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 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 = 0.34 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

Find the sample size necessary for an 80% confidence level with a maximal error of estimate E = 0.14 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

Explanation / Answer

Part-a

Mean xbar=3.15 standard deviation sigma = 0.34 and sample size n=11

As normal distriutio is assumed and standard deviation is known we caluclate confidnce inteverval bas on z.

For 80% CI, critical Z=1.28

So, margin of Error E=1.28* /sqrt(n)=1.28*0.34/sqrt(11) =0.13

Hence, 80% confidence interval =xbar-E, xbar+E

=(3.15- 0.13 , 3.15+0.13)

=(3.02 , 3.28)

Sample size necessary for 80% CI and E=0.14 is n>= (z*/E)2= (1.28*0.34/0.14)2=9.666

Hence, a minimum sample of size 10 is reQUIRED

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