Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 14 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 ? = 0.38 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) What conditions are necessary for your calculations? (Select all that apply.)

uniform distribution of weights? is unknownn is largenormal distribution of weights? is known

(c) Interpret your results in the context of this problem.

There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.     The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.

(d) Find the sample size necessary for an 80% confidence level with a maximal margin of error  E = 0.06 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
hummingbirds

Answer 1

A small group of 14 Allen’s hummingbirds has been under study in Arizona. The average weight for these birds is x-bar=3.15 grams. Based on previous studies, we can assume that the weights of Allen’s hummingbirds have a normal distribution, with sigma = 0.33 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?

In a confidence interval, the range of values above and below the sample statistic is called the margin of error.

The formula for calculating the margin of error is as follows:

To compute the margin of error, we need to find the critical value and the standard error of the mean. Since the population standard deviation is known, use the z-score.

1. Compute alpha

2. Find the critical probability

3. Find the degrees of freedom (df):

df = 14 - 1 = 13

4. Find the critical value. Since we know the population standard deviation, we'll express the critical value as a z statistic. For this problem, it will be the z statistic having a cumulative probability equal to 0.9. Using the standard normal table, we find that the critical value is 1.2816

And finally, we compute the margin of error (ME).

Lower Limit & upper limit of the average weights of Allen's hummingbirds will be given by

Lower Limit	3.02
Upper Limit	3.28
Margin of error	0.13

(b) What conditions are necessary for your calculations?

Normal distribution of weights. The population standard deviation is known.

c) Interpret your results in the context of this problem.

There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.

There is an 80% chance that the interval [3.02.3.28] contains the true average weight of Allen's hummingbirds in this region.

d) Find the sample size necessary for an 80% confidence level with a maximal margin of error  E = 0.06 for the mean weights of the hummingbirds.

65 hummingbirds sample weights are required to get 80% confidence level or the mean weights of the hummingbirds with a maximal margin of error  E = 0.06.