is the normal case.  

b) $-0.96-2.306\frac{0.359}{\sqrt{9}}=-1.24$  

$-0.96+2.306\frac{0.359}{\sqrt{9}}=-0.69$

The 95% confidence interval would be given by (-1.24;-0.69)

Step-by-step explanation:

Part a

If we design the experiment on this way we can check if we have an improvement with the method used.

We assume that we have the same individual and we take a value before with the normal impaired condition and the final condition is the normal case.

Part b

For this case first we need to find the differences like this :

Normal, Xi 4.47 4.24 4.58 4.65 4.31 4.80 4.55 5.00 4.79

Impaired, Yi 5.77 5.67 5.51 5.32 5.83 5.49 5.23 5.61 5.6

Let $d_i = Normal -Impaired$

$d_i : -1.3, -1.43, -0.93, -0.67,-1.52, -0.69, -0.68, -0.61, -0.81$

The second step is calculate the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}=-0.96$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =0.359$

The confidence interval for the mean is given by the following formula:  

$\bar d \pm t_{\alpha/2}\frac{s_d}{\sqrt{n}}$ (1)  

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:  

$df=n-1=9-1=8$  

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that $t_{\alpha/2}=2.306$  

Now we have everything in order to replace into formula (1):  

$-0.96-2.306\frac{0.359}{\sqrt{9}}=-1.24$  

$-0.96+2.306\frac{0.359}{\sqrt{9}}=-0.69$  

So on this case the 95% confidence interval would be given by (-1.24;-0.69)