Q:

# 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 = 525s = 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.)

Accepted Solution

A:
Answer:a) The 99% confidence interval would be given by (24.409;24.979)  b) n=464Step-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 aThe 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 bThe 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