The authors of the paper “Driven to Distraction” (Psychological Science [2001]:

Question

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 task while 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:
n 48 x 5 530 s 70
a. Construct and interpret a 95% confidence interval
for m, 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.

Explanation / Answer

Here,

n = 48

x = 5530

s = 70

a)

We assume that the population of reaction times are approximately normally distributed.

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    5530          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    70          
n = sample size =    48          
              
Thus,              
Margin of Error E =    19.80275035          
Lower bound =    5510.19725          
Upper bound =    5549.80275          
              
Thus, the confidence interval is              
              
(   5510.19725   ,   5549.80275   ) [ANSWER]

******************************

b)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.025  
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
s = sample standard deviation =    70  
E = margin of error =    5  
      
Thus,      
      
n =    752.9259289  
      
Rounding up,      
      
n =    753   [ANSWER]

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