Question

A sample of n = 25 n=25 diners at a local restaurant had a mean lunch bill of $16 with a standard deviation of σ = $ 4 σ=$4 . We obtain a 95% confidence interval as ( 14.43 , 17.57 ) (14.43,17.57) . Which choice correctly interprets this interval?

Answer 1

Answer:

( 14.43 , 17.57 ) correctly interprets this interval.

Step-by-step explanation:

mean = x`= $ 16

Standard deviation= S= $4

sample size= n= 25

Using the formula

x`± zₐ/₂  S/ √n

= 16 ± (1.96) 4/ √25

= 16 ± (1.96) 4/5

= 16 ± (1.96)0.8

= 16 ± (1.568)

= 16+ 1.568         ,       16-1.568

= 17.568             , 14.432

From 14.43 to 17.57

( 14.43 , 17.57 ) correctly interprets this interval.

If two samples are taken  and two standard deviations are considered then the answer would be completely different from 13.78 to 18.22

Answer 2

Answer:

$16-1.96\frac{4}{\sqrt{25}}=14.43$    

$16-1.96\frac{4}{\sqrt{25}}=17.57$    

So on this case the 95% confidence interval would be given by (14.43;17.57)

And the best interpretation would be:

We are 95% confident that the true mean for the luch bill in the local restaurant is between 14.43$ and 17.57 $    

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".

$\bar X=16$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=4$ represent the population standard deviation

n=25 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 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

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

$16-1.96\frac{4}{\sqrt{25}}=14.43$    

$16-1.96\frac{4}{\sqrt{25}}=17.57$    

So on this case the 95% confidence interval would be given by (14.43;17.57)

And the best interpretation would be:

We are 95% confident that the true mean for the luch bill in the local restaurant is between 14.43$ and 17.57 $