Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.t Suppose a small group of 18 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is ; = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with a = 0.37 gram. (a) Find an 30% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (b) What conditions are necessary for your calculations? (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 = 0.15 for the mean weights of the hummingbirds. Step 1 (a) Find an 80% condence interval for the average weights of Allen's hummingbirds in the study region. what is the margin of error? To find a confidence interval for the average weight of Allen's hummingbirds in the study region, you will need a point estimate for u, and you will either need to know the population standard deviation or estimate it. The point estimate for u is}? = ; = - ' 3.15 . The distribution of the weights is assumed to be normal with a known population standard deviation of a = - E: 0.37 . Step 2 To compute the margin of error for the 80% confidence interval for [4, you will need a critical value from the standard normal distribution. This number, 26, is the number such that an area equal to c under the standard normal curve falls between 72c and 2:. Use the table below to nd the critical value for a confidence level of 80%. Confidence Interval Critical Values zc Level of Condence c Critical Value zc 0.70, or 70% 1.04 0.75, or 75% 1.15 0.80, or 80% 1.28 0.85, or 85% 1.44 0.90, or 90% 1.645 0.95, or 95% 1.96 0.98, or 98% 2.33 0.99, or 99% 2.58 20.80 = v ' 115 Furthermore, you will need to estimate E, the maximal margin of error which can be approximated as a E=z Approximate the maximal margin of error, E, by using zc from above, the given population standard deviation, and the sample size, n Recall that n = 18 and a = 0.37. (Round your answer to two decimal places.) 0' E = 20.30%; 0.37 Ev '118 ~ = - 0.11 Find an 80% confidence interval for 11. (Round your answers to two decimal places.) Recall that}? = l = 3.15 and E = 0,11, > lower limit [4 E = upper limit [2+ E = Submit Ea (you cannot come back)