Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 17 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 = 0.26 gram.
Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. (Round your answers to two decimal places.)
Lower limit:
upper limit:
margin of error:
80% = mean ± 1.28 SD
To find the confidence interval for the average weights of Allen's hummingbirds, we will use the formula:
CI = x ± z * (σ / √n)
Where:
- CI represents the confidence interval
- x is the sample mean (average weight in this case)
- z is the z-score corresponding to the desired confidence level
- σ is the population standard deviation
- n is the sample size
Given:
- x = 3.15 grams
- σ = 0.26 gram
- n = 17
- Desired confidence level = 80%
First, let's find the z-score for an 80% confidence level. We'll use the standard normal distribution table or a calculator to determine this value.
The z-score for an 80% confidence level is approximately 1.28.
Now we can substitute the values into the formula:
CI = 3.15 ± 1.28 * (0.26 / √17)
Calculating this value:
CI = 3.15 ± 1.28 * (0.26 / √17) = 3.15 ± 0.14
Therefore, the 80% confidence interval for the average weights of the Allen's hummingbirds in the study region is:
Lower limit: 3.15 - 0.14 = 3.01 grams
Upper limit: 3.15 + 0.14 = 3.29 grams
Margin of error: 0.14 grams
To find the 80% confidence interval for the average weights of Allen's hummingbirds, we can use the formula:
Confidence interval = x ± (Z * σ/√n)
Where:
- x is the sample mean (average weight)
- Z is the z-score corresponding to the desired confidence level (80% in this case)
- σ is the population standard deviation
- n is the sample size
First, let's calculate the z-score for an 80% confidence level. The z-score can be found using a standard normal distribution table or a calculator. In this case, the z-score for an 80% confidence level is approximately 1.28.
Next, we can substitute the given values into the formula:
Confidence interval = 3.15 ± (1.28 * 0.26/√17)
Calculating the standard error:
Standard error (SE) = σ/√n = 0.26/√17 ≈ 0.063
Now, substituting the values into the formula:
Confidence interval = 3.15 ± (1.28 * 0.063)
To find the lower limit of the interval:
Lower limit = 3.15 - (1.28 * 0.063) ≈ 3.067
To find the upper limit of the interval:
Upper limit = 3.15 + (1.28 * 0.063) ≈ 3.233
Finally, the margin of error is determined by taking half the width of the interval:
Margin of error = (Upper limit - Lower limit) / 2
Substituting the values:
Margin of error ≈ (3.233 - 3.067) / 2 ≈ 0.083
Therefore, the 80% confidence interval for the average weights of Allen's hummingbirds in the study region is approximately (3.07, 3.23) grams. The margin of error is approximately 0.083 grams.