8.1.15Suppose a simple random sample of size n 49 is obtained from a population with u 79 and a =14.(a) Describe the sampling distribution of x.(b) What is P (x> 82.9)?(c) What is P (xs74.7)?(d) What is P (76.3 82.9)=Enter your answer In the answer box and then click Check Answer.Clear Allpartsremaining

(a)From the given information, sample size is n = 49. From the central limit theorem if the sample…

Answered: 8.1.15 Suppose a simple random sample…

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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How many possible of size n = 4 can be drawn from a population of size 15? (a) 210 (b) 1365 (c) 3003 (d) 32760 What is the shape of the sampling distribution of the means if random sample of size n become larger? (a) left skewed (b) right skewed (c) normal (d) linear Which sample size will give a smaller standard error of the mean? (a) 10 (b) 13 (c) 30 (d) 46 What is the finite population correction factor if the size of the population is 250 and the sample size is 20? (a) 0.9210 (b) 0.9361 (c) 0.9365 (d) 0.9610
Suppose a simple random sample of sizen3200 is obtained from a population whose size is N=20,000 and whose population proportion with a specified characteristic is p= 0.4. Complete parts (a) through (c) below. ... (a) Describe the sampling distribution of p. Choose the phrase that best describes the shape of the sampling distribution below. OA Not normal becausens0.05N and np(1-p) 10. OC. Not normal becausens0.05N and np(1-p) 10. O D. Approximately normal because n<0.05N and np(1-p) < 10. Determine the mean of the sampling distribution of p. HA= 0.4 (Round to one decimal place as needed.) Determine the standard deviation of the sampling distribution of p. on = 0.034641 (Round to six decimal places as needed.) (b) What is the probability of obtaining x = 90 or more individuals with the characteristic? That is, what is P(p20.45)? P(p20.45) = (Round to four decimal places as needed.) Clear all Check answer
A population of values is normally distributed with μ=58.8 and σ=24.3a) Find P61 for a single randomly selected value. This answer represents the score separating the bottom 61% scores from the top 39% scores. P61 = b) Now draw a random sample of size n=180 . Find P61 for the sampling distribution. This answer represents the score separating the bottom 61% of the sample means from the top 39% of the sample means. P61 = Enter your answers as numbers accurate to 1 decimal place.
5,5 Suppose a random variable xx is normally distributed with μ=20 and σ=6.7. According to the Central Limit Theorem, for samples of size 8: (a) The mean of the sampling distribution for x¯ is: (b) The standard deviation of the sampling distribution for x¯ is
Suppose a simple random sample of size n=64 is obtained from a population with μ=83 and σ=16. (a) Describe the sampling distribution of x. (b) What is P x>86.7? (c) What is P x≤78.3? (d) What is P 80.8<x<86.9?
Suppose a simple random sample of size n = 125 is obtained from a population whose size is N = 30,000 and whose population proportion with a specified characteristic is p = 0.6
Suppose a simple random sample of size na 150 is obtained from a population whose size is N 25,000 and whose population proportion with a specifed characteristic is p=0.6. Complete parts (a) through (c) below, (a) Describe the sampling distribution of p. Choose the phrase that best describes the shape of the sampling distribution below QA Not normal because ns0.05N and np(1-p) 10. OB Approximately normal becausens0.05N and np(1-p)< 10. OC. Not nomal because ns0.05N and np(1- p) 1o. o. Approximalely nomal becausens0.0SN and ne(t-pa 10 Determine the mean of the sampling distribuion of p H Round to one decimal place as needed)
Q1(10). Manatees are one creature found in coastal waterways. To study this creature, a random sample of 12 manatees were selected, the sample manatees were released before their weights (in pounds) were carefully recorded: 956, 1012, 954, 988, 973, 1048, 1075, 1064, 856, 1026, 1031, 1064. Assuming the underlying distribution of manatee weights is normal. 1(5). Find a 95% confidence interval for the true mean weight. 2(5). In a similar study 10 years ago, researchers concluded that the true mean weight of manatees was approximately 1000 pounds. Implement a hypothesis test to verify if there is any evidence to suggest the true mean weight is less than 1000 pounds now. Solve the hypothesis using P-value method. Let a = 0.05.
Suppose a simple random sample of size n=10 is obtained from a population with μ=68 and σ=19. Answer a-c (a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities regarding the sample mean? Assuming the normal model can be used, describe the sampling distribution x. (b) Assuming the normal model can be used, determine P(x<71.1). (c) Assuming the normal model can be used, determine P(x≥69.2).
A population consists of 1, 1, 1, 2, 2, 3 and 4 (i) List all possible samples of size n= 3 without replacement. (ii) Obtain the sampling distribution of X. (iii) Verify that: uz =u and N - n N - 1
How do you solve this statistics question without using code? Find P(X1 > X2)
A simple random sample of size n=81is obtained from a population with μ=82 and σ=36. (a) Describe the sampling distribution of x. (b) What is P x>86.4? (c) What is P x≤73.6? (d) What is P 77.4<x<90.2?

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