Question

Allen�s hummingbird (Selasphorus sasin ) has been studied by zoologist Bill 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 ? = 0.33 gram. Find an 80% confidence interval for the average weight of Allen�s hummingbirds in the study region. What is the margin of error? .109 -1.17 < u < 1.39 (Is this correct?) What conditions are necessary for your calculations? Compare your results in the context of this problem. Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.08 for the mean weight of the hummingbirds. 3.88 (Is this correct?)

Answer 1

Answer:

1) 80% CI: [3.04; 3.26]gr

d= 0.11

2) n= 28 hummingbirds

Step-by-step explanation:

Hello!

The study variable of this experiment is:

X: the weight of a hummingbird. (gr)

And it has a normal distribution, symbolically: X~N(μ;σ²)

And (I hope I got it correctly) its population standard deviation is σ= 0.33

There was a sample of n= 15 hummingbirds taken, its sample mean X[bar]= 3.15 gr

1) You need to construct an 80% Confidence Interval for the population mean of the hummingbird's weight.

Since the study variable has a normal distribution, you can use either the standard normal distribution or the Student's t distribution. Both are useful to estimate the population mean. Since the population standard variance is known, the best choice is the Standard normal.

Z= X[bar] - μ ~ N(0;1)

       σ/√n

The formula for the interval is:

X[bar] ± $Z_{1- \alpha /2}$ * (σ/√n)

$Z_{1- \alpha /2}= Z_{0.90} = 1.28$

3.15 ± 1.28 * (0.33/√15)

[3.04; 3.26]gr

The margin of error (d) of a confidence interval is hal its amplitude (a)

a= Upper bond - Lower bond

d= (Upper bond - Lower bond)/2

d= $\frac{(3.26-3.04)}{2}$ = 0.11

2) You need to calculate a sample size for a 80% Confidence interval for the average weight of the hummingbirds with a margin of error of d= 0.08

As I said before, the margin of error is half the amplitude of the interval, the formula you use to estimate the population mean has the following structure:

"point estamator" ± "margin of error"

Then the margin of error is:

d= $Z_{1- \alpha /2}$ * (σ/√n)

Now what you have to do is rewrite the formula based on the sample size

d= $Z_{1- \alpha /2}$ * (σ/√n)

$\frac{d}{Z_{1- \alpha /2}}$= σ/√n

√n * $\frac{d}{Z_{1- \alpha /2}}$= σ

√n = σ * $\frac{Z_1- \alpha /2}{d}$

n = (σ * $\frac{Z_1- \alpha /2}{d}$)²

n=  (0.33 * $\frac{1.28}{0.08}$)²

n= 27.8784 ≅ 28 hummingbirds.

I hope it helps!