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

False. The CLT concerns the shape of $\bar{X}$ becoming Normal, but the z-test also needs a known $\sigma$ in the denominator. With unknown $\sigma$ and small $n$ the estimated $s$ injects extra noise, so Student's t-distribution is required.

False. A large p-value means the data are consistent with $H_0$ — you failed to find evidence against it. Absence of evidence is not evidence of absence; many other hypotheses are also consistent.

True. As $n$ grows, $s\to \sigma$, so the denominator stops being noisy and the fatter tails of Student's t-distribution shrink until it coincides with the standard Normal Distribution.

False. Each term is $(O_i-E_i{)}^2/E_i$: the numerator is squared and $E_i>0$, so every term is $\ge 0$ and the sum can never be negative. Over- and under-prediction both add positively.

True (by convention). Putting the larger $s^2$ on top forces $F\ge 1$, so we only ever consult the upper tail of the F-distribution, which is what the standard critical-value tables list.

False. $\mathrm{SE}=\sigma /\sqrt{n}$, so it shrinks with $\sqrt{n}$, not $n$. Doubling $n$ multiplies SE by $1/\sqrt{2}\approx 0.71$; you need to quadruple $n$ to halve the SE.

Partly false. Bessel's Correction makes the variance $s^2$ unbiased ($\mathbb{E}[s^2]={\sigma }^2$). Because the square root is a nonlinear function, $s$ itself is still slightly biased — but $s^2$ is the quantity the correction is designed for.

False. Each sample variance is (up to scale) a ${\chi }^2$ divided by its degrees of freedom; an $F$ is by definition the ratio of two independent such quantities. The F-test is literally "two chi-squareds racing."

The raw difference is meaningless until compared to spread. You must divide by the standard error to get "how many SEs away"; $3$ could be enormous or trivial depending on $\sigma /\sqrt{n}$.

The ${\chi }^2$ approximation relies on each cell being roughly Normal, which needs $E_i\gtrsim 5$. Tiny expected counts inflate the statistic and give false rejections; pool small categories first.

Every parameter estimated from the same data costs a degree of freedom: $\nu =k-1-m=6-1-2=3$. Forgetting the $m$ makes your critical value too large and biases you toward failing to reject. See Degrees of Freedom.

The critical value $1.96$ is from the standard Normal Distribution, not the $t$. For $\nu =4$ the two-sided $5\%$ critical value is $2.78$; using $1.96$ ignores the fatter tails and rejects $H_0$ too readily.

Failing to reject can still be a Type II error — missing a real effect because the sample was too small or noisy. See Type I and Type II Errors. Non-rejection is not a guarantee of correctness.

$\alpha$ is the Type I error rate — the chance of rejecting a true $H_0$. Setting $\alpha =0.5$ means you'd wrongly reject half the time. Lowering the miss rate (Type II) is done by increasing $n$ or effect size, not by inflating $\alpha$.

The variance of the mean is ${\sigma }^2/n$, not ${\sigma }^2$. Averaging $n$ independent values shrinks the variance by a factor $n$, giving $\mathrm{SE}=\sigma /\sqrt{n}$. Using ${\sigma }^2$ hugely overstates the noise.

Dividing standardizes the difference into "number of standard errors," a unitless quantity comparable across problems and directly readable off a known reference distribution. See Hypothesis Testing.

Replacing $\sigma$ by the estimate $s$ adds a second source of randomness (the denominator now wobbles too). More ways to land far from centre means more probability in the tails — exactly Student's t-distribution.

For counts, a cell is roughly Poisson with mean $E_i$, so its variance is also $\approx E_i$. Dividing by $E_i$ turns $(O_i-E_i)$ into a standardized $z$-like quantity; squaring and summing gives a Chi-squared Distribution.

Once you know $n-1$ of the deviations, the last one is forced by the sum-to-zero rule. Only $n-1$ deviations are free to vary, so $\nu =n-1$ in a one-sample $t$.

An $F$ is a ratio of two independent variance estimates, each with its own noisiness. Both ${\nu }_1=n_1-1$ and ${\nu }_2=n_2-1$ shape how far the ratio can wander from 1 by luck. See F-distribution.

We need a fixed yardstick for "what luck alone produces." Assuming $H_0$ gives the statistic a known distribution; we then ask how extreme our observed value is against that yardstick to compute the p-value.

The counts are tied by $\sum O_i=\sum E_i=n$: fix the first $k-1$ cells and the last is determined. That single constraint removes one of the $k$ degrees of freedom, giving $\nu =k-1$.

The $t$-test assumes $\bar{X}$ is roughly Normal. With small $n$ the Central Limit Theorem hasn't "kicked in," so heavy skew breaks the reference distribution and the p-value is untrustworthy; consider a larger $n$ or a nonparametric test.

$F=s_1^2/s_2^2=1$, sitting at the centre of the F-distribution — the least surprising value possible. You would never reject $H_0$ there; equal spreads are exactly what $H_0$ predicts.

$Z=(\bar{x}-{\mu }_0)/\mathrm{SE}=0$. Zero standard errors away means perfect agreement with $H_0$ and the largest possible p-value; there is no evidence against the claim.

It converges to the standard Normal Distribution $\mathcal{N}(0,1)$. The fatter tails, caused by the noisy $s$, vanish as $s\to \sigma$ with growing sample size.

That cell contributes $(O_i-E_i{)}^2/E_i=0$ — zero surprise. Perfectly matched cells add nothing to the statistic, which is why only mismatches build up the total.

No — the two-variance F-test compares spreads ($s_1^2/s_2^2$), not centres. Comparing several means uses the F-statistic differently, inside ANOVA; the two-variance test is blind to a shift in mean.

You are on the boundary: the observed statistic sits exactly at the critical value. By the usual "reject if p-value $<\alpha$" rule you would not reject; conventions vary, but this borderline signals the evidence is precisely at the chosen threshold.

Only if the underlying population is itself Normal. With known $\sigma$ the denominator is fixed, but for small $n$ the CLT can't rescue a non-Normal $\bar{X}$; Normality of the raw data is then essential.