Question

A random sample of ten 2011 sports cars is taken and their city mileage is recorded. The results are as follows: 20 21 25 21 21 23 31 32 28 26 Assuming the population distribution is normal, compute E, the margin of error for the t interval, for a 90% confidence interval for m, the population mean of the city mpg for 2011 sports cars.

Answer 1

`ME = 1.833 * (4.367)/(√(10))= 2.531`

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 confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X))/(n-1)}` (3)

The mean calculated for this case is
`\bar X=24.8`

The sample deviation calculated
`s=4.367`

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=10-1=9`

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.005`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,9)".And we see that
`t_(\alpha/2)=1.833`

And the margin of error would be given by:

`ME = 1.833 * (4.367)/(√(10))= 2.531`