Confused with z-statistic and t-statistic

A researcher has gathered data on the daily returns on a portfolio of call options over a recent 250-day period. The mean daily return has been 0.1%, and the sample standard deviation of daily portfolio returns is 0.25%. The researcher believes that the mean daily portfolio return is not equal to zero.

  1. Construct a 95% confidence interval for the population mean daily return over the 250-day sample period.


  1. Given a sample size of 250 with a standard deviation of 0.25%, the standard error can be computed as .
  2. At the 5% level of significance, the critical z-values for the confidence interval are z0.025 = 1.96 and −z0.025 = –1.96. Thus, given a sample mean equal to 0.1%, the 95% confidence interval for the population mean is:
    0.1 – 1.96(0.0158) ≤ µ ≤ 0.1 + 1.96(0.0158), or
    0.069% ≤ µ ≤ 0.131%

In the second step, why do we use the z-statistic instead of the t-statistic?
The promblem doesn’t show population variance, and the standard error is calculated by using sample standdard deviation.

I think for these kinds of problems, you assume a normal distribution (an application of the Central Limit Theorem), but even if you used the t-distribution, because the sample size is so large, the difference in results is quite small.

for example, with the T-distribution, you have a size of 250, so there are 249 degrees of freedom. interpolating the critical values for 200 and 300 gives an approximate critical value of 1.970, and the error is about .0311484% using the T-distribution, resulting in an interval of .0688516% to .1311484%, compared to the interval .0690103% to .1309897% using the Z-distribution.

If you only take the first 3 significant digits, you get the same results with either approach, but using the Normal distribution is easier.

(note that i used more significant digits than is justified, but the conclusion remains the same)

Because the sample size is large. Huge, actually.

Thank you for answering

My pleasure.