Question 1193836
since you are using the standard deviation of the sample, the use of the t-score is indicated.
in this case, it probably doesn't make much difference.
i'll show you why.
standard error = 75 / sqrt(64) = 75/8 = 9.375
z = (x - m) / s
x is the sample mean
m is the assumed population mean.
s = standard error.
the formula becomes:
z-score = (975 - 1000) / 9.375 = -2.67
t-score is the same.
the probability that a z-score less than -2.67 will occur is equal to .00379.
the probability that a t-score with 63 degrees of freedom is less than -2.67 is equal to .00482.
a two tailed confidence interval of 95% has an area of .025 to the left of it and .025 to the right of it.
.0038 and.0048 are both less than that, making the results significant whether or not you use the z-score or the t-score.