Question 1206353
sample size is 13.
sample mean is 35.6
standard deviation is 3.3


tk-score formula is t = (x-m)/s
t is the t-score
x is the critical sample mean
m is the sample mean
s is the standard error


at 99% two tail confidence interval with 12 degrees of freedom (sample size minus 1 = degrees of freedom in this problem), the critical t-score is equal to plus or minus t = 3.054540 rounded to 6 decimal places.


s = standard error = standard deviation / sqrt(sample size) = 3.3 / sqrt(13) = .915255 rounded to 6 decimal places.


t-score formula becomes plus or minus 3.054540 = (x - 35.6) / .915255.


on the high side of the confidence interval, solve for x to get x = 3.054540 * .915255 + 35.6 = 38.395683 rounded to 6 decimal places.


on the low side of the confidence interval, solve for x to get x = -3.054540 * .915255 = 35.5 = 32.804317 rounded to 6 decimal places.


your solution is that the a typical parent would spend from 32.804 to 38.396 on their child's birthday, rounded to 3 decimal places, at 99% confidence interval.