Question

0e9e, E increases, resulting in a longer confidence interval, Need Help? Pos 4/6 points 1 Previous Answers 2 7.1015 Notes O Ask Your T Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther. Suppose a small group of hummingbirds has been under study in Arizona. The average weight for these birds is x 3.15 grams. Based on previous studies with a 0.36 gram. ingbirds in the study region what is the we can assume that the weights of Allen's hummingbirds have a normal distribution, (a) Find an 80% confidence interval for the average weights of Allen's hurn margin of etror? (Round your answers to two decimal places.) ower limit 3.03 3 26 , 109 upper limt 320 margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) σ is unknown @ ơis known normal distribution of weights O uniform distribution of weights O n is large (c) Interpret your results in the context of this problem. O There is a 20% chance that the interval is one of the inter als containing the tn e average weight of Allen's hummingbirds in this region. O The probability that this interval contains the true average weight of Allen's hummingbirds is o.20. O The probability to the true average weight of Allen's hummingbirds is equal to the sample mean. O 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. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80. (d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E-0.11 for the mean weights of the hummingbirds. (Round up to the nearest whole number.) 28 X hummingbirds My Notes Ask Your Teacher plasma volume is important in determining the required plasma component in blood replacement therapy for a person

Answer 1

a)

margin of error = z * sd/sqrt(n)

= 1.282 * 0.36/sqrt(16)

= 0.11538

=0.12

d)

n= (z * sd/e)^2

= (1.282 *0.36/0.11)^2

= 17.603

hence

n = 18

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