8.1 Confidence Interval1. Crop researchers are interested in the productivity of a new variety of corn. They plant 25 plots withrandomly selected seeds of the new variety, record the yield in bushels per acre, and find that a 99%confidence interval for the true mean yield is 118 to 130 bushels per acre.(a) Interpret the confidence interval.(b) What is the point estimate that was used to create the interval? What is the margin of error?(c) Interpret the confidence level.(d) Based on the research, a marketer tells a possible customer that with this new variety of corn they canexpect to receive more than 130 bushels per acre, on average. Use the confidence interval to evaluatethis statement.

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Answered: 1. Crop researchers are interested in…

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 researcher surveyed college students in the United States on the typical amount of time each day that they spend interacting with different types of media (television, social media, Internet-connected devices, game consoles, etc.) The researcher found that the mean amount of time that college students spent watching television each day is 135 minutes with a 95% confidence interval of (105, 165). a. State the conclusion the researcher can make from this confidence interval. b. What is the margin of error for the confidence interval?
A random sample of 16 employees taken from the management level of the company gives a mean monthly salary of $17,000 with a standard deviation of $5500. (a) Calculate the 99% confidence interval for the population mean of management level employees? (b) Write an interpretation for the interval. (c) How much wider/narrower will the 98% confidence interval be?
(a) A study is interested in the proportion of postgraduate students who have met the criteria of depression during their study periods. In a random sample of 50 postgraduate students, 15 are found to be afflicted. Use the normal distribution table to construct a 98% confidence interval for the proportion of postgraduate students who have experienced depression. What can we assert with 98% confidence about the possible size of our error if we estimate the proportion to be 0.30?
Suppose a 90 percent confidence interval to estimate a population proportion was calculated from a sample proportion of 18 percent and a margin of error of 4 percent. What is the width of the confidence interval?
You would like to construct a 99% confidence interval to estimate the population mean price of milk (per gallon) in your city. You select a random sample of prices from different stores. The sample has a mean of 3.66 dollars and a standard deviation of 0.20 dollars. (a) What is the best point estimate, based on the sample, to use for the population mean? O dollars (b) For each of the following sampling scenarios, determine which distribution should be used to calculate the critical value for the 99% confidence interval for the population mean. (In the table, Z refers to a standard normal distribution, and t refers to at distribution.) Could use Sampling scenario Unclear either Z or t The sample has size 17, and it is from a normally distributed population with a known standard deviation of 0.26. The sample has size 80, and it is from a non-normally distributed population. The sample has size 13, and it is from a normally distributed population with an unknown standard deviation.
CEO compensation A sample of 20 CEOs from theForbes 500 shows total annual compensations rangingfrom a minimum of $0.1 to $62.24 million. The averagefor these 20 CEOs is $7.946 million. Here’s a histogram:Based on these data, a computer program found that a 95%confidence interval for the mean annual compensationof all Forbes 500 CEOs is (1.69, 14.20) $ million. Whyshould you be hesitant to trust this confidence interval?
Suppose that you are interested in the number of people in each household in the U.S. Your sample includes 10 households, with a mean 2.5 and a standard deviation 1.2. Construct a 95% confidence interval (CI) of the mean number of household members. (A) 1.2-3.8 (B) 1.4-3.6 (C)1.6-3.4 (D) 1.7-3.3 (E) 1.8-3.2
Margin of error and the confidence interval. A study of stress on the campus of your university using the College Undergraduate Stress Scale (CUSS) reported an average stress level of 1123 (a higher score indicating more stress) with a margin of error of 43 for 95% confidence. The study was based on a random sample of 400 undergraduates. (a) Give the 95% confidence interval. (b) If you wanted 99% confidence for the same study, would your margin of error be greater than, equal to, or less than 43?
The mortgage department of a large bank is interested in the nature of loans of first-time borrowers. This information will be used to tailor their marketing strategy. They believe that the probability that first-time borrowers take out smaller loans than other borrowers is 0.37. They perform a hypothesis test to determine if the probability is the same or different from 0.37. They sample 104 first-time borrowers and find 84 of these loans are smaller than the other borrowers. What is the test statistic to test the null hypothesis H0: p = 0.37 against the alternative hypothesis Ha: p ≠ 0.37?
A physical therapist wanted to know whether the mean step pulse of men was less than Two sample T for Men vs Women the mean step pulse of women. She randomly selected 51 men and 71 women to Mean StDev SE Mean participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained. Men Women 95% Cl for mu Men - mu Women (-11.44, - 1.96) T-Test mu Men = mu Women (vs <) 112.2 118.9 51 11.8 1.7 71 14.4 1.7 T= - 2.82 P = 0.0028 DF = 117
A flower shop owner wants to estimate the average retail value of the best-selling flower in his inventory. A random sample of 10 flowers indicates an average value of 15 pesos and a standard deviation of 3.5 pesos. What are the point estimate and the 99% confidence interval in the average value of the flower? What is the confidence interval estimate of the problem? a. (11.40 , 18.60) b. (7.29 , 12.71) c. (11.68 , 18.32) d. (7.06 , 12.94)

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