Question

Bob Nale is the owner of Nale’s Quick Fill. Bob would like to estimate the mean number of gallons of gasoline sold to his customers. Assume the number of gallons sold follows the normal distribution with a population standard deviation of 1.90 gallons. From his records, he selects a random sample of 45 sales and finds the mean number of gallons sold is 5.70.
(a) The point estimate of the population mean is

Answer 1

Answer:

a) The best estimator for the population mean is given by the sample mean $\hat \mu = \bar X = 5.70$

b) The 99% confidence interval would be given by (4.969;6.431)    

c) We are 99% confident that the true mean for th number of gallons of gasoline  sold to his customers is between 4.969 and 6.431.

Step-by-step 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".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=1.9$ represent the population standard deviation

n=45 represent the sample size  

2) Part a

The best estimator for the population mean is given by the sample mean $\hat \mu = \bar X = 5.70$

3) Part b. Develop a 99 percent confidence interval for the population mean.

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

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

Since the Confidence is 0.95 or 95%, 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: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

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

$5.7-2.58\frac{1.9}{\sqrt{45}}=4.969$    

$5.7+2.58\frac{1.9}{\sqrt{45}}=6.431$

So on this case the 99% confidence interval would be given by (4.969;6.431)    

4) Part c. Interpret the meaning of part (b).

On this case we can say this: We are 99% confident that the true mean for th number of gallons of gasoline  sold to his customers is between 4.969 and 6.431.