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

n=126

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)(s)/(√(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:

`√(n)=(t_(\alpha/2) s)/(Me)`

`n=((t_(\alpha/2) s)/(Me))^2`

And we have everything to replace into the formula:

`n=((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.