I randomly selected 50 text books and found their mean cost to be $248.47 with a standard deviation of $27.43. I computed a95% confidence interval of ($240.67, $256.27). What is the appropriate interpretation?8.For the instructor, this was question 13.X I am 95% confident that text books cost between $240.67 and $256.27.X There is a 95% probability that the true mean cost of text books is between $240.67 and $256.27./ We are 95% confident that the true mean cost of text books is between $240.67 and $256.27.X We are 95% confident that the true population proportion of text books is between $240.67 and $256.27.

Given information is :n=50x=$248.47S=$27.43And also given that 95% confidence interval is…

Answered: T randomly selected 50 text books and…

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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suppose you obtained a mean of $167 spent on transportation and a standard deviation of $40. Calculate a 95% confidence interval for the mean and select the values that come closest to those that would fill the spaces in the following interpretation: we can be 95% confident that they mean amount of money spent on transportation lies between _____ and _____ .
A teacher administers a standardized math .4 test to his class of 75 students. the mean score (out of 300 possible points) is 235. from previous studies, you know the population standard deviation is 28. using the sample data given, calculate a 95% confidence interval for the population mean (235.9 ,234.1) O (243.3 ,226.7) O (241.3 ,228.7) O (237 ,233) O (300 ,200) O
Sally is analyzing wheat production. She sampled 700 wheat farmers from across the state. She found a sample mean of 39 bushels per acre, with a standard deviation of 4.5 bushels. Using a 95% confidence interval, what inference can she make? O The yield per acre of wheat farms in the state ranges from 36.32 to 37.35 bushels. O The yield per acre of wheat farms in the state ranges from 38.67 to 39.33 bushels. O The yield per acre of wheat farms in the state ranges from 42.64 to 46.31 bushels. The yield per acre of wheat farms in the state ranges from 43.86 to 49.52 bushels.
.A study of 100 patients with back pain reported that the mean duration of pain was 12 months. It is known that the standard deviation of duration of back pain is 6.8 months. Assuming that the duration of this problem is normally distributed in the population; determine a 95% confidence interval for the mean duration of back pain in the population.
answer d
A sample of seven laptop computers had a variance of 25.88 years when the lifespan of the computer was studied. Find the 90% confidence interval for the true variance of the lifespan of laptop computers. Show all work leading up to your answer below
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The distribution of study times for first-year students follow a Normal distribution with a standard deviation σ = 78 minutes. Suppose a sample of 431 students found that the mean study time for these students was x¯= 195 minutes per week. Use this to find a 95% confidence interval to estimate the mean study time for all students. We are 95% confident that the mean study time for first-year students is between (Answer) and minutes ( Answer) minutes.
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)
If we want to provide a 95% confidence interval for the mean of a population, the Z value is, .95 1.645 .485 1.96
MY Question is attached as an image below. I have ansswered it below, but I think I've got (b) wrong, not sure how to get to the right answer. Exercise 5.12 (a) We are 95% sure that the true proportion of survey respondents’ answers of “not good” are between 3.4% and 4.24% with a mean of 3.82%. The mean is the midpoint of the confidence interval (3.4% + .42%). (b) 95% confidence means that about 95% of the intervals contain the parameter, p. The Standard Error is .0011 or 1.1%. > (sqrt(.0382*(1-.0382)/1151))*1.96 [1] 0.01107369 = Standard Error > 1.96*.011 [1] 0.02156 > .0382+0.02156 [1] 0.05976 > .0382-0.02156 [1] 0.01664 (c) The interval at 99% will be wider. At 99% the standard deviation from the mean will be -2.58 to +2.58 , versus -1.96 to +1.96 at the 95% confidence interval. (d) The standard error of estimate will be larger. A bigger sample tends to provide a more precise point estimate than a smaller sample. In this case there is a “larger net” with this…

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