Statistics for Data Science II · End Term · May 2026 · FN

Let S be the sum of ten Bernoulli(p) observations. The estimator is (S+1)/12. At p=0.2, E(S)=2 and Var(S)=10(0.2)(0.8)=1.6. Thus its expectation is 3/12=0.25 and bias is 0.05; its variance is 1.6/144. Mean squared error equals variance plus squared bias: \[1000E[(\hat p-p)^2]=1000\left(\frac{1.6}{144}+0.05^2\right)=13.611111\ldots.\] Round to 13.61. The added constant changes bias but does not add variance. **Answer: 13.61.**

With shape-rate prior Gamma(2,1), the Poisson likelihood contributes a factor lambda to the power of the observed sum and an exponential rate equal to the sample size. The counts sum to 3+6+2+4+2+3=20 across six hours. The posterior is Gamma(2+20,1+6)=Gamma(22,7). Its mean is shape/rate=22/7≈3.142857, rounding to 3.14. Do not use the posterior mode (21/7=3) in place of the mean. **Answer: 3.14.**

Power is the rejection probability when the specified alternative mean 304 is true. The sample mean has standard deviation 8/sqrt(16)=2. Therefore \[P_{304}(\bar X>303.29)=P\left(Z>\frac{303.29-304}{2}\right)=P(Z>-0.355)\approx0.639.\] Using the supplied rounded table for -0.35 gives 1-0.3613=0.6387. Either rounds to 0.64. Standardise under the alternative mean, not the null 300. **Answer: 0.64.**

Use the one-sided variance test under the normal-population assumption. The hypotheses are sigma=2.5 versus sigma>2.5. Under the null, \(T=(n-1)S^2/\sigma_0^2=19(3^2)/(2.5^2)=27.36\) has a chi-square distribution with 19 degrees of freedom. The supplied 95th percentile is 30.14. Since 27.36<30.14, do not reject at 5%. This means insufficient evidence for increased variability, not proof that the standard deviation equals 2.5. **Answer: B.** A: **Incorrect:** The statistic does not exceed the upper critical value. B: **Correct:** The apparent increase is not statistically significant at the stated level.

Each Xi is Poisson(4), so its mean and variance are 4. For ten independent observations, E(Xbar)=4 and Var(Xbar)=4/10=0.4; SD(Xbar)=sqrt(0.4)≈0.632. Independence also makes the sum Poisson(40), not Poisson(4). i.i.d. explicitly means identically distributed and mutually independent. **Answer: A, C, D.** A: **Correct:** States the meaning of i.i.d. B: **Incorrect:** The variance of the average is 0.4, not 4. C: **Correct:** sqrt(0.4) rounds to approximately 0.632. D: **Correct:** Averaging preserves the common expectation 4. E: **Incorrect:** The sum's Poisson rate is 10 times 4=40.

In the frequentist model used by this question, p is a fixed unknown parameter. An estimator is a statistic of the random sample, so it is itself a random variable (a constant statistic is a degenerate example). Estimators need not be unbiased. For Bernoulli data the likelihood p^S(1-p)^(n-S) is maximised at p-hat=S/n=Xbar, including the natural boundary cases S=0 or n. **Answer: A, C, D.** A: **Correct:** Describes the frequentist interpretation of p. B: **Incorrect:** Biased estimators such as (S+1)/(n+2) are valid counterexamples. C: **Correct:** A statistic of random observations has a sampling distribution. D: **Correct:** The Bernoulli maximum-likelihood estimate is the sample proportion.

The first 40 seeds have 0.65×40=26 successes, giving p-hat=0.65. The next batch adds 8 successes out of 10, so the combined estimate is (26+8)/(40+10)=34/50=0.68. Therefore 100(p-hat-new-p-hat)=100(0.68-0.65)=3. Combine the counts rather than giving the two unequal batches equal weight. **Answer: 3.**

The observed ordered sequence H,H,T,H,T contains three heads and two tails. Its conditional likelihood is p³(1-p)²; no binomial coefficient is needed because S is this specific sequence, not the event of any three heads. Total probability is \[P(S)=0.3(0.4^3\,0.6^2)+0.7(0.6^3\,0.4^2)=0.006912+0.024192=0.031104.\] Round to three decimals: 0.031. **Answer: 0.031.**

Multiply each candidate p's likelihood by its prior mass. For p=0.4 the unnormalised posterior weight is 0.3×0.4³×0.6²=0.006912. For p=0.6 it is 0.7×0.6³×0.4²=0.024192. The larger weight is at p=0.6, so that is the posterior mode (MAP estimate). A common normalising denominator does not change which candidate is largest. **Answer: 0.6.**

The density is x on (0,1) and 2-x on [1,2). To obtain F(1.5), integrate both pieces: \[F(1.5)=\int_0^1x\,dx+\int_1^{1.5}(2-x)\,dx=0.5+0.375=0.875.\] Equivalently subtract the right triangular tail of base and height 0.5 from 1, giving 1-0.125. **Answer: 0.875.**

Split the interval at the density's breakpoint 1: \[P(0.5<X<1.5)=\int_{0.5}^1x\,dx+\int_1^{1.5}(2-x)\,dx=0.375+0.375=0.75.\] The distribution is continuous, so endpoints have zero probability. Both tails outside this symmetric interval each have area 0.125. **Answer: 0.75.**

A joint PMF must sum to 1. Here \(\sum_{i=1}^3\sum_{j=1}^3(i+j)=3(1+2+3)+3(1+2+3)=36\). Consequently 36/c=1 and c=36. The nine cells have numerators 2,3,4,3,4,5,4,5,6, providing the same check. **Answer: 36.**

The event 1<X<=3 and 1<Y<=3 includes four cells: (2,2),(2,3),(3,2),(3,3). Their numerators i+j are 4,5,5,6. With normalising constant 36, probability=(4+5+5+6)/36=20/36=5/9. **Answer: B.** A: **Incorrect:** Does not equal the sum of the four required cell probabilities. B: **Correct:** 5/9 is the summed event probability. C: **Incorrect:** 7/14 equals 1/2 rather than the required 20/36. D: **Incorrect:** 2/3 overstates the sum of the four event cells.

The two normal samples are X=(96,98,100,102,104) and Y=(49,50,50,50,51), each of size 5. Their sample means are 100 and 50. Unbiased sample variances divide squared deviations by n-1=4. Under equal population variances, T=Sx²/Sy² follows F(4,4). The two-sided test at alpha=0.05 splits its error probability equally between two tails. For X, squared deviations sum to 16+4+0+4+16=40. Thus Sx²=40/4=10, option 10. Enter the option label 10; it happens to equal the numerical variance. **Answer: 10.**

The two normal samples are X=(96,98,100,102,104) and Y=(49,50,50,50,51), each of size 5. Their sample means are 100 and 50. Unbiased sample variances divide squared deviations by n-1=4. Under equal population variances, T=Sx²/Sy² follows F(4,4). The two-sided test at alpha=0.05 splits its error probability equally between two tails. For Y, squared deviations sum to 1+0+0+0+1=2. Thus Sy²=2/4=0.5, labelled option 4. Enter 4 rather than the variance 0.5. **Answer: 4.**

The two normal samples are X=(96,98,100,102,104) and Y=(49,50,50,50,51), each of size 5. Their sample means are 100 and 50. Unbiased sample variances divide squared deviations by n-1=4. Under equal population variances, T=Sx²/Sy² follows F(4,4). The two-sided test at alpha=0.05 splits its error probability equally between two tails. The upper rejection threshold is 1+cR, labelled option 3. Rejecting above this threshold assigns alpha/2 to the upper tail. This is a threshold for the variance ratio, not for either sample variance alone. **Answer: 3.**

The two normal samples are X=(96,98,100,102,104) and Y=(49,50,50,50,51), each of size 5. Their sample means are 100 and 50. Unbiased sample variances divide squared deviations by n-1=4. Under equal population variances, T=Sx²/Sy² follows F(4,4). The two-sided test at alpha=0.05 splits its error probability equally between two tails. The lower rejection threshold is 1-cL, labelled option 11. It captures unusually small Sx²/Sy² ratios in the other tail. Enter the label 11. **Answer: 11.**

The two normal samples are X=(96,98,100,102,104) and Y=(49,50,50,50,51), each of size 5. Their sample means are 100 and 50. Unbiased sample variances divide squared deviations by n-1=4. Under equal population variances, T=Sx²/Sy² follows F(4,4). The two-sided test at alpha=0.05 splits its error probability equally between two tails. P(T>1+cR)=alpha/2=0.025. Therefore the CDF at that threshold is 1-0.025=0.975, labelled option 6. Enter 6. **Answer: 6.**

The two normal samples are X=(96,98,100,102,104) and Y=(49,50,50,50,51), each of size 5. Their sample means are 100 and 50. Unbiased sample variances divide squared deviations by n-1=4. Under equal population variances, T=Sx²/Sy² follows F(4,4). The two-sided test at alpha=0.05 splits its error probability equally between two tails. The supplied F(4,4) 97.5th percentile is 9.605. Thus 1+cR=9.605, labelled option 13. The two-sided test needs the 97.5th percentile, not the 95th percentile. **Answer: 13.**

The two normal samples are X=(96,98,100,102,104) and Y=(49,50,50,50,51), each of size 5. Their sample means are 100 and 50. Unbiased sample variances divide squared deviations by n-1=4. Under equal population variances, T=Sx²/Sy² follows F(4,4). The two-sided test at alpha=0.05 splits its error probability equally between two tails. The observed statistic is 10/0.5=20. This is greater than 9.605, so the missing comparison is >, labelled option 12. Enter 12. **Answer: 12.**

The two normal samples are X=(96,98,100,102,104) and Y=(49,50,50,50,51), each of size 5. Their sample means are 100 and 50. Unbiased sample variances divide squared deviations by n-1=4. Under equal population variances, T=Sx²/Sy² follows F(4,4). The two-sided test at alpha=0.05 splits its error probability equally between two tails. The observed ratio 20 exceeds the upper critical threshold 9.605, so reject the equal-variance null hypothesis at 5%. Reject is labelled option 8. The evidence concerns population variance, not equality of sample means. **Answer: 8.**