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?

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

And the margin of error is given by:

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".

represent the sample mean

population mean (variable of interest)

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:

  (1)

Since the Confidence is 0.80 or 80%, the value of and , 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

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

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

And the margin of error is given by:

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