Construct a 90% confidence interval estimate for thepopulation mean given the following values:x = 20.2, o 11.6 and n=16X=O a. 12.73

confidence level = 90 % sample mean = 20.2, σ = 11.6 and n=16

Answered: Construct a 90% confidence interval…

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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To decide what personnel requirements should be enforced, a hospital manager wants to estimate the mean number of people who are admitted to the emergency room during a 24-hour period. The director randomly selects 48 different 24-hour periods and counts the number of admissions for each. In his sample, the sample mean is 296 and the sample standard deviation is 89. Estimate the mean number of admissions per 24-hour period with a 95% confidence interval.
A Gallup poll of 1487 adults showed that 43% of the respondents have Facebook pages. Construct a 95% confidence interval for the proportion of adults who have a Facebook page.
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Assume that the 95% confidence interval for mean=9 and standard devaition=20. For a particular data, the standardized value z = 2.73. We can say that: Select one: a. The data with standardized value 2.73 does not belong to the confidence interval and the value is less than the average. b. The data with standardized value 2.73 does not belong to the confidence interval and the value is greater than the average. c. Cannot interpret the z-value information provided. d. The data with standardized value 2.73 if it belongs to the confidence interval and the value is less than the average. e. The data with standardized value 2.73 if it belongs to the confidence interval and the value is greater than the average.
which of the following sample data: sample size= 100 sample mean= 340z/hr sample variance= 90z/ hr construct a 98% confidence interval for μ. the population mean is what?
Find the critical value for a 98% confidence interval. Then calculate the interval.
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Calculate the 95% confidence interval and record the lower and upper limits from the following: Mean= 24.74 Standard Deviation= 14.37 Sample Size= 8
An online retailer has identified the maximum willingness of 10 consumers to pay for the same product. The 10 prices are: $10.00, $11.50, $9.50, $10.20, $11.20, $10.50, $9.80, $10.10, $9.90, $11.00 The retailer seeks a 95% confidence interval for the mean price consumers are willing to pay. The sample mean is calculated as $10.37, and the sample standard deviation is $0.66. Also consider the following t-values: Degrees of freedom 8 a. (9.99, 10.75) 9 10 90% confidence 1.86 1.833 1.812 95% confidence 2.306 2.262 2.228 99% confidence 3.355 3.250 3.169 Selecting the correct t-value, calculate the limits of the confidence interval.
Calculate a 90% lower confidence bound for the population mean using the following sample statistics: x=118.7 , s= 136.0 , n=37. Use one-decimal place accuracy.

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Construct a 90% confidence interval estimate for the population mean given the following values: