Question

The authors of a paper 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. The reaction time (in msec) was recorded. The following summary statistics are based on a graph that appeared in the paper.
n = 47
x = 525
s = 75

(a) Construct a 95% confidence interval for μ, the mean time to react to a red light while talking on a cell phone. (Round your answers to three decimal places.)
(b) Suppose that the researchers wanted to estimate the mean reaction time to within 7 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. (Round your answer up to the nearest whole number.)

Answer 1

Answer:

a) The 99% confidence interval would be given by (24.409;24.979)  

b) n=464

Step-by-step explanation:

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".  

Part a

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

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

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

$df=n-1=47-1=46$  

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(0.025,46)".And we see that $t_{\alpha/2}=2.01$  

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

$525-2.01\frac{75}{\sqrt{47}}=503.01$  

$525+2.01\frac{75}{\sqrt{47}}=546.99$  

So on this case the 95% confidence interval would be given by (503.01;546.99)

Part b

The margin of error is given by this formula:  

$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)  

And on this case we have that ME =7 msec, we are interested in order to find the value of n, if we solve n from equation (1) we got:  

$n=(\frac{t_{\alpha/2} s}{ME})^2$ (2)  

The critical value for 95% of confidence interval is provided, $t_{\alpha/2}=2.01$ from part a, replacing into formula (2) we got:  

$n=(\frac{2.01(75)}{7})^2 =463.79 \approx 464$  

So the answer for this case would be n=464 rounded up to the nearest integer