Question

Aller's hummingbird (Selasphorus sasin) has been studied by the suppose a small group of 16 le's hummingbirds has been under study in Aurora. The average weight for these birds is x 3.15 grams. Based on previous studies, we can assume that the weights of Alen's hummingbirds have a normal trbution, with (a) Fund an 80% confidence terval for the average weight of te hummingbirds in the region. What is the margin of error (Round your answers to the decimal places) lower limt 301 uppert margin of error (b) What conditions are necessary for your calculations et at that is known form distribution of weights We are 30% confident that were average weg of 'shuningdeals with The pretty that the worvalcontains the true rage will offers humingbirds is 0.00 The probably that this nervat contains the true average weight of Alessis is 0.20 • We went that the average weige's most interval (c) Pad the sample se necessary for an 80chance level wina mama margin of 10 For the news of human and up to the Xums

Answer 1

$\\Given,\\ a.)Level~of~significance(\alpha)=1-0.80=0.20\\\\ sample~Size(n)=16\\\\ Standard~Deviation(\sigma)=0.24\\\\ Sample~Mean(\bar{x})=3.15\\\\ Critical~Value(Z_{\alpha/2})=1.28\\\\ (use~z~table~)\\\\ Confidence~Interval:\bar{x} \pm \frac{Z_{\alpha/2}*\sigma}{\sqrt{n}}\\\\ =3.15 \pm 1.28*\frac{0.24}{\sqrt{16}}\\\\ =3.15 \pm 0.0768\\\\ =(3.0732, 3.2268)\\\\ Ans:Lower~limit:3.07\\\\ Upper~limit:3.23\\\\ Margin~of~Error:0.08$

$\\\\ b.)Ans:Normal~distribution~of~weight\\\\ \sigma~is~known\\\\ c.)we~are~80\% ~confident~that the true average weight of .............\\\\ d.)Margin~of~Error(E)=0.10\\\\ Critical~Value(Z_{\alpha/2})=1.28\\\\ Sample~Size(n)=[\frac{Z_{\alpha/2}*\sigma}{E}]^2\\\\ =[\frac{1.28*0.24}{0.10}]^2\\\\ =9.437\\\\ Ans:10$