Allen\'s hummingbird (Selasphorus sasin) has been studied by zoologist. Bill. Su

Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist. Bill. 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 sigma = 0.40 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? lower limit 3.0485 upper limit 275 margin of error 0 (b) What conditions are necessary for your calculations? (Select all that apply.) sigma is unknown uniform distribution of weights sigma is known n is large normal distribution of weights (c) Interpret 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 weigh of Allen's hummingbirds in this region 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 interval's 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. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.

Explanation / Answer

a. Here mean=3.15 and sd=0.40

Here it is given that distribution is normal so we will use z table

Now P(-1.282<z<1.282)=0.80, so z=1.282

Hence E=z*sd/sqrt(n)=1.282*0.40/sqrt(11)=0.15

So CI=mean+/-E=3.15+/-0.15=(3,3.3)

d. E=0.07

As per formula of E=z*sd/sqrt(n)

So n=(z*sd/E)^2=53.67=54

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