Question

The Highway Safety Department wants to study the driving habits of individuals. A sample of 40 cars traveling on a particular stretch of highway revealed an average speed of 69 miles per hour with a standard deviation of 7.8 miles per hour. Round to 4 decimal places.
What sample size is needed to estimate the true average speed to within 2 mph at 99% confidence?

Answer 1

Answer:

We need a sample size of at least 101

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.005 = 0.995$, so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

What sample size is needed to estimate the true average speed to within 2 mph at 99% confidence?

We need a sample of at least n.

n is found when M = 2. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$2 = 2.575*\frac{7.8}{\sqrt{n}}$

$2\sqrt{n} = 2.575*7.8$

$\sqrt{n} = \frac{2.575*7.8}{2}$

$(\sqrt{n})^{2} = (\frac{2.575*7.8}{2})^{2}$

$n = 100.85$

Rounding up

We need a sample size of at least 101

Answer 2

Answer:

$n=(\frac{2.58(7.8)}{2})^2 =101.24 \approx 102$

So the answer for this case would be n=102 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

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".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (a)

The best estimator for the population variance is the sample variance $\hat \sigma^2 = s^2$. And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.005;0;1)", and we got $z_{\alpha/2}=2.58$, replacing into formula (b) we got:

$n=(\frac{2.58(7.8)}{2})^2 =101.24 \approx 102$

So the answer for this case would be n=102 rounded up to the nearest integer