Out of 85 adults selected randomly from one town, 63 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. a) find sample proportion . b) use to find margin of Error for 90% confidence interval. c) Find the confidence interval. d) Interpret your interval show work. round all numbers to at least 3 decimal places. you don't need to explain in words where you get the numbers(unless it is easier for you than writing in math notations). you can only write in math notations.it should be clear which function you are using.(normalcdf, invnorm. invt,..)
Out of 85 adults selected randomly from one town, 63 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. a) find sample proportion . b) use to find margin of Error for 90% confidence interval. c) Find the confidence interval. d) Interpret your interval show work. round all numbers to at least 3 decimal places. you don't need to explain in words where you get the numbers(unless it is easier for you than writing in math notations). you can only write in math notations.it should be clear which function you are using.(normalcdf, invnorm. invt,..)
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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Out of 85 adults selected randomly from one town, 63 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance.
a) find sample proportion .
b) use to find margin of Error for 90% confidence interval.
c) Find the confidence interval.
d) Interpret your interval
show work. round all numbers to at least 3 decimal places. you don't need to explain in words where you get the numbers(unless it is easier for you than writing in math notations). you can only write in math notations.it should be clear which
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