Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.t Suppose a small group of 19 Allen's hummingbirds has been under study in Arizona. The average weight for thesebirds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with o =0.22 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.)lower limitupper limitmargin of error(b) What conditions are necessary for your calculations? (Select all that apply.)o is knowno is unknownuniform distribution of weightsn is largenormal distribution of weights(c) Interpret your results in the context of this problem.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.We are 80% confident that the true average weight of Allen's hummingbirds falls within this interval.We are 20% confident that the true average weight of Allen's hummingbirds falls within this interval.(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.)hummingbirds

Given that Number of humming birds (n) = 19 Average weight of birds (x¯) = 3.15 grams Standard…

Answered: Allen's hummingbird (Selasphorus sasin)…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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A new drug to treat high cholesterol is being tested by pharmaceutical company. The cholesterol levels for 8 patients were recorded before administering the drug and after. The mean difference in total cholesterol levels (after - before) was 15.91 mg/dL with a standard deviation of 15.2714 mg/dL. Create a 95% confidence interval for the true average difference in cholesterol levels by the drug. Question 9 options: 1) (-3.1428, 28.6772) 2) (3.1428, 28.6772) 3) (3.4593, 28.3607) 4) (10.5107, 21.3093) 5) (13.5454, 18.2746)
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 ? = 0.28 gram. a) When finding an 80% confidence interval, what is the critical value for confidence level? (Give your answer to two decimal places.)zc = b) 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.)lower limit =upper limit =margin of error = c) 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. (Round up to the nearest whole number.)
We have developed a new drug used to reduce fever and want to test its efficacy. Ten ER patients who are running a fever agree to take the new drug instead of the traditional treatment. Their temperatures are taken before the drug is administered and 30 minutes after taking the drug. The results are recorded in the Excel file below. Use the data to construct and interpret a 99% confidence interval for the mean change in temperatures of patients after taking this drug using Excel. Be sure to include any explanation necessary along with a sentence that explains what the interval means. First Temp Second Temp 100.1 98.9 101.3 99.1 102.1 99.2 102.7 99 101.9 98.7 100.8 98.6 103.1 99.4 102.5 99.2 103.5 100.1 101.7 99.2
In order to estimate the average GPA of all students at SCC, we selected a random sample of 36 students and found that their average GPA is 2.1 with a s. d. of 0.8.Construct a 95% confidence interval for the average GPA of all students at SCC.
The design of controls and instruments affects how easily people can use them. A student project investigated this effect by asking 25 right handed students to turn a knob with their right hands that moved an indicator by screw action. There were two identical instruments, one with a right-hand thread and the other with a left-hand thread. The table given has the data from the student project.1. Calculate and interpret a 90% confidence interval for the mean time advantage of right hand over left hand threads in the setting above.
A study of 22 goats reveals that the average length of a goat’s tail is 6.3 inches. If the standarddeviation is known to be 1.2 inches find a 95% confidence interval for the mean tail length.
A researcher wants to see if birds that build larger nests lay larger eggs. She selects two random samples of nests: one of small nests and the other of large nests. Then she weighs one egg (chosen at random if there is more than one egg) from each nest. A 95% confidence interval for the difference (Large - Small) between the mean mass (in grams) of eggs in small and large nests is 1.6 ± 2.0. Does the interval provide convincing evidence of a difference in the true mean egg mass of birds with small nests and birds with large nests? Explain your answer. No, the interval does not include a difference of 0 (no difference) as a plausible value. No, because the interval includes a difference of 0 (no difference) as a plausible value. Yes, because random samples were selected. Yes, because the interval includes a difference of 0 (no difference) as a plausible value.

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