A population has standard deviation o = 17.4.Part 1 of 2(a) How large a sample must be drawn so that a 99% confidence interval for u will have a margin of error equal to 4.1?Round the critical value to no less than three decimal places.Round the sample size up to the nearest integer.A sample size ofis needed to be drawn in order to obtain a 99% confidence intervaldowith a margin of error equal to 4.1.Part 2 of 2(b) If the required confidence level were 99.9%, would the necessary sample size be larger or smaller?(Choose one) ▼because the confidence level is |(Choose one) ▼higherlower

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Answered: (a) How large a sample must be drawn so…

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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A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 19.6, and the sample standard deviation, s, is found to be 5.6. (a) Construct a 98% confidence interval about u if the sample size, n, is 32. (b) Construct a 98% confidence interval about µ if the sample size, n, is 58. How does increasing the sample size affect the margin of error, E? (c) Construct a 99% confidence interval about u if the sample size, n, is 32. How does increasing the level of confidence affect the size of the margin of error, E? (d) If the sample size is 21, what conditions must be satisfied to compute the confidence interval? (a) Construct a 98% confidence interval about µ if the sample size, n, is 32. Lower bound:; Upper bound: (Round to two decimal places as needed.) C...
What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 farming regions gave a sample mean of x = Assume that o is known to be $1.92 per 100 pounds. $6.88 per 100 pounds of watermelon. (a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop (in dollars). What is the margin of error (in dollars)? (For each answer, enter a number. Round your answers to two decimal places.) lower limit $ upper limit 2$ margin of error $ (b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.27 for the mean price per 100 pounds of watermelon. (Enter a number. Round up to the nearest whole number.) farming regions (c) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop (in dollars). What is the margin of error (in dollars)? Hint: 1 ton is 2000 pounds. (For each…
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…
Answer question 4
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.t Suppose a small group of 20 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 = 0.32 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 limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) O normal distribution of weights Oo is known O n is large O o is unknown O uniform distribution of weights (c) Interpret your results in the context of this problem. O The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20. O we are 80% confident that the true average weight of Allen's hummingbirds falls…
Question 12 Astudent was asked to find a 90% confidence interval for the proportion of students who take notes using data from a random sample of size n = 86. Which of the following is a correct interpretation of the interval 0.13 p < 0.27? Check all that are correct. With 90% confidence, the proportion of all students who take notes is between 0.13 and0.27. U There is a 90% chance that the proportion of the population is between 0.13 and 0.27. O There is a 90% chance that the proportion of notetakers in a sample of 86 students wilL be between 0.13 and 0.27. O The proprtion of all students who take notes is between 0.13 and 0.27, 90% of the time. OWith 90% confidence, a randomly selected student takes notes in a proportion of their classes that is between 0.13 and 0.27.
Part 1 of 3: A car company takes a random sample of 40 cars and finds a mean mileage of 28.1 mpg. If the company knows the population standard deviation is 4.7 mpg, Find the 3-decimal small value of the 92% confidence interval. Part 2 of 3: Find the 3-decimal large value of the 92% confidence interval. Part 3 of 3: Find the 3-decimal margin of error
A confidence interval is desired for the true mean stray-load loss (Watts) for a certain type of induction motor. Assume that the stray-load loss is normally distributed with =3.3Watts. Situation 3: You take 100 measurements of stray-load loss from induction motors and determine that the mean of your sample is 53.6 Watts. When you calculate the 95% confidence interval for this situation, you get: (52.95 Watts, 54.25 Watts) is a 95% confidence interval for M. Compare this confidence interval to the interval you calculated in Situation 1. They are both 95% confidence intervals, but they have different sample sizes. What effect did sample size have on the interval estimate endpoints? Question 1: 25 sample size, Confidence Interval: (52.3064,54.8936)
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.t Suppose a small group of 12 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 a = 0.22 gram. When finding an 80% confidence interval, what is the critical value for confidence level? (Give your answer to two decimal places.) (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 limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) Oo is known Oo is unknown O uniform distribution of weights O normal distribution of weights On is large (c) Interpret your results in the context of this problem. We are 80% confident that the true average weight of…
A sample of n = 64 scores has a mean of M = 45 and an estimated standard error of 3 points. What is the sample variance? Select one: a. s2 = 113 b. s2 = 576 c. s2 = 498 d. s2 = 502
Boys and girls: The National Health Statistics Reports reported that a sample of 310 one-year-old boys had a mean weight of 25,3 pounds with a standard deviation of 3.3 pounds. In addition, a sample of 286 one-year-old girls had a mean weight of 24.3 pounds with a standard deviation of 3.6 pounds. Part: 0/2 Part 1 of 2 Construct a 98% confidence interval for the difference between the mean weights. Let u, denote the mean weight of one-year-old boys. Use the TI-84 plus calculator and round the answers to one decimal place. A 98% confidence interval for the difference between the mean weights is
In each case, use the information given to compute a confidence interval for the population mean ?. Assume that all necessary conditions are present for using the method described in Section 11.2. (Round your answers to two decimal places.) (a) x = 79, s = 9, n = 25, 95% confidence level to (b) x = 102, s = 9, n = 25, 95% confidence level ___to___ (c) x = 102, s = 19, n = 81, 99% confidence level ___to___ (d) x = 102, s = 19, n = 81, 90% confidence level ___to___

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