BSc CSIT (TU) Science Statistics II (BSc CSIT, STA210) Question Paper 2082 Nepal

Multiple Linear Regression of Current Gain on Two Variables

We fit the model $\hat{y} = b_0 + b_1 x_1 + b_2 x_2$ relating current gain $y$ to diffusion time $x_1$ and sheet resistance $x_2$.

Data and Required Sums ($n=6$)

Totals:

Corrected Sums of Squares and Products

Normal Equations (Solving for Slopes)

Denominator: $S_{11}S_{22}-S_{12}^2 = 3.74\times4338.83-3.70^2 = 16227.22-13.69 = 16213.53$.

Intercept: with $\bar{x_1}=1.3,\;\bar{x_2}=99.167,\;\bar{y}=9.167$,

Fitted Regression Equation

Interpretation: Holding sheet resistance fixed, every additional hour of diffusion time raises current gain by about 0.97 units; holding diffusion time fixed, each extra ohm-cm of sheet resistance raises current gain by about 0.064 units. Both processing variables have a positive effect on the transistor's current gain.

Kruskal-Wallis H-Test for Three Water Conditions

This is a non-parametric test of whether $k=3$ independent samples come from identical populations (equal medians of paint-life).

Step 1 — Hypotheses

Level of significance $\alpha = 0.01$.

Step 2 — Rank all 15 observations jointly (smallest = 1)

Step 3 — Rank sums ($n_i=5$ each, $N=15$)

• Saltwater: $3+4+10+12+15 = R_1 = 44$
• Freshwater no weeds: $1+7+9+11+14 = R_2 = 42$
• Freshwater with weeds: $2+5+6+8+13 = R_3 = 34$

Check: $44+42+34 = 120 = N(N+1)/2 = 15\times16/2$. ✓

Step 4 — Test statistic

Step 5 — Critical value and decision

$H$ follows $\chi^2$ with $k-1 = 2$ d.f. The critical value $\chi^2_{0.01,2} = 9.21$.

Since $H_{cal}=0.56 < 9.21$, we fail to reject $H_0$.

Conclusion

At the 1% level of significance there is no evidence that the lasting quality of the paint differs among saltwater, freshwater without weeds and freshwater with weeds. The paint life may be regarded as the same across the three water conditions.

Part A — Difference between Z-test and t-test

Part B — Test of Dissolved Oxygen Content

Sample: 4.5, 5.1, 4.8, 4.9, 5.1, 4.9, 5.0, 4.7 ($n=8$, small sample, $\sigma$ unknown → use t-test).

Step 1 — Hypotheses (one-tailed, left)

$\alpha = 0.05$.

Step 2 — Sample mean and s.d.

Deviations from $\bar{x}$ and squares:

Step 3 — Test statistic

Step 4 — Critical value. For a left-tailed test, $t_{0.05,\,7}= -1.895$.

Step 5 — Decision. Since $t_{cal}=-1.722 > -1.895$ (it does not fall in the rejection region), we fail to reject $H_0$.

Conclusion

At the 5% level there is insufficient evidence to conclude that the mean dissolved oxygen content is less than 5 ppm. The oxygen level may be regarded as adequate ($\mu = 5$ ppm) to support aquatic life.

Sampling Distribution of the Mean (Without Replacement)

Population: $4, 7, 10, 11$ with $N=4$, sample size $n=2$.

Population mean: $\mu = \dfrac{4+7+10+11}{4}=\dfrac{32}{4}=8$.

(i) All Possible Samples of Size 2 (WOR)

Number of samples $= \binom{4}{2}=6$:

(ii) Mean of the Sampling Distribution of Means

Since $\mu_{\bar{x}}=8=\mu$, the mean of the sampling distribution of means equals the population mean. ✓

(iii) Variance of the Sampling Distribution of Means

Deviations from 8: $-2.5,\,-1.0,\,-0.5,\,0.5,\,1.0,\,2.5$; squares: $6.25,1.0,0.25,0.25,1.0,6.25$; sum $=15.0$.

(Verification via the WOR formula: population variance $\sigma^2=\dfrac{\sum(x-\mu)^2}{N}=\dfrac{16+1+4+9}{4}=7.5$, and $\sigma^2_{\bar{x}}=\dfrac{\sigma^2}{n}\cdot\dfrac{N-n}{N-1}=\dfrac{7.5}{2}\cdot\dfrac{2}{3}=2.5$. ✓)

(iv) Standard Error of the Sample Mean

Probability for the Sample Mean (Central Limit Theorem)

Given: population mean $\mu = 76$ kg, population s.d. $\sigma = 5.2$ kg, sample size $n = 25$. Weight is normally distributed, so $\bar{x}$ is exactly normal.

Step 1 — Standard error of the mean

Step 2 — Standardize for $\bar{x}=78$

Step 3 — Required probability

From the standard normal table, $P(0<Z<1.92)\approx 0.4726$.

Conclusion

There is about a 2.74% chance that the average weight of a random sample of 25 participants exceeds 78 kg.

Part A — Point and Interval Estimation

Point estimation gives a single value (statistic) as the estimate of an unknown population parameter, e.g. the sample proportion $p$ estimates the population proportion $P$.

Interval estimation gives a range (confidence interval) within which the parameter is expected to lie with a stated confidence level $(1-\alpha)$, e.g. $p \pm Z_{\alpha/2}\,\text{S.E.}(p)$. Unlike a point estimate, it indicates the precision/uncertainty of the estimate.

Part B — Estimation for a Proportion

Given: $n=1000$, sample proportion $p=0.58$, so $q=1-p=0.42$.

(i) Standard error of the sample proportion

(ii) 99% Confidence limits

For 99% confidence, $Z_{\alpha/2}=2.58$.

Conclusion

We are 99% confident that the true proportion of all voters favoring the candidate lies between about 54.0% and 62.0%.

Two-Sample Z-Test for Difference of Means (Large Samples)

We test whether the Treatment (oxygen) gives a higher mean Ankle Function Score than the Control (air). Since $n_1=n_2=100$ are large, use the Z-test.

Given:

Step 1 — Hypotheses (one-tailed)

$\alpha=0.05$.

Step 2 — Standard error of the difference

Step 3 — Test statistic

Step 4 — Critical value and decision

For a right-tailed test at 5%, $Z_{0.05}=1.645$.

Since $Z_{cal}=11.0 \gg 1.645$, we reject $H_0$.

Conclusion

At the 5% level of significance there is strong evidence that the treatment (oxygen at 1.2 ATA) produces a significantly higher Ankle Function Score than the control. The treatment is better than the control.

Randomized Block Design (RBD)

A Randomized Block Design is an experimental design in which the experimental units are first grouped into homogeneous blocks (so that units within a block are similar), and then the treatments are allotted at random to the units within each block. Each treatment appears once (or an equal number of times) in every block.

It is a two-way classification: variation is controlled by two factors — treatments and blocks. By isolating the block (e.g. soil-fertility strip, batch, day) variation from the error, RBD increases the precision of treatment comparisons compared with a completely randomized design.

Layout idea (3 treatments, 4 blocks):

          T-randomly assigned within each block
Block 1 : T2  T1  T3
Block 2 : T1  T3  T2
Block 3 : T3  T2  T1
Block 4 : T2  T3  T1

Linear Model

where $\mu$=overall mean, $\tau_i$=$i^{th}$ treatment effect, $\beta_j$=$j^{th}$ block effect, $e_{ij}$=random error.

Assumptions in the Analysis of RBD

1. The model is additive — treatment and block effects add up linearly with no interaction between blocks and treatments.
2. The errors $e_{ij}$ are independent.
3. The errors are normally distributed with mean zero, $e_{ij}\sim N(0,\sigma^2)$.
4. The errors have a common (constant) variance $\sigma^2$ — homoscedasticity.
5. Units within a block are homogeneous, while different blocks may differ.
6. Treatments are assigned randomly within each block.

ANOVA Table for Multiple Regression (2 Independent Variables)

Given: $k=2$ independent variables, total SS $=190$, total d.f. $=20$ (so $n=21$ observations), regression SS $=60$.

(i) Filling in the Blanks

Degrees of freedom:

• Regression d.f. $= k = 2$
• Total d.f. $= n-1 = 20$
• Error d.f. $= 20 - 2 = 18$

Sum of squares:

• Error SS $=$ Total SS $-$ Regression SS $= 190 - 60 = 130$

Mean squares:

• $MSR = \dfrac{SSR}{2}=\dfrac{60}{2}=30$
• $MSE = \dfrac{SSE}{18}=\dfrac{130}{18}=7.222$

F-value:

Completed ANOVA Table

(ii) Test of Overall Significance ($\alpha = 0.05$)

Hypotheses:

Critical value: $F_{0.05,(2,18)}=3.55$.

Decision: Since $F_{cal}=4.154 > 3.55$, we reject $H_0$.

Conclusion: At the 5% level of significance the overall regression model is statistically significant — the two independent variables together explain a significant portion of the variation in $y$.

Chi-Square Test of Independence (Attachment vs Child-Care Time)

We test whether attachment pattern is independent of the amount of time spent in child care. This is a $2\times3$ contingency table, $N=90$.

Observed Frequencies

Step 1 — Hypotheses

$\alpha=0.05$.

Step 2 — Expected frequencies $E=\dfrac{\text{row total}\times\text{col total}}{N}$

Step 3 — Chi-square statistic $\;\chi^2=\sum\dfrac{(O-E)^2}{E}$

Step 4 — Critical value and decision

d.f. $=(r-1)(c-1)=(2-1)(3-1)=2$; $\chi^2_{0.05,2}=5.991$.

Since $\chi^2_{cal}=6.567 > 5.991$, we reject $H_0$.

Conclusion

At the 5% level of significance there is sufficient evidence of a difference in attachment pattern depending on the amount of time spent in child care; attachment pattern and child-care time are not independent.

Stochastic Process

A stochastic process is a collection of random variables $\{X_t : t \in T\}$ indexed by a parameter $t$ (usually time), describing how a random system evolves. At each time point the state of the system is random, and the process specifies the probability law governing transitions between states. A Markov chain is a stochastic process in which the future state depends only on the present state (not the past).

Weather Forecast (Markov Chain)

States: Sunny (S), Rainy (R). Given transition probabilities:

• $P(S\to S)=0.8 \Rightarrow P(S\to R)=0.2$
• $P(R\to S)=0.4 \Rightarrow P(R\to R)=0.6$

Transition matrix (rows = today, columns = tomorrow):

Initial state (Sunday): it rains, so $\pi_0=(S,R)=(0,\;1)$.

Forecast for Monday (one step)

Starting from Rainy on Sunday, use the R-row:

So Monday: 40% sunny, 60% rainy (more likely rainy).

Forecast for Tuesday (two steps)

So Tuesday: 56% sunny, 44% rainy (more likely sunny).

(i) Parametric and Non-Parametric Tests

Parametric tests are statistical tests that assume the data come from a population following a specific distribution (usually normal) and involve population parameters (mean, variance). They require quantitative data measured on interval/ratio scales.

• Examples: Z-test, t-test, F-test, ANOVA.
• Merits: more powerful when assumptions hold; use all information in the data.

Non-parametric (distribution-free) tests make no assumption about the form of the population distribution and are based on ranks/signs/counts. They suit ordinal/nominal data and small samples.

• Examples: Chi-square test, Mann-Whitney U, Kruskal-Wallis, sign test.
• Merits: fewer assumptions, applicable to qualitative data; Demerit: less powerful than parametric tests when normality holds.

(ii) Type I and Type II Errors

In testing a hypothesis, two kinds of wrong decisions are possible:

Type I error (α): Rejecting the null hypothesis $H_0$ when it is actually true.

It is the level of significance; e.g. convicting an innocent person.

Type II error (β): Accepting (failing to reject) $H_0$ when it is actually false.

The quantity $1-\beta$ is the power of the test; e.g. acquitting a guilty person.

Reducing $\alpha$ generally increases $\beta$; both can be lowered by increasing the sample size.