Question

In: Statistics and Probability

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

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 12 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 = 0.28 gram.

(a) 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    


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

σ is known

normal distribution of weights

uniform distribution of weights

n is large

σ is unknown


(c) Give a brief interpretation of your results in the context of this problem.

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.

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 0.20.

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.

The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.


(d) Find the sample size necessary for an 80% confidence level with a maximal error of estimate E = 0.06 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
___________ hummingbirds

Solutions

Expert Solution

b) Conditions necessary for calculations-

σ is known

normal distribution of weights

c) Interpretation : 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)   = 0.28 , E = 0.06 , C = 0.80

Sample size n = [( Z0.80 * ) / E]2

= [ (( -1.28) * 0.28 ) / 0.06 ]2

= 35.77

36

Hence the required sample size n = 36 hummingbirds

Hope this will help you. Thank you :)


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