Question

An automobile club reported that the average price of regular gasoline in a certain state was ​$2.61 per gallon. The following data show the price per gallon of regular gasoline for 15 randomly selected stations in the state. Complete parts a through c below. 2.69 2.51 2.54 2.72 2.55 2.57 2.59 2.73 2.67 2.72 2.66 2.59 2.66 2.52 2.55 a. Construct a 99​% confidence interval to estimate the average price per gallon of gasoline in the state. The 99​% confidence interval to estimate the average price per gallon of gasoline in the state is from ​$ nothing to ​$ nothing. ​(Round to the nearest​ cent.)

Answer 1

Answer:

$2.618-2.98\frac{0.078}{\sqrt{15}}=2.56$    

$2.618+2.98\frac{0.078}{\sqrt{15}}=2.68$    

So on this case the 99% confidence interval would be given by (2.56;2.68)

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 $\bar X=2.618$

The sample deviation calculated $s=0.078$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=15-1=14$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ 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(0.005,14)".And we see that $t_{\alpha/2}=2.98$

Now we have everything in order to replace into formula (1):

$2.618-2.98\frac{0.078}{\sqrt{15}}=2.56$    

$2.618+2.98\frac{0.078}{\sqrt{15}}=2.68$    

So on this case the 99% confidence interval would be given by (2.56;2.68)