Consider the following procedure:

Take a random sample of size n = 13 from a Normally Distributed population
Compute the point estimates of the mean of the population and standard deviation of the population.
Estimate the standard error of the standard mean.

Assuming that you will follow this procedure in the future, what is the probability that the sample mean will be within 1 (estimated) standard error of the population mean?

Question

Consider the following procedure:

Take a random sample of size n =  13 from a Normally Distributed population
Compute the point estimates of the mean of the population and standard deviation of the population.
Estimate the standard error of the standard mean.

Assuming that you will follow this procedure in the future, what is the probability that the sample mean will be within 1 (estimated) standard error of the population mean?

BlackTom AI · Accepted Answer

Begin by restating the problem core: we take a random sample of size n = 13 from a Normal population, compute the sample mean Xbar and the sample standard deviation S, and consider the standard error of the mean as SE_hat = S / sqrt(n). We are asked for the probability that the sample mean lies within one estimated standard error of the population mean, i.e., P(|Xbar - mu| <= SE_hat).

Key fact: for a Normal population, when we standardize the sample mean by the estimated standard error, the resulting statistic follows a t-distribution with df = n - 1 = 12. Specifically, (Xbar - mu) / (S / sqrt(n)) ~ t_{12}.

Therefore, the probability in question can be expressed in terms of the t-distribution as P(|T| <= 1) where T ~ t_{12}.

To evaluate this, we use the symmetry of the t-distribution and the cumulative distribution function F_t of t_{12}:
P(|T| <= 1) = F_t(1; 12) - F_t(-1; 12) = 2*F_t(1; 12) - 1.

Because the t-distribution has heavier tails than the standard normal, the value of 2*F_t(1; 12) - 1 will be less than the corresponding normal-calculated value P(|Z| <= 1) ≈ 0.6827. With df = 12, the CDF at 1 is about 0.8315, giving 2*0.8315 - 1 ≈ 0.663. Numerical evaluation yields a result in this vicinity, reflecting the heavier tails of t_{12}.

So, conceptually, we’re leveraging the exact distribution of the standardized sample mean under normality with an estimated standard error, rather than a fixed-se standard error, which leads to a t-distribution with 12 degrees of freedom and the corresponding probability about 0.66. This aligns with the intuition that using S in place of the true sigma inflates the variability of the standardized mean slightly, compared to the normal case with known sigma.

Notes on potential confusion: if one mistakenly treated SE_hat as a fixed known value and used a normal z framework, the probability would be computed with a standard normal, yielding about 0.683, which is larger than the t-based result. Recognizing the dependency structure between Xbar and S in the normal model is crucial to justify the use of t_{12} rather than Z.

In summary, the calculation relies on the statistic (Xbar - mu) / (S / sqrt(n)) following a t-distribution with 12 degrees of freedom, and the probability of being within 1 unit of that random standard error is given by P(|T| <= 1) for T ~ t_{12}, which evaluates to approximately 0.66.

BTRY3010/BTRY5010/STSCI2200/STSCI5200 Lecture #20 ( 92 - 93, 99 - 101 )

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