Allens hummingbird Selasphorus sasin has been studied by zoo

Allen\'s hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 11 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 = 0.26 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.)

(b) What conditions are necessary for your calculations? (Select all that apply.)

1.uniform distribution of weights

2. is known

3. is unknown

4.n is large

5. normal distribution of weights

(c) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.13 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

Solution

Answer to the question)

Part a)

Since the population standard deivation is given , we make use of the z confidence intervals

x bar -z * s/ sqrt(n) , x bar + z* s / sqrt(n)

.
we got n = 11

xbar = 3.15

s = 0.26

z for 80% confidence level is: 1.28

.

On pluggign the values we get

3.15 - 1.28 * 0.26/sqrt(11) , 3.15 + 1.28 * 0.26 / sqrt(11)

3.04966 , 3.2503

.

Part b)

the condition that helped us is sigma is known

.

Part c)

z = 1.28

e = 0.13

s = 0.26

.

n = (z*s/e)^2

n = (1.28 * 0.26 / 0.13)^2

n = 6.5536 ~ 7