Question 1173384: Question :
A sample of 49 is drawn from a normal population with unknown mean and variance. If the sample means is 131.8 and standard deviation 6.25,setup a 90% confidence interval for the population mean?
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Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sample size is 49
mean of sample is 131.8
standard deviation of sample is 6.25
you want to set up a 90% confidence interval.
degrees of freedom is 49 - 1 = 48.
standard error = standard deviation of sample divided by square root of sample size.
you will be using t-scores with 48 degrees of freedom.
standard error = 6.25 / sqrt(49) = 6.25 / 7 = .8928571429.
i rounded this to .892857 which is probability a little more than close enouogh.
in the t-score table, with 90% confidence interval, one tailed alpha would be 5% and two tailed alpha would be 10% (5% on each end).
here's the table i used.
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
that table only showed degrees of freedom of 40 and 60.
48 degrees of freedom is somewhere in between.
fortunately, there is a t-score calculator online that can be used to get more accurate critical t-score.
that t-score calculator can be found at https://stattrek.com/online-calculator/t-distribution.aspx
the calculator gets you the area to the left of the t-score.
to find the two tailed 10% conidence interval, i entered .05 in the p(T < t) box and a mean of 0 and a standard deviatiion of 1 and degrees of freedom of 48 and clicked on calculate.
calculator said the t-score was -1.677
since the normal distribution table is symmetric, my confidence interval would be between a t-score of -1.677 and a t-score of 1.677
that would be 5% area to the left of -1.677 and 5% area to the right of -1.677.
in the table, two tailed of 10% with 40 degrees of freedom gave a t-score of -1.684 and 60 degrees of freedom gave a t-score of 1.671.
about halfway between those t-score would yield a t-score of -1.6775 which is pretty close to -1.677 found by the calculator.
i used a critical t-score of plus or minus 1.677 for my 10% two tailed confidence interval.
once this is done, you need to use the t-score formula to find the raw score.
t-score formula is t = (x - m) / s
t is the t-score
x is the raw score
m is the mean
s is the standard error.
for t = -1.677, i got:
-1.677 = (x - 131.8) / .892857
solve for x to get:
x = -1.677 * .892857 + 131.8 = 130.3026788
for t = 1.677, i got:
1.677 = (x - 131.8) / .892857
solve for x to get:
x = 1.677 * .892857 + 131.8 = 133.2973212
your answer should be:
90% confidence interval is between 130.3026788 and 133.2973212.
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