Question 1131566: . A population has variance σ^2=100. A random sample of 25 from this population had a mean equal
17 and an estimated standard deviation equal 16. Can we conclude that the population mean is less
than 25? Let α=5%. Assume x is normally distributed. Hint, find σ_x .
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! population variance is 100
population standard deviation is sqrt(100) = 10
sample size is 25.
sample mean is 17
sample standard deviation is 16.
if you know the population standard deviation, then you use that.
otherwise you use the sample standard deviation and then the t-score rather than the z-score.
at 95% confidence level, the z-score will be plus or minus 1.96.
z-score formula is z = (x-m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard error.
standard error = population stnadard deviation / sqrt (sample size).
that becomes s - 10 / sqrt(25) = 10 / 5 = 2
on the high side for z, the formula becomes 1.96 = (x-17) / 2.
solve for x to get x = 2 * 1.96 + 17 = 3.92 + 17 = 20.92
on the low side for z, the formula becomes -1.96 = (1-17) / 2.
solve for x to get x = -2 * 1.96 + 17 = 13.08.
at 95% confident level, the true population mean is expected to be between 13.08 and 20.92.
based on this, you can conclude that the true population mean is less than 25.
that's my take.
here's some info on when to use t-score versus z-score.
https://www.statisticshowto.datasciencecentral.com/probability-and-statistics/hypothesis-testing/t-score-vs-z-score/
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