SOLUTION: Dear tutor, I have a problem with CI section. Any help is appreciated. Thank you. Q. You take a sample of 22 from a population of test scores, and the mean of your sample is 60

Question 1003683: Dear tutor,
I have a problem with CI section. Any help is appreciated. Thank you.
Q. You take a sample of 22 from a population of test scores, and the mean of your sample is 60.
(a) You know the standard deviation of the population is 10. What is the 99% confidence interval on the population mean.
I got standard error of the mean which is 10/ square root of 22 =2.13
mean +- (Standard deviation of 99%)(2.13)= ??
If this formula is right how do I get standard deviation for 99%? I used normal distribution calculator the answer was between 34.2417 and 85.7583. It doesn't seem right.

(b) Now assume that you do not know the population standard deviation, but the standard deviation in your sample is 10. What is the 99% confidence interval on the mean now?
I still have standard error of the mean which is 2.13
I assume I use t table so I looked at 99% for df 21. the value was 2.831
so I tried to put these number to the formula above but I can't because I don't know how to get SD of 99%. .04)
The book says answer for (b) is (53.96, 6
Please help. thank you in advance.

Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
To calculate a confidence interval we need the confidence level, statistic, and margin of error.
***************************************************************************
a) margin of error = critical value * standard deviation of statistic
we want the 99% confidence interval on the population mean,
note that we use the t-distribution because our sample size is < 30
the degrees of freedom(df) = 22 - 1 = 21
t(alpha/2) is determined by consulting the t-distribution table
t(alpha/2) for 99% cf is 2.831
99% cf for population mean = 60 + or - (t(alpha/2) * (10/sqrt(22)) =
60 + or - (2.831 * 2.13) = 60 + or - 6.03
our 99% cf is (53.97, 66.03)
note that the sample mean is also the population mean
***************************************************************************
b) the std dev of the sample = st dev of population / sqrt(sample size)
for this part of the problem
std dev of the population = 10 * sqrt(22) = 46.9
99% cf for population mean = 60 + or - 2.831 * (46.9/sqrt(22)) =
60 + or - 28.3 = (31.7, 88.3)

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