A data set includes 108 body temperatures of healthy adult humans having a mean of 98.2°F and a standard deviation of 0.65°F. Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6°F as the mean body temperature? Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean μ? °F<<°F (Round to three decimal places as needed.) What does this suggest about the use of 98.6°F as the mean body temperature? A. This suggests that the mean body temperature is lower than 98.6°F. OB. This suggests that the mean body temperature could very possibly be 98.6°F. OC. This suggests that the mean body temperature is higher than 98.6°F. CXLE
A data set includes 108 body temperatures of healthy adult humans having a mean of 98.2°F and a standard deviation of 0.65°F. Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6°F as the mean body temperature? Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean μ? °F<<°F (Round to three decimal places as needed.) What does this suggest about the use of 98.6°F as the mean body temperature? A. This suggests that the mean body temperature is lower than 98.6°F. OB. This suggests that the mean body temperature could very possibly be 98.6°F. OC. This suggests that the mean body temperature is higher than 98.6°F. CXLE
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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Expert Solution
Step 1
Given information-
Sample size, n = 108
Sample mean, x-bar = 98.2 ℉
Sample standard deviation, s = 0.65 ℉
Significance level, α = 0.01
Population mean, μ = 98.6℉
Since here population standard deviation is unknown so we will use t-distribution.
Now, degree of freedom is given by-
Degree of freedom, df = n - 1
Degree of freedom, df = 108 - 1 = 107
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