z-test, t-test, chi-squared goodness of fit, F-test: Exercises

Quick symbol refresher (nothing new is assumed):

Rejection rule everywhere: compute a test statistic, compare its size to a critical value from a table. Bigger than critical ⇒ ==reject $H_{0}$ ; smaller ⇒ fail to reject==. See Hypothesis Testing and p-value.

Figure — z-test, t-test, chi-squared goodness of fit, F-test

The picture above is the whole game: the bell/hump is "what luck alone does under $H_{0}$ "; the shaded ends are the critical region. If your statistic lands in the shade, luck-alone becomes implausible.

Level 1 — Recognition

L1·Q1 — Pick the test

For each situation, name z, t, χ², or F. (a) Testing if a coin's true mean bias equals 0.5, with the population standard deviation known and $n = 200$ . (b) Testing if a mean fill volume is 250 ml from $n = 6$ bottles, $σ$ unknown. (c) Testing whether the tally of eye colours in a class matches a claimed ratio. (d) Testing whether two machines produce parts with equal variance.

Recall Solution

(a) $σ$ known, large $n$ ⇒ z-test. (b) $σ$ unknown, small $n$ , a mean ⇒ t-test. (c) Comparing observed counts in categories to expected ⇒ chi-squared goodness of fit. (d) A ratio of two variances ⇒ F-test. Mnemonic from the parent: Z Sure, T Tries, Chi Counts, F Fights.

L1·Q2 — Read the critical value

At $α = 0.05$ , two-sided, state the critical cutoff for: (a) a z-test, (b) a t-test with $ν = 4$ , (c) a χ² GOF with $ν = 5$ (one-sided upper).

Recall Solution

(a) $z_{0.025} = 1.96$ . (b) $t_{0.025, 4} = 2.78$ (fatter tails than z because $s$ adds uncertainty — Student's t-distribution). (c) $χ_{0.05, 5}^{2} = 11.07$ (χ² is one-tailed on the upper end because "surprise" only accumulates as a positive sum — Chi-squared Distribution).

Level 2 — Application

L2·Q1 — One-sample z

A factory claims wire breaks at $μ_{0} = 120$ N. Known $σ = 15$ N. Sample $n = 25$ , $\overset{x}{ˉ} = 114$ . Test at $α = 0.05$ , two-sided.

Recall Solution

WHAT: compute the standardized distance from the claim. $SE = \frac{σ}{n} = \frac{15}{25} = \frac{15}{5} = 3$ . Why $25$ ? averaging 25 values shrinks the wobble by $25 = 5$ . $Z = \frac{x ˉ - μ _{0}}{SE} = \frac{114 - 120}{3} = \frac{- 6}{3} = - 2.0$ . Compare: $∣ - 2.0∣ = 2.0 > 1.96$ ⇒ ==reject $H_{0}$ ==. The wire is significantly weaker than claimed.

L2·Q2 — One-sample t

$H_{0} : μ = 100$ . Data: $103, 98, 101, 99, 104$ ( $n = 5$ ), $σ$ unknown. Test at $α = 0.05$ , two-sided.

Recall Solution

Mean: $\overset{x}{ˉ} = \frac{103 + 98 + 101 + 99 + 104}{5} = \frac{505}{5} = 101$ . Deviations from $\overset{x}{ˉ}$ : $2, - 3, 0, - 2, 3$ . Squares: $4, 9, 0, 4, 9$ , sum $= 26$ . Variance (Bessel's $n - 1$ ): $s^{2} = \frac{26}{5 - 1} = 6.5$ , so $s = 6.5 \approx 2.55$ . Why $n - 1$ ? deviations are from $\overset{x}{ˉ}$ , which sits closest to its own data, so we lose 1 d.f. to the constraint $\sum (x_{i} - \overset{x}{ˉ}) = 0$ — see Bessel's Correction. SE: $s / 5 = 2.55/2.236 \approx 1.140$ . Statistic: $t = \frac{101 - 100}{1.140} \approx 0.877$ , $ν = 4$ . Compare: $∣0.877∣ < 2.78$ ⇒ fail to reject. No evidence the mean differs from 100.

L2·Q3 — χ² goodness of fit

A bag should hold red:blue:green in ratio $2 : 1 : 1$ . Sample of $n = 100$ gives $O = (46, 28, 26)$ . Test the ratio at $α = 0.05$ .

Recall Solution

Expected counts from the claimed ratio (total $2 + 1 + 1 = 4$ parts): $E_{red} = 100 \cdot \frac{2}{4} = 50$ , $E_{blue} = 100 \cdot \frac{1}{4} = 25$ , $E_{green} = 25$ . Each cell's surprise $(O - E)^{2} / E$ : red $\frac{( 46 - 50 ) ^{2}}{50} = \frac{16}{50} = 0.32$ ; blue $\frac{( 28 - 25 ) ^{2}}{25} = \frac{9}{25} = 0.36$ ; green $\frac{( 26 - 25 ) ^{2}}{25} = \frac{1}{25} = 0.04$ . Sum: $χ^{2} = 0.32 + 0.36 + 0.04 = 0.72$ . d.f.: $k = 3$ categories, no parameters estimated ( $m = 0$ ), so $ν = k - 1 - m = 2$ . Critical $χ_{0.05, 2}^{2} = 5.99$ . Compare: $0.72 < 5.99$ ⇒ fail to reject. Data are consistent with $2 : 1 : 1$ .

Level 3 — Analysis

L3·Q1 — Why does the same data reject with z but not t?

A sample of $n = 5$ gives $\overset{x}{ˉ} = 101$ , $s = 2.55$ , testing $H_{0} : μ = 100$ (this is L2·Q2). Suppose a colleague wrongly treats $s$ as a known $σ$ and runs a z-test. Show the two conclusions and explain the mechanism.

Recall Solution

Wrong (z) path: $Z = \frac{101 - 100}{2.55/ 5} = 0.877$ . Same number as the $t$ statistic — because the arithmetic is identical! The difference is the yardstick, not the statistic.

z uses cutoff $1.96$ : $0.877 < 1.96$ ⇒ fail to reject here too. Let's make it bite: raise $\overset{x}{ˉ}$ to $103$ . Then $t = \frac{3}{1.140} = 2.63$ .
t (correct): cutoff $2.78$ ; $2.63 < 2.78$ ⇒ fail to reject.
z (wrong): cutoff $1.96$ ; $2.63 > 1.96$ ⇒ reject. Mechanism: with small $n$ the estimate $s$ is itself noisy. Sometimes $s$ happens to come out small, making the ratio look large by luck. The $t$ distribution's fatter tails widen the cutoff to absorb exactly this extra luck. Using $1.96$ ignores it and manufactures a false "significant" result. As $n \to \infty$ , $s \to σ$ and the two cutoffs merge.

L3·Q2 — Degrees of freedom bookkeeping

You fit a Poisson model to counts in $k = 6$ bins, but you estimated the Poisson rate $λ$ from the same data. You get $χ^{2} = 9.5$ . Test the fit at $α = 0.05$ .

Recall Solution

d.f. accounting: start with $k = 6$ cells. Subtract 1 for the fixed total ( $\sum O = \sum E = n$ ). Subtract 1 more because $λ$ was estimated from the data ( $m = 1$ ). $ν = k - 1 - m = 6 - 1 - 1 = 4.$ Critical $χ_{0.05, 4}^{2} = 9.49$ . Compare: $9.5 > 9.49$ (just barely) ⇒ reject the fit. Why it matters: had you forgotten the estimated parameter and used $ν = 5$ , the cutoff would be $11.07$ , and $9.5 < 11.07$ would wrongly say "fits fine." Every parameter you steal from the data to build $E_{i}$ costs one degree of freedom, tightening the cutoff.

Level 4 — Synthesis

L4·Q1 — The F-test as a gatekeeper

Machine A: $s_{1}^{2} = 45$ , $n_{1} = 8$ . Machine B: $s_{2}^{2} = 15$ , $n_{2} = 10$ . First decide whether their variances are equal ( $α = 0.05$ , two-sided). Then say what test you'd use next to compare their means.

Recall Solution

Step 1 — F-test. Put the larger variance on top so $F \geq 1$ and we use one upper tail: $F = \frac{s _{1}^{2}}{s _{2}^{2}} = \frac{45}{15} = 3.0, ν_{1} = n_{1} - 1 = 7, ν_{2} = n_{2} - 1 = 9.$ Two-sided at $α = 0.05$ uses $F_{0.025, 7, 9} \approx 4.20$ . $3.0 < 4.20$ ⇒ fail to reject equality of variances. Why an F? Each $s^{2}$ is (up to scale) a $χ^{2} / ν$ , and the ratio of two such is by definition $F$ — see F-distribution and Chi-squared Distribution. Step 2 — next test. Since we could not reject equal variances, a two-sample t-test with pooled variance (equal-variance assumption satisfied) is the appropriate follow-up for the means. Had F rejected, we'd use the unequal-variance (Welch) t-test instead. This chaining is exactly the logic behind ANOVA: F-ratios decide whether group spreads/means differ.

L4·Q2 — From p-value to decision, carefully

A two-sided z-test gives $Z = 1.7$ . (a) Is it significant at $α = 0.05$ ? (b) At $α = 0.10$ ? (c) A student concludes "the p-value is large-ish, so $H_{0}$ is true." Correct them.

Recall Solution

(a) Cutoff $z_{0.025} = 1.96$ . $1.7 < 1.96$ ⇒ fail to reject at $5%$ . (b) Cutoff $z_{0.05} = 1.645$ . $1.7 > 1.645$ ⇒ reject at $10%$ . The very same data flips conclusion when $α$ changes — significance is a chosen threshold, not a fact of nature. (c) Failing to reject means we found insufficient evidence against $H_{0}$ , not proof of it. Absence of evidence ≠ evidence of absence. A larger sample might well have crossed the line. See p-value and Type I and Type II Errors.

Level 5 — Mastery

L5·Q1 — Design and catch the trap

You have $n = 4$ measurements $12, 15, 11, 14$ of a quantity claimed to be $μ_{0} = 10$ , with $σ$ unknown. A colleague reports a z-test giving "reject at 5% since $Z = 2.5 > 1.96$ ." Reproduce their number, then correct the analysis.

Recall Solution

Reproduce (their flawed path). $\overset{x}{ˉ} = \frac{12 + 15 + 11 + 14}{4} = \frac{52}{4} = 13$ . Deviations $- 1, 2, - 2, 1$ ; squares $1, 4, 4, 1$ , sum $= 10$ . $s^{2} = \frac{10}{4 - 1} = 3.333$ , $s = 1.826$ . $SE = 1.826/ 4 = 0.913$ . Statistic $= \frac{13 - 10}{0.913} = 3.286$ . (The colleague likely mis-divided by $n$ not $n - 1$ : with $s^{2} = 10/4 = 2.5$ , $s = 1.581$ , SE $= 0.79$ , stat $= 3.80$ — or simply rounded to "2.5." Either way the arithmetic is not the crime.) The real error: with $σ$ unknown and $n = 4$ , this is a t-test, $ν = n - 1 = 3$ , cutoff $t_{0.025, 3} = 3.18$ . Correct comparison: statistic $3.286 > 3.18$ ⇒ ==reject $H_{0}$ == — but only just, and using the honest, wider cutoff $3.18$ rather than the dishonest $1.96$ . The z cutoff would have rejected on far weaker evidence. Mastery point: the same data can be genuinely significant, yet the reason the colleague is wrong is that they under-stated the uncertainty. Right conclusion for the wrong reason is still a methodological error — with slightly weaker data ( $\overset{x}{ˉ} = 12.5$ , statistic $\approx 2.74$ ) the flawed z-test rejects while the correct t-test does not.

L5·Q2 — Full χ² with estimated proportion

A quality line records defects per shift in $k = 5$ severity classes: $O = (20, 15, 10, 8, 7)$ , total $n = 60$ . A geometric-type model with one estimated parameter predicts $E = (18, 14, 11, 9, 8)$ . Test the fit at $α = 0.05$ .

Recall Solution

Cell surprises $(O - E)^{2} / E$ : $\frac{( 20 - 18 ) ^{2}}{18} = \frac{4}{18} = 0.2222$ ; $\frac{( 15 - 14 ) ^{2}}{14} = \frac{1}{14} = 0.0714$ ; $\frac{( 10 - 11 ) ^{2}}{11} = \frac{1}{11} = 0.0909$ ; $\frac{( 8 - 9 ) ^{2}}{9} = \frac{1}{9} = 0.1111$ ; $\frac{( 7 - 8 ) ^{2}}{8} = \frac{1}{8} = 0.1250$ . Sum: $χ^{2} = 0.2222 + 0.0714 + 0.0909 + 0.1111 + 0.1250 = 0.6206$ . d.f.: $k = 5$ , total-count constraint $- 1$ , one estimated parameter $m = 1$ , so $ν = 5 - 1 - 1 = 3$ . Critical $χ_{0.05, 3}^{2} = 7.81$ . Compare: $0.62 < 7.81$ ⇒ fail to reject. The model fits comfortably. Check the size rule: every $E_{i} \geq 5$ (smallest is $8$ ), so the χ² approximation is valid — no pooling needed.

Recall One-line self-check

Every test here is (signal) ÷ (noise), compared to "what luck does" ::: yes — z/t standardize a mean, χ² sums standardized squared count-errors, F ratios two variance estimates; each has its own reference distribution and its own degrees-of-freedom bookkeeping.