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

Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 15 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 = 0.40 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? (Round your answers to two decimal places.)

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

(c) Interpret your results in the context of this problem.

(d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.08 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

Explanation / Answer

standard error = standard deviation/ sqrt(n) = 0.4/(sqrt(15)) = 0.1033

t-score for 0.1 alpha (for 2 tailed test, 0.1 on either side) and 14 df (n-1 df) is 1.345

CI = mean +/- t-score * std error

= 3.15+/- 1.345*0.1033

(3.29,3.01)

b)

The distribution has to be normal.

The observations have to be independent.

c)

The 80% confidence interval is 3.29, 3.01

signifies, for 80% of the samples, the mean weight will be between 3.29 and 3.01

d)

margin of error = tscore* std. devn/ sqrt(n)

n = (tscore* std devn/ moe) ^2 = (1.345*0.4/0.08)^2

= 45.23

Hence approximately 46.

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