4. The following sample of numbers is chosen from a normally distributed population.22, 25, 28, 30, 27The 95% confidence interval for the population mean is(A) 26.4 + 1.0033(B) 26.4 ± 1.3638(C) 26.4 + 2.2435(D) 26.4 + 2.6731(E) 26.4 + 3.7860

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Answered: 4. The following sample of numbers is…

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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Assuming that the sample statistics are n=36, mean = 24.7 and s=19.8. Then, the resulting 99% confidence interval for population mean is: O a. 20.38 < μ< 29.02 O b. 18.00 < μ<31.40 O c. 15.71 <μ< 33.69 O d. 19.12 μ< 30.28
A large fishing operation takes a sample of 40 lobster pots and finds an average of 35 in each with a sample standard deviation of 10. What is the 95% confidence interval for the mean number of lobsters in each of the lobster pots? a. (86.34, 92.04) b. (53.06, 58.94) c. (175.69, 182.31) d. (31.90, 38.10) (89.86,94.14) е.
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Is the mean length of adult invasive Lionfish in the Atlantic coast of South Carolina greater than 355.6mm? (This is just the research question I using please just answer question 2 and 3 please.) Lower limit: 429.737 Upper limit: 445.163 Question 2. Interpret the above confidence intervals: Question 3. If you have one sample group, determine whether the value in your research question (the value which is compared to the population mean) is within the confidence interval or not. Then describe what it implies. If you have two groups of data, determine whether the confidence intervals of the two groups overlap or not. Then describe what that implies.
What z score would be used to calculate the margin of error for a 98% confidence interval for the mean? 2.33 1.75 2.05 correct answer is not given
1. The strength of a fiber is normally distributed with a known population standard deviation of 12 kg. An engineer sampled random pieces of fiber, and got the following data: 43, 54, 74.2, 47, 61.6, 80, 38, 40, 32 Find the 96% confidence interval for the mean strength of the fiber.
20. In order to estimate the percentage of adults who file their taxes at least two weeks before the April 15th deadline, a 99% confidence interval is constructed. The interval ends up being from 0.776 to 0.884. Which of the following could be a 95% confidence interval for the same sample data? I. 0.789 to 0.871 II. 0.764 to 0.896 III. 0.754 to 0.894 IV. 0.746 to 0.844 1. IV only 2. III only 3. I only 4. None of the answer options are reasonable 95% confidence intervals. 5. II only
A simple random sample of 5 recently hospitalized migraine sufferers were given a new treatment; the results are listed below. Construct the 90% confidence estimate of the mean difference in the number of migraines per month before and after the treatment. Summary of # of migraines before and after treatment (5/3, 9/9, 7/4, 5/2, 3/4) Subject A Subject B Subject C Subject D Subject E # of migraines before 5 9 7 5 3 # of migraines after 3 9 4 2 4 **Let the first population be of the "before" results and the second population be of the "after" results.** The right hand critical value is . Round to 2 decimal places. The error is . Round to 2 decimal places. The mean difference in migraines per month is between and .
If 64 out of 320 people received fruit in addition to candy what are the 95% confidence interval values? (0.14863, 0.15137) (0.14825, 0.15175) (0.19902, 0.20098) (0.15617, 0.24383) (0.09063, 0.20937) (0.09638, 0.20362) 0000
Construct a confidence interval for the ratio of the population variances given the following sample statistics. Round your answers to four decimal places. n1=9, n2=18, s1^2/s2^2=1.17, a=0.1
A random sample of 49 statistics examinations was taken. The average score, in the sample, was 85 with a variance of 12.96. What is the 95% confidence interval for the average examination score of the population of the examinations? O 79.00 to 91.00 O 81.40 to 88.60 O 83.97 to 86.03 84.15 to 85.85
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4. The following sample of numbers is chosen from a normally distributed population. 22, 25, 28, 30, 27 The 95% confidence interval for the population mean is (A) 26.4 ± 1.0033 (B) 26.4 ± 1.3638 (C) 26.4 ± 2.2435 (D) 26.4 ± 2.6731 (E) 26.4 ± 3.7860