Question

A sample of 16 eggs yields a sample mean weight of 3.00 ounces and a sample standard deviation = 0.360 ounces, find the margin of error in estimating a confidence interval estimate at the 90% level of confidence. Round your answer to three decimal places. You may assume that egg weights are normally distribute.

Answer 1

Answer:

$ME =1.753\frac{0.360}{\sqrt{16}}= 0.158$

$3-1.753\frac{0.360}{\sqrt{16}}=2.842$    

$3+1.753\frac{0.360}{\sqrt{16}}=3.158$    

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".

$\bar X=3$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=0.360 represent the sample standard deviation

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

$df=n-1=16-1=15$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,15)".And we see that $t_{\alpha/2}=1.753$

The margin of error is given by:

$ME =1.753\frac{0.360}{\sqrt{16}}= 0.158$

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

$3-1.753\frac{0.360}{\sqrt{16}}=2.842$    

$3+1.753\frac{0.360}{\sqrt{16}}=3.158$    

So on this case the 90% confidence interval would be given by (2.842;3.158)