SOLUTION: Find the margin of error and confidence interval.
95% confidence, n=100, xbar=9500, standard deviation = 12,345
Algebra.Com
Question 1001734: Find the margin of error and confidence interval.
95% confidence, n=100, xbar=9500, standard deviation = 12,345
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The SE is s/sqrt(n)=1234.5
n=100 is large enough to use the z-distribution, and z(0.95)-1.96
1.96*SE=2419.62, That is the half-width of the CI and the margin of error when said +/-.
(9500-2419.6, 9500+2419.6)
(7080.4, 11919.6)
RELATED QUESTIONS
Find the margin of error (E) of a confidence estimate interval, when the confidence level (answered by Boreal)
Finding confidence intervals.
Find the margin of error and the confidence interval for... (answered by Boreal)
Find the sample size.
Margin of error = $125, confidence level = 95%, standard deviation (answered by Boreal)
please help me im stuck and my time is running out i only have 55 minutes left to answers (answered by lynnlo)
In a survey, 26 people were asked how much they spent on their child's last birthday... (answered by Boreal)
FINDING SAMPLE SIZE.
Q1. margin of error:$125, confidence level:95%,standard... (answered by stanbon)
When 1460 subjects were asked, "On how many days in the past 7 days have you... (answered by rothauserc)
A simple random sample of 40 items resulted in a sample mean of 25. The population... (answered by stanbon)
In a random sample of 26 computers, the mean repair cost was $130 with a sample standard... (answered by stanbon)