UPHESC Assistant Professor Statistics 2025 PYQs

Concept: Bartlett: ${\chi }_0^2=\frac{q}{c}$ with natural logs; with common logs ${\chi }_0^2=2.3026\frac{q}{c}$.

1. $q=(N-a){\log }_{10}{S}_p^2-\sum (n_i-1){\log }_{10}{S}_i^2$.
2. $c=1+\frac{1}{3(a-1)}[\sum \frac{1}{n_i-1}-\frac{1}{N-a}]$.
3. $2.3026=\ln 10$ converts ${\log }_{10}$ to $\ln$; ${\chi }_0^2\sim {\chi }_{a-1}^2$ under $H_0$.

Short trick: 2.3026 = ln 10 – the usual base-change constant.

Concept: LTPD (Lot Tolerance Percent Defective) $p_t$ is the worst lot quality the consumer is willing to tolerate.

1. AQL = quality level that is acceptable (satisfactory) – better than LTPD.
2. The probability of accepting a lot of quality $p_t$ is the consumer's risk $\beta$.
3. AOQL = maximum average outgoing quality after rectification; producer's risk = rejecting a good (AQL) lot.

Short trick: AQL ↔ producer's risk α; LTPD ↔ consumer's risk β.

Concept: $\text{Var}({\hat{\beta }}_j)=\frac{{\sigma }^2}{S_{jj}}\cdot \frac{1}{1-R_j^2}$; the factor $\frac{1}{1-R_j^2}$ is the variance inflation factor.

1. If $X_j$ is unrelated to the other regressors, $R_j^2=0$ and VIF = 1.
2. As $R_j^2\to 1$ (strong collinearity), VIF $\to \mathrm{\infty }$; VIF > 10 is the usual warning sign.
3. The square root (option C) inflates the standard error, not the variance.

Short trick: VIF = 1/tolerance, tolerance = $1-R_j^2$.

Concept: In partial confounding of a $2^3$ design with 4 replications, each interaction (AB, AC, BC, ABC) is confounded in one replicate only.

1. Each partially confounded effect is estimated from the 3 replicates in which it is not confounded.
2. Relative information = (replicates unconfounded)/(total replicates) = $\frac{3}{4}$.
3. Yates (1937) called this ratio the relative information of the confounded effects.

Short trick: Confounded in 1 of 4 replicates → information 3/4.

Concept: Regression coefficients are independent of change of origin but NOT of scale.

1. If $u=\frac{x-a}{h}$, $v=\frac{y-b}{k}$ then $b_{yx}=\frac{k}{h}{b}_{vu}$ – depends on the scales $h,k$. So (A) is false.
2. $r=\pm \sqrt{b_{yx}{b}_{xy}}$ – $r$ is the geometric mean of the two regression coefficients. So (R) is true.
3. (It is $r$, not $b$, that is independent of both origin and scale.)

Short trick: r: origin ✗ scale ✗ (independent of both); b: origin ✗, scale ✓ (depends).

Concept: $D^2=(x-\mu {)}^{\prime }{\mathrm{\Sigma }}^{-1}(x-\mu )$ measures how far a point lies from the centre of a multivariate distribution, allowing for variances and correlations.

1. It generalises the standardised distance $\frac{(x-\mu {)}^2}{{\sigma }^2}$ to many variables.
2. Uses: classification/discriminant analysis, outlier detection, distance between populations (P.C. Mahalanobis, 1936).
3. Testing variances (Bartlett/Box M) and means (Hotelling $T^2$) are other procedures; $T^2$ merely uses a $D^2$-type form.

Short trick: D² = scale-free, correlation-adjusted distance from a distribution.

Concept: Median $M$ satisfies $F(M)=\frac{1}{2}$; for $f(x)=\lambda e^{-\lambda x}$, $F(x)=1-e^{-\lambda x}$.

1. $1-e^{-2M}=\frac{1}{2}$ ⇒ $e^{-2M}=0.5$.
2. $-2M=\ln 0.5$ ⇒ $M=-\frac{1}{2}\ln 0.5=\frac{\ln 2}{2}\approx 0.347$.
3. ½ is the mean, not the median.

Short trick: Exponential median = $\frac{\ln 2}{\lambda }$ < mean $\frac{1}{\lambda }$.

Concept: Each graph has a characteristic use.

1. Box plot – displays the five-number summary and flags outliers (a–IV).
2. Ogive – cumulative frequency curve used to locate the median $Q_2$ graphically (b–I).
3. Histogram – for a continuous quantitative variable (c–II).
4. Bar chart – for a categorical variable (d–III).

Short trick: Box→outliers, Ogive→median, Histogram→numbers, Bar→categories.

Concept: Standard procedure of a significance test.

1. Set up $H_0$ and $H_1$ (I).
2. Fix the level of significance $\alpha$ before seeing the data (IV).
3. Compute the test statistic from the sample (III).
4. Compare with the critical value / p-value and decide (II).

Short trick: Hypothesis → α → statistic → decision.

Concept: Probability that a person aged 20 survives to age 25: ${}_5{p}_{20}=\frac{l_{25}}{l_{20}}$.

1. $\frac{l_{25}}{l_{20}}=\frac{93393}{95510}=0.9778\approx 0.98$.
2. 0.02 is the complementary probability of dying ${}_5{q}_{20}$.
3. Other ratios (e.g. $l_{30}/l_{20}=0.944$) do not correspond to the 20 → 25 interval.

Short trick: p = l(later)/l(earlier); q = 1 − p.

Concept: Translate each relation into a probability statement.

1. Mutually exclusive: $A\cap B=\phi$ ⇒ $A^C\cup B^C=(A\cap B{)}^C=S$ ⇒ probability 1 (a–III).
2. Independent: $P(A\cap B)=P(A)P(B)$ (b–I).
3. $A\supset B$ ⇒ $A\cup B=A$ ⇒ $P(A\cup B)=P(A)$ (c–IV).
4. $A\subset B$ ⇒ $P(A)\le P(B)$ (d–II).

Short trick: De Morgan: $A^C\cup B^C=(A\cap B{)}^C$.

Concept: Fisher's linear discriminant function $Y={\lambda }^{\prime }x$, $\lambda ={\mathrm{\Sigma }}^{-1}({\mu }_1-{\mu }_2)$, gives a score; the observation is allotted to a class by comparing the score with a cut-off.

1. (A) is true – classification of a new observation from its measured features is exactly the purpose of a discriminant function.
2. (R) is false – the function produces a discriminant score (a linear combination of variables), not a probability of class membership; probabilities come from Bayes' rule or logistic models.
3. Hence (A) true, (R) false.

Short trick: Discriminant function → score and cut-off, not probability.

Concept: Base-year weights overstate and current-year weights understate a price rise.

1. Paasche (current quantities) – downward bias (a–I).
2. Laspeyres (base quantities) – upward bias (b–II).
3. Factor reversal: price index × quantity index = value index $V_{01}$ (c–III).
4. Time reversal: $P_{01}\times P_{10}=1$ (d–IV). Fisher's ideal index satisfies both tests.

Short trick: L-Up, P-Down; FRT = P×Q = V; TRT = P₀₁×P₁₀ = 1.

Concept: $V_{WOR}(\overline{y})=\frac{N-n}{N-1}\cdot \frac{{\sigma }^2}{n}$, $V_{WR}(\overline{y})=\frac{{\sigma }^2}{n}$.

1. Condition: $\frac{N-n}{N-1}=\frac{1}{2}$ ⇒ $2N-2n=N-1$.
2. ⇒ $n=\frac{N+1}{2}$ (the options give the sample size in terms of N).
3. E.g. $N=9$: $n=5$, and $\frac{9-5}{8}=\frac{1}{2}$ ✓.

Short trick: Finite population correction $\frac{N-n}{N-1}=\frac{1}{2}$ ⇒ sample about half the population.

Concept: Sampling error falls as $n$ grows ($V(\overline{y})\propto 1/n$), but non-sampling errors (non-response, measurement, processing) grow with the scale of the survey.

1. (A) is false: a larger sample needs more investigators, more data handling and less supervision per unit, so non-sampling errors tend to increase (they are largest in a census).
2. (R) is true: larger $n$ means smaller standard error, i.e. more precise estimates.
3. Hence (A) false, (R) true.

Short trick: Big n: sampling error ↓, non-sampling error ↑.

Concept: $P=\frac{(\genfrac{}{}{0ex}{}{6}{3})(\genfrac{}{}{0ex}{}{9}{2})}{(\genfrac{}{}{0ex}{}{15}{5})}$.

1. $(\genfrac{}{}{0ex}{}{6}{3})=20$, $(\genfrac{}{}{0ex}{}{9}{2})=36$ ⇒ favourable $=720$.
2. $(\genfrac{}{}{0ex}{}{15}{5})=3003$.
3. $P=720/3003=0.2398\approx 0.239$.

Short trick: Choose from each group separately, divide by choosing from all.

Concept: Every student takes at least one class, so $P(A\cup D\cup S)=1$.

1. $P(A\cap D)=0.50+0.65-0.80=0.35$.
2. $1=0.50+0.65+0.55-0.35-0.25-0.25+P(A\cap D\cap S)$.
3. $1=0.85+P(ADS)$ ⇒ $P(ADS)=0.15$.

Short trick: Union of all = 1 when everyone takes at least one option.

Concept: $r_{12.3}=\frac{r_{12}-r_{13}{r}_{23}}{\sqrt{(1-r_{13}^2)(1-r_{23}^2)}}$.

1. Put all $r=\rho$: $r_{12.3}=\frac{\rho -{\rho }^2}{1-{\rho }^2}$.
2. $=\frac{\rho (1-\rho )}{(1-\rho )(1+\rho )}=\frac{\rho }{1+\rho }$.

Short trick: Equal correlations ρ ⇒ every first-order partial = ρ/(1+ρ).

Concept: Compare $(AB)$ with its expectation $\frac{(A)(B)}{N}$, or use Yule's $Q=\frac{(AB)(\alpha \beta )-(A\beta )(\alpha B)}{(AB)(\alpha \beta )+(A\beta )(\alpha B)}$.

1. $N=1216$, $(A)=304$, $(B)=400$ ⇒ expected $(AB)=\frac{304\times 400}{1216}=100$; observed 256 > 100.
2. $Q=\frac{196608-6912}{196608+6912}=\frac{189696}{203520}\approx0.93>0$.
3. So the attributes are (strongly) positively associated – the precise description.

Short trick: $(AB)(\alpha\beta)>(A\beta)(\alpha B)$ ⇒ positive association.

Concept: $\overline{X}$ chart from ranges: UCL $=\overline{\overline{X}}+A_2\overline{R}$.

1. $\overline{\overline{X}}=525/25=21$, $\overline{R}=90/25=3.6$.
2. UCL $=21+0.729\times 3.6=21+2.624=23.62$ (LCL $=18.38$).
3. $d_2$ is not needed here (it gives $\hat{\sigma }=\overline{R}/d_2$).

Short trick: Mean chart limits: $\overline{\overline{X}}\pm A_2\overline{R}$.