**Statistical Analysis of a Normally Distributed Population Sample****Sample Data:**A sample of 18 observations taken from a normally distributed population produced the following data:```28.1, 27.1, 25.2, 25.5, 31.2, 23.1, 26.2, 24.1, 28.4,37.1, 23.9, 28.9, 27.6, 25.5, 27.3, 25.4, 22.7, 22.9```**Instructions:**Round your answers to three decimal places.**a. Point Estimate of \(\mu\):**Calculate the point estimate (sample mean) of the population mean \(\mu\).\[\overline{x} = \boxed{\text{}} \]**b. 95% Confidence Interval for \(\mu\):**Construct a 95% confidence interval for the population mean \(\mu\).\[(\boxed{\text{}}, \boxed{\text{}})\]**c. Margin of Error for \(\mu\) in Part b:**Determine the margin of error for the estimate of \(\mu\) in part b.\[E = \boxed{\text{}}\]

Solution given data, 28.1 27.1 25.2 25.5 31.2 23.1 26.2 24.1 28.4 37.1 23.9 28.9 27.6 25.5 27.3…

Answered: A sample of 18 observations taken from…

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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You intend to estimate a population mean μμ with the following sample. 13.3 22.3 30.2 27.2 24.8 22.9 28.7 24.2 23.4 27 You believe the population is normally distributed. Find the 80% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).
Consider the following measures based on independently drawn samples from normally distributed populations: (You may find it useful to reference the appropriate table: chi-square table or F table) Sample 1: s = 241, and n1 = 26 Sample 2: s = 220, and n2 = 16 a. Construct the 95% interval estimate for the ratio of the population variances. (Round "F" value and final answers to 2 decimal places.) Confidence interval to b. Using the confidence interval from Part (a), test if the ratio of the population variances differs from 1 at the 5% significance level. The 95% confidence interval the value 1. Thus, we conclude that the population variances differ at the 5% significance level.
You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures: 29.4 50.7 45.5 30.3 40.1 57.3 44.3 39.4 39.9 46.2 Find the 95% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. Assume the data is from a normally distributed population.95% C.I. =
Tina catches a 14-pound bass. She does not know the population mean or standard deviation. So she takes a sample of five friends and they say the last bass they caught was 9, 12, 13, 10, and 10 pounds. Find the t and calculate a 95% (α = .05) confidence interval.
A hotel chain wants to estimate the mean number of rooms rented daily in a given month. The population of rooms rented daily is assumed to be normally distributed for each month with a standard deviation of 240 rooms. During February, a sample of 25 days has a sample mean of 370 rooms. The critical value for a 90% confidence interval for this sample is ________. Question 12 options: 1) 2.58 2) 1.96 3) 2.649 4) 1.645
A laboratory tested 85 chicken eggs and found that the mean amount of cholesterol was 191 milligrams. Assume that o = 16.2 milligrams. Construct a 95% confidence interval for the true mean cholesterol content, µ, of all such eggs. Round to one decimal place. O A. 186.9 to 195.1 milligrams B. 186.5 to 195.5 milligrams C. 188.1 to 193.9 milligrams D. 187.6 to 194.4 milligrams
Construct the indicated confidence interval for the population mean μ using the t-distribution.c = 0.95, x = 645, s = 31, n = 16 Question 3 options: (876.2, 981.5) (321.7, 365.8) (531.2, 612.9) (628.5, 661.5)
I need the right answer to a and b ASAP, please.
You must estimate…
Consider evidence from a random sample of 10 cars of the same model. The average gas mileage from this sample is 27 mpg. Assume that the gas mileage for all cars of this model is Normally distributed with a standard deviation of 5 mpg. Calculate a 95% confidence interval for the mean. Group of answer choices 23.84 to 30.16 17.20 to 36.80 23.90 to 30.10 26.02 to 27.98

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