SOLUTION: You are given the sample mean and the standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. A random sample of 32 ei

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Question 664090: You are given the sample mean and the standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. A random sample of 32 eight ounce drinks has a mean of 86.7 calories and a standard deviation of 43.9 calories.
The 90% CI is
The 95% CI is
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
sample size is equal to 32
sample mean is equal to 86.7 calories
sample standard deviation is equal to 43.9 calories.

if this is a distribution of means, you have to calculate the standard error.
standard error = standard deviation / square root of sample size.
se = sd / sqrt(n)
se = 43.9 / sqrt(32) = 7.76050

standard error is equal to 7.76050 calories.

since you are using the sample standard deviation rather than the population standard deviation, you have to use a T score rather than a Z score.

the degrees of freedom is equal to the sample size minus 1 which = 31.

The critical T value at 90% confidence interval with 31 degrees of freedom is equal to +/- 1.69552

The critical T value at 95% confidence interval with 31 degrees of freedom is equal to +/- 2.03951

you need to translate your T score to X score.

X score = T score * standard error of sample + mean of sample

at 90% confidence interval this becomes:

X score lower limit = -1.69552 * 7.76050 + 86.7 = 73.54192
X score upper limit = 1.69552 * 7.76050 + 86.7 = 99.85808

at 95% confidence interval this becomes:

X score lower limit = -2.03951 * 7.76050 + 86.7 = 70.87246
X score upper limit = 2.03951 * 7.76050 + 86.7 = 102.52762

you are 90% confident that the population score will be between 73.54192 and 99.85808.

you are 95% confident that the population score will be between 70.87246 and 102.52762

i'm reasonably confident this is correct.
if you were using the population standard deviation then you would have used a Z score instead of a T score.