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

Answer:

$ME = 1.833 * \frac{4.367}{\sqrt{10}}= 2.531$

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

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

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)  

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (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:

$ME = 1.833 * \frac{4.367}{\sqrt{10}}= 2.531$