Question 664090
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.