Afeed supply company has developed a special feed supplement to see if it will promote weight gainin livestock. Company researchers report that the 121 cows studied gained an average of 47 poundsand that a 95% confidence interval for the mean weight gain this supplement produces has a marginof error of ±8 pounds. Which of the following statements gives a valid interpretation of the 95%confidence interval?It is estimated that the mean weight gain of cows fed this supplement is between 39 and 55 pounds, and thisestimation process is correct 95% of the time.If this supplement is tested on another sample of 121 cows there is a 95% chance that their average weightgain will be between 39 and 55 pounds.95% of the 121 cows studied gained between 39 and 55 pounds.We are 95% sure that the average weight gain among the 121 cows in this study was between 39 and 55pounds.

Answered: Image /qna-images/answer/d17c2bec-b5ee-4c07-a8ae-0224fa230d30.jpg

Answered: A feed supply company has developed a…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Q. Hạ: µN – µs > 200 R. H3: µN – µs -200 V. Hạ: µN – µs smoker = 0 (sample of nonsmokers) Variable | Obs Mean Std. Dev. Min Max birthweightI 49 3279 57,8 3146 3405 -> smoker = 1 (sample of smokers) VariableI Obs Mean Std. Dev. Min Мак birthweight | 69.4 2866 3225 36 3060 invttail(34,0.005)=2.73 invttail (35,0.005)=2.72 invttail(36,0.005)=2.71 invttail(34,0.025)=2.03 invttail(35,0.025)=2.02 invttail (36,0.025)=2.01 invttail(34,0.05)=1.69 invttail (35,0.05)=1.68 invttail (36,0.05)=1.67 invttail(47,0.005)=2.69 invttail(48,0.005)=2.68 invttail(49,0.005)=2.67 invttail (47,0.025)=1.99 invttail(48,0.025)=1.98 invttail (49,0.025)=1.97 invttail (47,0.05)=1.65 invttail (48,0.05)=1.64 invttail(49,0.05)=1.63 invttail(83,0.005)=2.64 invttail (84,0.005)=2.63 invttail (85,0.005)=2.62 invttail(83,0.025)=1.95 invttail(84,0.025)=1.94 invttail(85,0.025)=1.93 invttail(83,0.05)=1.62 invttail(84,0.05)=1.61 invttail (85,0.05)=1.60
Allen's hummingbird has been studied by zoologist Bill Alther. Suppose a small group of 16 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 o=0.20 gram. find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round answer to two decimal places) lower limit=upper limit=
Allen's hummingbird has been studied by zoologist Bill Alther. Suppose a small group of 15 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x=3.15 grams. Based in previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with o=0.22 gram. 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=
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)
a. interpret the corresponding 95% confidence intervals for both Birth weight and Age. b. What is the predicted average SBP of an infant born with birth weight 8lbs (128oz) measured at 3 days old? c. Calculate the Pearson correlation between Age and Birth weight, and test if this correlation is significant at the 0.05 level. Does this answer surprise you, why or why not?
Assume that last year in a particular state there were 211 children out of 1990 who were diagnosed with Autism Spectrum Disorder. Nationally, 1 out of 88 children are diagnosed with ASD. It is believed that the incident of ASD is more common in that state than nationally. Calculate a 96% confidence interval for the percentage of children in that state diagnosed with ASD.P: Parameter What is the correct parameter symbol for this problem? What is the wording of the parameter in the context of this problem? A: Assumptions Since information was collected from each object, what conditions do we need to check? Check all that apply. n≥30n≥30 or normal population. N≥20nN≥20n n(pˆ)≥10n(p̂)≥10 σσ is known. σσ is unknown. n(1−pˆ)≥10n(1-p̂)≥10 Check those assumptions: 1. nˆpnp^= which is 2. n(1−ˆp)n(1-p^)= which is 3. NN= which is If no N is given in the problem, use 1000000N: Name the procedure The…
The birth weights of a sample of newborns were recorded for a group born to mothers who did not get regular prenatal care (Group 1) and for a group born to mothers who did (Group 2). A 93% confidence interval for 41 – 2 was computed as follows: (-2.12 pounds, -0.89 pounds) Interpret the endpoints of this interval in the context of this problem. Do not use the word "difference" in your answer.
Assume that the 95% confidence interval for mean=9 and standard devaition=20. For a particular data, the standardized value z = 2.73. We can say that: Select one: a. The data with standardized value 2.73 does not belong to the confidence interval and the value is less than the average. b. The data with standardized value 2.73 does not belong to the confidence interval and the value is greater than the average. c. Cannot interpret the z-value information provided. d. The data with standardized value 2.73 if it belongs to the confidence interval and the value is less than the average. e. The data with standardized value 2.73 if it belongs to the confidence interval and the value is greater than the average.
you conducted additional studies with these men and found the following: a. The RR for those who drank one drink per day was 1.50 (95% CI:1.32-1.67) in comparison to the abstainers. b. For those men who drank three or more drinks per day the RR compared to the abstainers was 4.20 (95% CI 2.68-5.85). Question: We can tell from the Confidence Intervals provided in #7 whether there are likely more people in the “one drink a day” group or the “3+ drinks per day” group. Which one likely has more people and how can we tell?
An insurance company collects data on seat-belt use among drivers in a country. Of 1500 drivers 30-39 years old, 33% said that they buckle up, whereas 458 of 1800 drivers 55-64 years old said that they did. Find a 95% confidence interval for the difference between the proportions of seat-belt users for drivers in the age groups 30-39 years and 55-64 years. ..... Construct a 95% confidence interval. The 95% confidence interval for p, - p2 is from to (Round to three decimal places as needed.)

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