Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.â  Suppose a small group of 11 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with Ï = 0.38 gram.
Find an 80% confidence interval for the average weight of Allen's hummingbirds in the study region. What is the margin of error?

Answer 1

Answer:

$3.15-1.28\frac{0.38}{\sqrt{11}}=3.003$    

$3.15+ 1.28\frac{0.38}{\sqrt{11}}=3.297$    

So on this case the 80% confidence interval would be given by (3.003;3.297)

And the margin of error is given by:

$ME = 1.28\frac{0.38}{\sqrt{11}}= 0.147$    

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".

$\bar X=3.15$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=0.38$ represent the population standard deviation

n=11 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 0.80 or 80%, the value of $\alpha=0.2$ and $\alpha/2 =0.1$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NOR.INV(0.1,0,1)".And we see that $z_{\alpha/2}=1.28$

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

$3.15-1.28\frac{0.38}{\sqrt{11}}=3.003$    

$3.15+ 1.28\frac{0.38}{\sqrt{11}}=3.297$    

So on this case the 80% confidence interval would be given by (3.003;3.297)

And the margin of error is given by:

$ME = 1.28\frac{0.38}{\sqrt{11}}= 0.147$