Question 1183724: Find 80%, 85% and 99.9% confidence interval for the sample mean of a population, if we know that in a random sample of 70 people from the population, the sample mean is 60 and the standard deviation is 5.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sample size is 70.
sample mean is 60.
sample standard deviation is 5.
use of t-score is indicated.
degrees of freedom is 69 (70 minus 1).
standard error is 5 / sqrt(70) = .5976 rounded to 4 decimal places.
at 80% confidence interval, alpha is .20/2 = .10.
t-score for that, with 69 degrees of freedom, is plus or minus t = 1.294, rounded to 3 decimal places.
raw score for that is 59.23 to 60.77.
at 85% confidence interval, alpha is .15/2 = .075.
t-score for that, with 69 degrees of freedom, is plus or minus t = 1.456, founded to 3 decimal places.
raw score for that is 59.13 to 60.87.
at 99.9% confidence interval, alpha is .01/2 = .005.
t-score for that, with 69 degrees of freedom, is plus of minus t = 2.649, rounded to 3 decimal places.
raw score for that is 58.42 to 61.58.
i used the ti-84 plus calculator.
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