The authors of the paper "Driven to Distraction" (Psychological Science [2001]: 462—466) describe an experiment to
evaluate the effect of using a cell phone on reaction time. Subjects were asked to perform a simulated driving taskwhile
talking on a cell phone. While performing this task, occasional red and green lights flashed on the computer screen. If a
green light flashed, subjects were to continue driving. but if a red light flashed. subjects were to brake as quickly as
possible and the reaction time [in msec) was recorded. The following summary statistics are based on a graph that
appeared in the paper:

is = 48 E = 530 x = 70
a. Construct and interpret a 95% confidence interval for u, the mean time to react to a red light while talking on a cell
phone. What assumption must be made in order to generalize this confidence interval to the population of all drivers?
b. Suppose that the researchers wanted to estimate the mean reaction time to within 5 msec with 95% confidence. Using
the sample standard deviation from the study described as a preliminary estimate of the standard deviation of reaction
times. compute the required sample size.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X=530[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=70 represent the sample standard deviation

n=40 represent the sample size

2) Part a

The assumptions in order to do a t confidence interval are:

1) Data selected from a random sample (Assumed)

2) Sample size large (n=48>30) (Satisfied)

3) Population deviation [tex]\sigma[/tex] unknown (Assumed)

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

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

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

[tex]df=n-1=48-1=47[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,47)".And we see that [tex]t_{\alpha/2}=2.012[/tex]

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

[tex]530-2.012\frac{70}{\sqrt{48}}=509.672[/tex]

[tex]530+2.012\frac{70}{\sqrt{48}}=550.329[/tex]

So on this case the 95% confidence interval would be given by (509.672;550.329)

3) Part b

On this case the margin of error is given by:

[tex]Me=t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]

And we want Me=5 and the same confidence level, and we have also the critical value [tex]t_{\alpha/2}[/tex] so we can solve for n like this:

[tex]n=(\frac{t_{\alpha/2} s}{Me})^2 =(\frac{2.012(70)}{5})^2 =793.436[/tex]

And rounded up to the nearest integer would be n=794