Question

The Highway Safety Department wants to study the driving habits of individuals. A sample of 32 cars traveling on the highway revealed an average speed of 60 miles per hour with a standard deviation of 11 miles per hour. What sample size should be chosen if we want a margin of error of less than 2 miles per hour for a 95% confidence interval?

Answer 1

Answer:

n=126

Step-by-step explanation:

Data given and previous concepts

$\bar X=60$ represent the sample mean

$s=11$ represent the sample deviation

n=32 represent the sample size

Confidence =0.95 or 95%

$\alpha =1-0.95=0.05$ represent the significance level

ME= 2 represent the margin of error required

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".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma)$

Solution to the problem

Since we don't have the population deviation. We know that the margin of error for a confidence interval is given by:

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

The next step would be find the value of $\t_{\alpha/2}$, $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$. The degrees of freedom are given by:

$df=n-1=32-1=31$

Using the normal standard table, excel or a calculator we see that:

$t_{\alpha/2}=2.04$

And the code in excel is this one:  "=T.INV(1-0.025,31)"

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{t_{\alpha/2} s}{Me}$

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

And we have everything to replace into the formula:

$n=(\frac{2.04(11)}{2})^2 =125.88$

And if we round up the answer we see that the value of n to ensure the margin of error required $\pm=2 miles$ is n=126.