Question

An automobile engineer would like to test his newly invented anti-lock braking system. Specifically, he would like to calculate the average number of feet it takes the brakes to stop a car going 50 miles per hour. The engineer took a sample of 6 cars and from this sample calculated a mean stopping distance of 15 feet with a standard deviation of 2.28 feet. Assuming stopping distances are normally distributed, what is the 80% confidence interval for the population mean?

Answer 1

The 80% confidence interval is given by (13.622;16.378)

Explanation:

1) 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".

Assuming the X the random variable that represent the number of feet it takes the brakes to stop a car going 50 miles per hour follows a normal distribution

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

`\bar X =15` represent the sample mean

`s=2.28` represent the sample deviation

n=6 sample size selected

Confidence =0.8 or 80%

2) Confidence interval

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 80% of confidence, our significance level would be given by
`\alpha=1-0.80=0.2` and
`\alpha/2 =0.1`. The degrees of freedom are given by:

`df=n-1=6-1=5`

We can find the critical values in excel using the following formulas:

"=T.INV(0.1,5)" for
`t_(\alpha/2)=-1.48`

"=T.INV(1-0.1,5)" for
`t_(1-\alpha/2)=1.48`

The critical value
`tc=\pm 1.48`

Now we can calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

`m=t_c (s)/(√(n))`

`m=1.48 (2.28)/(√(6))=1.378`

The interval for the mean is given by this formula:

`\bar X \pm t_(c) (s)/(√(n))`

And calculating the limits we got:

`15 -1.48 (2.28)/(√(6))=13.622`

`15 +1.48 (2.28)/(√(6))=16.378`

The 80% confidence interval is given by (13.622;16.378)