Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 18 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 ? = 0.20 gram.

(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.)

lower limit
upper limit
margin of error

b. Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.07 for the mean weights of the hummingbirds. (Round up to the nearest whole number.) ______ hummingbirds.

Answer 1

Given data

number of bird sample size (n)=18

Sample mean

population Standard daviation

a) Since population standard daviation is given

confidence level =80%

hence

Now at from the Z distribution table the Z criticle value is

Now

wher ME= margional error

So the confidence interval will be

So the lower limit =

And the upper limit will be

and margin of error is

b)  

Now given that margin of error

confidence level =80%

and we have to find the sample size

Now at    from the Z distribution table the Z criticle value is

so the number of hummingbird shoul be (n)=13