x-3.15 grams. Based on previous studes, we can assume that theweights of Allen\'

Question

x-3.15 grams. Based on previous studes, we can assume that theweights of Allen's hummingbirds have a normal distribution, with -G 34 gram. (a) Find an 80% confidence interval for the average weights of Allen's hummingbird. the st udy region. what is the margin of error? (Round your answers to two decimal places.) lower limit upper limit margin of error (c) Interpret your results in the context of this problem O The probability to the true average weight of Allen's hummingbirds is equal to the sample mean. O 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 o.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. 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 d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E 0.07 for the mean weights of the hummingbirds Round up to the nearest

Explanation / Answer

Part a

Confidence interval = Xbar -/+ Z*/sqrt(n)

We are given

Xbar = 3.15

Confidence level = 80%

Z = 1.2816 (by using z-table)

= 0.34

n = 19

Margin of error = Z*/sqrt(n)

Margin of error = 1.2816*0.34/sqrt(19)

Margin of error = 0.09996653

Confidence interval = 3.15 -/+ 1.2816*0.34/sqrt(19)

Confidence interval = 3.15 -/+ 0.09996653

Lower limit = 3.15 - 0.09996653 = 3.05003347

Upper limit = 3.15 + 0.09996653 = 3.249967

Lower limit = 3.05

Upper limit = 3.25

Margin of error = 0.10

Part b

Answer:

is known

Normal distribution of weights

(As per assumptions of z confidence interval for population mean)

Part c

The probability that this interval contains the true average weight of Allen’s hummingbirds is 0.80.

(As per interpretation of confidence interval: We are 80% confident that the population mean will lies between given confidence interval.)

Part d

We are given

Confidence level = 80%

E = 0.07

Z = 1.2816

= 0.34

Sample size = n = (Z*/E)^2 = (1.2816*0.34/0.07)^2 = 38.74956

Required sample size = 39

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