Critical Value vs Z-Score

A critical value is a quantile of a reference distribution, chosen so that a stated area falls beyond it. For a confidence interval at level $C$ it is the number $z^*$ (z-star) or $t^*$ (t-star) with the middle $C$ of the curve lying between $-z^*$ and $z^*$. For a significance test it is the point cutting off $\alpha$ (alpha) in whichever tail the alternative hypothesis points to. One idea, two jobs: it sets a width, or it sets a boundary.

The standard normal values, listed in full on the z-table:

At 95%, 1.960 is the value leaving 0.025 in each tail, so 0.95 of the standard normal curve sits between $-1.960$ and $1.960$. With the population standard deviation unknown you switch to $t^*$ at the right degrees of freedom, and it is larger than the matching $z^*$ at every finite df. At 95% the t-table gives 2.262 at 9 df, 2.064 at 24 df, and 1.984 at 99 df, closing on 1.960 from above without reaching it.

The error worth naming is "the critical value for 95% confidence is 1.96." It is 1.96 only when the reference curve is the standard normal. A mean from a sample of 12 uses $t$ with 11 df, where the critical value is 2.201, and with a normal population, reaching out only 1.96 standard errors captures the true mean 92.4% of the time rather than 95%. The interval is about 11% too narrow and it undercovers by more than two points.

Not every critical value comes in a $\pm$ pair. The chi-square distribution is right skewed and its tests use the upper tail alone, so there is a single cutoff: 7.815 at 3 degrees of freedom and $\alpha = 0.05$, from the chi-square table. A critical value also marks only the boundary. How far past it you landed is what a p-value reports, which is why two tests can share a critical value and carry very different evidence.

A z-score rescales one value into a count of standard deviations. For a population it is $z = \frac{x - \mu}{\sigma}$, where $\mu$ (mu) is the mean and $\sigma$ (sigma) is the standard deviation; for a sample it is $z = \frac{x - \bar{x}}{s}$, with $\bar{x}$ (x-bar) and $s$ in those roles. The units cancel in the division, so a z-score has no units of its own and values from different scales become comparable. The sign carries the direction and the size carries the distance.

A student scores 88 on a test with mean 80 and standard deviation 4, then 92 on a test with mean 86 and standard deviation 8. The raw scores say the 92 was the better performance. The z-scores say otherwise: $z = (88 - 80)/4 = 2.00$ against $z = (92 - 86)/8 = 0.75$. Relative to the class each score came from, the 88 is much the stronger result, and that comparison is the whole reason to standardize.

The wrong sentence to watch for is "her z-score of 1.5 means she beat 93 percent of the class." That 93.3 percent is the area to the left of 1.5 under the standard normal curve, so it is a fact about the normal model rather than about z-scores. A z-score is defined for any distribution that has a mean and a standard deviation, including badly skewed ones, and only when the distribution is roughly normal does the z-table turn it into a percentile.

Standardizing cannot repair shape. Subtracting and dividing by constants shifts and stretches the values without reordering them, so a right-skewed set of z-scores is still right-skewed and its outliers are still outliers.

Small samples put a hard ceiling on the answer. Using the sample standard deviation, no value among $n$ observations can have $|z|$ larger than $\frac{n-1}{\sqrt{n}}$. For $n = 5$ that ceiling is 1.79, so a z-score of 2 computed from five data points is an arithmetic error and not a rare event. The formula also breaks down when $s = 0$, which happens only when every value is the same.