Suppose we are interested in studying a population to estimate its mean. The population is normal and has a standard deviation of \(\sigma = 9\). We have taken a random sample of size \(n = 60\) from the population. This is Sample 1 in the table below. (In the table, Sample 1 is written "S1", Sample 2 is written "S2", etc.) As shown in the table, the sample mean of Sample 1 is \(\bar{x} = 81.1\). Also shown are the lower and upper limits of the 75% confidence interval for the population mean using this sample, as well as the lower and upper limits of the 95% confidence interval. Suppose that the true mean of the population is \(\mu = 80\), which is shown on the displays for the confidence intervals.Press the "Generate Samples" button to simulate taking 19 more random samples of size \(n = 60\) from this same population. (The 75% and 95% confidence intervals for all of the samples are shown in the table and graphed.) Then complete parts (a) through (c) below the table.### Table of Sample Means and Confidence Intervals| | \(\bar{x}\) | 75% lower limit | 75% upper limit | 95% lower limit | 95% upper limit ||-------|-------------|----------------|----------------|----------------|----------------|| S1 | 81.1 | 79.8 | 82.4 | 78.8 | 83.4 || S2 | 80.8 | 79.5 | 82.1 | 78.5 | 83.1 || S3 | 80.1 | 78.8 | 81.4 | 77.8 | 82.4 || S4 | 79.3 | 78.0 | 80.6 | 77.0 | 81.6 || S5 | 81.0 | 79.7 | 82.3 | 78.7 | 83.3 || S6 | 81.8 | 80.5 | 83.1 (a) How many of the 75% confidence intervals constructed from the 20 samples contain the population mean, μ = 80? □(b) How many of the 95% confidence intervals constructed from the 20 samples contain the population mean, μ = 80? □(c) Choose ALL that are true.- □ It is not surprising that some 75% confidence intervals are different from other 75% confidence intervals. Each confidence interval depends on its sample, and different samples may give different confidence intervals.- □ All of the 95% confidence intervals should be the same as each other. Since they are not all the same, there must have been errors due to rounding.- □ The center of the 75% confidence interval for Sample 1 is 80, because the center of any confidence interval for the population mean must be the population mean.- □ We would expect to find more 75% confidence intervals that contain the population mean than 95% confidence intervals that contain the population mean. Given a sample, a higher confidence level results in a narrower interval.- □ None of the choices above are true.

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Answered: Suppose we are interested in studying 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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Question 23 A test had 12 items. A new version of a test was made by adding a set of 12 items which resembles the original 12-item test. The new test having 24-items was administered a group of students. The mean and the standard deviation of the new 24-item test scores were 18 and 6, respectively. The difference score between the original 12-item test and the new set of 12 items was calculated. The standard deviation of the difference scores was 3. Estimate reliability.
• Question 6 Suppose the standard deviation of a sample is 5.8. Find the variance. Do not round. Submit Question
QUESTION 8 Nabisco claims that Chip Ahoy Cookies have an x ̅=28.67 chocolate chips per cookie with a SD=2.67. You randomly select 15 cookies from a new bag and carefully remove all the chocolate chips from each of the 36 cookies. You find the mean of the chocolate chips for the 15 cookies is 24.27 chocolate chips per cookie. Calculate the standard error for the chocolate chips in your experiment. (round to the nearest hundredth).
Question 4
Question #11
Example 4: The average cholesterol level for a random sample of 25 adult women is 263 units with a standard deviation of 43 units. From previous study, it is known that the population is normally distributed with a standard deviation of 40 units. Find a 90% C.I for the mean cholesterol level for all adult women in the population.
Question 7 Use the following information to answer Question 7 and Question 8.The marks of six students in a Mathematics test are represented by the variable x. The following data was produced: n=6, Σx = 279 and Σx2= 13 093.Find the mean and standard deviation of the following data.A. mean =46.5 , standard deviation = 5.82B. mean = 52.5, standard deviation = 4.89C. mean =46.5 , standard deviation = 4.89D. mean =52.5 , standard deviation = 5.82 Question 8 It was later discovered that one of the values was incorrectly recorded as 43. The correct value, in fact, is 34. Find the new values of the mean and standard deviation.A. mean = 45 , standard deviation = 7.07B. mean = 51, standard deviation = 6.84C. mean = 45 , standard deviation = 6.84D. mean = 51 , standard deviation = 7.07
Question 10
A Moving to another question will save this response. Question 1 Consider a sample with a mean of 30 and a standard deviation of 2.5. Use Chebyshev's theorem to determine the percentage of the data within the ranges from 23 to 37. O a. 61% O b. 75% O c. 87% Od. 83%
Solve question #9 for part D only.
Question 4. a. On a random day, a store can sell a mean value of 32 pasta packages and a standard deviation of 4. What is the probability that the total number of pasta packages sold by the store for a random sample of 58 days is less than 1862? Explain. Let b. X₁, X2, X3, and X4 represent the weight of shipment packages at a certain shipment facility. Suppose they are independent normal random variables with means M₁ = 0.8, M₂ = 2.6, μ3 = 8.4, µ4 = 2.3 pounds and variances . Find o² = 0² = 0} = 0} = 1 P(3X₁ + 1X2 + 1X3 + 2X4 ≤ 16.8)
Question 7: give the test statistic

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