Question

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to check out was 4.0 minutes. It is known that the standard deviation of the checkout time is one minute. The 98% confidence interval for the average checkout time of all customers is Group of answer choices 3.02 to 4.98 3.00 to 5.00 3.795 to 4.205 3.767 to 4.233

Answer 1

Answer:

$4-2.326\frac{1}{\sqrt{100}}=3.767$    

$4+2.326\frac{1}{\sqrt{100}}=4.233$    

So on this case the 98% confidence interval would be given by (3.767;4.233)

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".

$\bar X=4$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=1$ represent the population standard deviation

n=100 represent the sample size  

Solution to the problem

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 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.01,0,1)".And we see that $z_{\alpha/2}=2.326$

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

$4-2.326\frac{1}{\sqrt{100}}=3.767$    

$4+2.326\frac{1}{\sqrt{100}}=4.233$    

So on this case the 98% confidence interval would be given by (3.767;4.233)