Question

Allen’s hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther (Reference: Hummingbirds by K. Long and W. Alther). A small group of 15 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 σ= 0.33 gram.

1.Is it appropriate to use a normal distribution to compute a confidence interval for the population mean weights?

2.Find an 90% confidence interval for the mean weights of Allen’s hummingbirds in the study region. What is the margin of error?

3.Interpret the confidence interval in the context of this problem.

4.Find the sample size necessary for an 90% confidence level with a maximal margin of error 0.08 for the mean weights of the hummingbirds.

Answer 1

Solution:

1) Since, the weights of Allen's humming birds have a normal distribution with know population standard deviation of 0.33 grams, therefore it is appropriate to use a normal distribution to compute a confidence interval for the population mean weights.

2) The 90% confidence interval for population mean is given as follows:

Where, x̄ is sample mean, σ is population standard deviation, n is sample size and Z_(0.10/2) is critical z-value to construct 90% confidence interval.

We have, x̄ = 3.15 grams, σ = 0.33 gram and n = 15

Using Z-table we get, Z_(0.10/2) = 1.645

Hence, the 90% confidence interval for the mean weights of Allen’s hummingbirds is,

Hence, the 90% confidence interval for the mean weights of Allen’s hummingbirds is, (3.0098 grams, 3.2902 grams).

3) Interpretation : We are 90% confident that true population mean weight of Allen's hummingbirds in the region Arizona lies within the calculated 90% confidence interval.

4) The sample size needed to obtain interval estimate of population mean at 90% confidence level is given as follows:

Where, E is margin of error, σ is population standard deviation, n is sample size and Z_(0.10/2) is critical z-value to construct 90% confidence interval.

We have, σ = 0.33 gram, E = 0.08 and n = 15

Using Z-table we get, Z_(0.10/2) = 1.645

The needed sample size is 46.

Please rate the answer. Thank you.