Question

# 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 = 0.36 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.) lower limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) σ is known uniform distribution of weights σ is unknown n is large normal distribution of weights (c) Interpret your results in the context of this problem. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80. 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 that this interval contains the true average weight of Allen's hummingbirds is 0.20. The probability to the true average weight of Allen's hummingbirds is equal to the sample mean. 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 = 0.10 for the mean weights of the hummingbirds. (Round up to the nearest whole number.) hummingbirds

Ans(a)- Note:

 CI z -value 80%     z= 1.2816 90%     z =1.645 92%     z = 1.751 95%     z = 1.96 98%     z = 2.326 99%     z = 2.576

Ans(B)- According to the condition there is need of is known and normal distribution. They go together, and when I have them, n can be anything, so it is not required to be large.

Ans(C)-

If the weights of 100 hummingbirds was taken, 80 of them would cross µ. This means that we are 80% confidence that the µ lies between 3.017 gm and 3.283 .

20% of the cases will not cross µ.

Ans(D)-

width of one sided = 1.28*0.36/sqrt(n) = 0.1
0.2123 = .01 n,
squaring in both sided
21.23 is the sample size required, or

n=22

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