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

`n=((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

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)(\sigma)/(√(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=((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=((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

Answer 2

We need a sample size of at least 101

Explanation:

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

`\alpha = (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*(\sigma)/(√(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*(\sigma)/(√(n))`

`2 = 2.575*(7.8)/(√(n))`

`2√(n) = 2.575*7.8`

`√(n) = (2.575*7.8)/(2)`

`(√(n))^(2) = ((2.575*7.8)/(2))^(2)`

`n = 100.85`

Rounding up

We need a sample size of at least 101