Statistics II

Hypotheses. $H_0$: type of e-mail account and hacking status are independent. $H_1$: they are associated.

Observed table with totals:

Expected frequencies $E=\dfrac{\text{row total}\times\text{col total}}{N}$:

• Yahoo, Yes: $75\times80/215=27.91$
• Yahoo, No: $75\times135/215=47.09$
• Gmail, Yes: $140\times80/215=52.09$
• Gmail, No: $140\times135/215=87.91$

Test statistic:

$\chi^2=\sum\frac{(O-E)^2}{E}=\frac{(32.09)^2}{27.91}+\frac{(-32.09)^2}{47.09}+\frac{(-32.09)^2}{52.09}+\frac{(32.09)^2}{87.91}$

$=36.90+21.87+19.77+11.71\approx 90.3$

Critical value. $df=(2-1)(2-1)=1$; at $\alpha=0.01$, $\chi^2_{0.01,1}=6.635$.

Decision. $90.3>6.635$, so reject $H_0$.

Conclusion. At the 1% level there is sufficient evidence that the type of e-mail account and hacking status are associated. Yahoo accounts show a much higher hacked proportion ($60/75=80\%$) than Gmail ($20/140\approx14\%$).

Let $p$ be the proportion of people in Kathmandu having a Samsung mobile.

$H_0: p = 0.5$ (equally popular) $H_1: p \ne 0.5$ (two-tailed)

Sample proportion: $\hat p = 300/500 = 0.6$, with $n=500$.

$z=\frac{\hat p-p_0}{\sqrt{p_0q_0/n}}=\frac{0.6-0.5}{\sqrt{0.5(0.5)/500}}=\frac{0.1}{0.02236}=4.47$

At the 5% level the critical values are $\pm1.96$.

Since $4.47 > 1.96$, reject $H_0$. The claim that Samsung and Redmi are equally popular is not supported; Samsung is significantly more popular in the sample.

Type I and type II errors

• Type I error: rejecting the null hypothesis $H_0$ when it is actually true. Its probability is $\alpha$, the level of significance.
• Type II error: accepting (failing to reject) $H_0$ when it is actually false. Its probability is $\beta$; the power of the test is $1-\beta$.

Test of the claim

$H_0: p=0.5$ (Samsung and Huawei equally popular) $H_1: p\ne0.5$

$n=600$, $\hat p=\dfrac{350}{600}=0.5833$.

$z=\frac{\hat p-p_0}{\sqrt{p_0q_0/n}}=\frac{0.5833-0.5}{\sqrt{0.25/600}}=\frac{0.0833}{0.02041}=4.08$

Critical values at 5%: $\pm1.96$.

Since $4.08>1.96$, reject $H_0$. The claim is not supported; Samsung is significantly more popular than Huawei in Kathmandu.

Multiple linear regression (MLR). MLR is a statistical technique that models the linear relationship between one dependent (response) variable $Y$ and two or more independent (explanatory) variables $X_1, X_2, \dots, X_k$:

$Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \dots + \beta_k X_k + \varepsilon,$

where $\beta_0$ is the intercept, $\beta_j$ the partial regression coefficients and $\varepsilon$ the random error. The coefficients are estimated by least squares.

Fitting $\hat Y = b_0 + b_1X_1 + b_2X_2$ with $n = 10$.

Means: $\bar X_1 = 27.2$, $\bar X_2 = 44.1$, $\bar Y = 14.7$.

Corrected sums of squares and products:

$S_{11} = 7428 - \frac{272^2}{10} = 29.6, \qquad S_{22} = 19461 - \frac{441^2}{10} = 12.9,$ $S_{12} = 12005 - \frac{272\times441}{10} = 9.8,$ $S_{1y} = 4013 - \frac{272\times147}{10} = 14.6, \qquad S_{2y} = 6485 - \frac{441\times147}{10} = 2.3 .$

Normal equations:

$29.6\,b_1 + 9.8\,b_2 = 14.6, \qquad 9.8\,b_1 + 12.9\,b_2 = 2.3 .$

Determinant $= 29.6\times12.9 - 9.8^2 = 381.84 - 96.04 = 285.8$.

$$b_1 = \frac{14.6\times12.9 - 9.8\times2.3}{285.8} = \frac{165.8}{285.8} = 0.580, \qquad b_2 = \frac{29.6\times2.3 - 9.8\times14.6}{285.8} = \frac{-75}{285.8} = -0.262 .$$

$b_0 = \bar Y - b_1\bar X_1 - b_2\bar X_2 = 14.7 - 0.580(27.2) + 0.262(44.1) = 10.49 .$

Fitted regression equation:

$\hat Y = 10.49 + 0.580\,X_1 - 0.262\,X_2 .$

(Check: $SSR = b_1S_{1y} + b_2S_{2y} = 7.87$, $SST = 2173 - 147^2/10 = 12.1$, so $R^2 = 0.65$.)

Interpretation of the coefficients

• $b_0 = 10.49$: the expected value of $Y$ when $X_1 = X_2 = 0$ (a baseline value).
• $b_1 = 0.580$: holding $X_2$ constant, a one-unit increase in $X_1$ increases $Y$ by 0.580 units on average (positive effect).
• $b_2 = -0.262$: holding $X_1$ constant, a one-unit increase in $X_2$ decreases $Y$ by 0.262 units on average (negative effect).
• About 65% of the variation in $Y$ is explained jointly by $X_1$ and $X_2$.

Latin Square Design (LSD). In a Latin square design, $t$ treatments are arranged in a $t\times t$ square so that each treatment appears exactly once in every row and exactly once in every column. It therefore controls two sources of variation (rows and columns) at the same time, in addition to the treatment effect, and needs $t^2$ experimental units. The model is

$y_{ijk} = \mu + \alpha_i + \beta_j + \tau_k + \varepsilon_{ijk} .$

Merits

• Eliminates two sources of extraneous variation (row and column), so it is more precise than CRD and RBD when both are present.
• Needs fewer units than a full factorial arrangement; the analysis is simple.
• Missing values can be estimated, and the design is easily extended.

Demerits

• The number of rows = columns = treatments, so it is unsuitable for many treatments (more than about 12) and impractical for very few (fewer than 4, because the error degrees of freedom are too small).
• Needs the assumption of no interaction among rows, columns and treatments.
• Randomisation is restricted, and a missing observation complicates the analysis.

Analysis of the given data ($t = 3$, $N = 9$). Layout (treatment and yield):

Treatment totals: A = 10 + 15 + 20 = 45, B = 15 + 25 + 15 = 55, C = 20 + 10 + 25 = 55.

Hypotheses: $H_0$: the treatment means (A, B, C) are equal, against $H_1$: at least one differs. (Row and column effects are tested similarly.)

$CF = \frac{155^2}{9} = 2669.44$

$\sum y^2 = 100+225+400+625+100+225+625+400+225 = 2925$, so

$TSS = 2925 - 2669.44 = 255.56$ $SSR = \frac{45^2+50^2+60^2}{3} - CF = \frac{8125}{3} - 2669.44 = 38.89$ $SSC = \frac{60^2+45^2+50^2}{3} - CF = \frac{8125}{3} - 2669.44 = 38.89$ $SST_r = \frac{45^2+55^2+55^2}{3} - CF = \frac{8075}{3} - 2669.44 = 22.22$ $SSE = 255.56 - 38.89 - 38.89 - 22.22 = 155.56$

ANOVA table

Tabulated $F_{0.05}(2,2) = 19.00$. All calculated F values (0.25, 0.25, 0.14) are less than 19.00, so $H_0$ is accepted: there is no significant difference among treatments, and the row and column effects are also not significant at the 5% level.

Hypothesis. A hypothesis is a statement or assumption about a population parameter (or distribution) that is tested on the basis of sample information.

• Null hypothesis $H_0$: the statement of "no difference" or "no effect" (for example $\mu_1 = \mu_2$) that is assumed true and tested; it always contains the equality sign. We either reject it or fail to reject it.
• Alternative hypothesis $H_1$: the statement accepted when $H_0$ is rejected (for example $\mu_1 > \mu_2$, $\mu_1 < \mu_2$ or $\mu_1 \ne \mu_2$). It decides whether the test is one-tailed or two-tailed.

Test of the claim. Company: $n_1 = 40$, $\bar x_1 = 628$, $s_1 = 27$. Competitor: $n_2 = 30$, $\bar x_2 = 619$, $s_2 = 25$. Both samples are large, so a Z test is used.

1. $H_0: \mu_1 = \mu_2$ against $H_1: \mu_1 > \mu_2$ (the company's bulbs are superior); one-tailed test.
2. Level of significance $\alpha = 0.05$; critical value $Z_{0.05} = 1.645$.
3. Test statistic:

$Z = \frac{\bar x_1 - \bar x_2}{\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}} = \frac{628 - 619}{\sqrt{\dfrac{729}{40} + \dfrac{625}{30}}} = \frac{9}{\sqrt{18.225 + 20.833}} = \frac{9}{6.25} = 1.44$

4. Decision: since $Z = 1.44 < 1.645$, $H_0$ is not rejected at the 5% level.

Conclusion: the sample does not provide sufficient evidence that the company's bulbs have a longer average life than the competitor's. The company's claim is not justified at the 5% level of significance.

Given $n = 7$, $k = 2$ independent variables, $SST = 1493$ and $SSE = 91$, so

$SSR = SST - SSE = 1493 - 91 = 1402 .$

i) Coefficient of determination

$R^2 = \frac{SSR}{SST} = \frac{1402}{1493} = 0.939 .$

Interpretation: about 93.9% of the total variation in the dependent variable is explained by the regression on $X_1$ and $X_2$ together; only 6.1% is unexplained (due to error). The fit is very good.

ii) Overall significance (F test)

$H_0: \beta_1 = \beta_2 = 0$ against $H_1$: at least one $\beta_j \neq 0$; $\alpha = 0.05$.

$F = \frac{SSR/k}{SSE/(n-k-1)} = \frac{1402/2}{91/4} = \frac{701}{22.75} = 30.81 .$

Degrees of freedom $(2, 4)$; tabulated $F_{0.05}(2,4) = 6.94$.

Since $30.81 > 6.94$, $H_0$ is rejected. The regression model is statistically significant overall: at least one of $X_1$ and $X_2$ has a significant linear effect on the dependent variable.

Chi-square test of independence.

$H_0$: type of feeding and nature of teeth are independent. $H_1$: they are dependent. $\alpha = 0.05$.

Observed table with totals:

Expected frequencies $E = \dfrac{\text{row total}\times\text{column total}}{N}$:

$E_{11} = \dfrac{30\times20}{45} = 13.33$, $E_{12} = \dfrac{30\times25}{45} = 16.67$, $E_{21} = \dfrac{15\times20}{45} = 6.67$, $E_{22} = \dfrac{15\times25}{45} = 8.33$.

$\chi^2 = \sum\frac{(O-E)^2}{E} = \frac{(4.67)^2}{13.33} + \frac{(4.67)^2}{16.67} + \frac{(4.67)^2}{6.67} + \frac{(4.67)^2}{8.33} = 1.63 + 1.31 + 3.27 + 2.61 = 8.82$

Degrees of freedom $= (2-1)(2-1) = 1$; tabulated $\chi^2_{0.05}(1) = 3.841$.

Since $8.82 > 3.841$, $H_0$ is rejected. There is sufficient evidence at the 5% level that the type of feeding and the nature of teeth are dependent: bottle-fed children show a much higher share of defective teeth (13 of 15) than breast-fed children (12 of 30).

The sample estimate must lie within 10% of the true value, i.e. the allowable relative error is $r = 0.10$, with 95% confidence ($Z = 1.96$) and coefficient of variation $CV = \sigma/\mu = 60\% = 0.60$.

The condition is $Z\dfrac{\sigma}{\sqrt n} \le r\mu$, so

$n = \left(\frac{Z\cdot CV}{r}\right)^2 = \left(\frac{1.96\times0.60}{0.10}\right)^2 = (11.76)^2 = 138.3 .$

Rounding up, the minimum sample size required is $n = 139$.

Given: $\sigma = 0.01$ inch (known), $n = 100$, sample mean $\bar x = 10.68$ inches. The standard error is

$\sigma_{\bar x} = \frac{\sigma}{\sqrt n} = \frac{0.01}{10} = 0.001 .$

The confidence interval for the population mean is $\bar x \pm Z_{\alpha/2}\,\dfrac{\sigma}{\sqrt n}$.

95% interval: $Z_{0.025} = 1.96$

$10.68 \pm 1.96(0.001) = 10.68 \pm 0.00196 = (10.67804,\; 10.68196) \text{ inches.}$

90% interval: $Z_{0.05} = 1.645$

$10.68 \pm 1.645(0.001) = 10.68 \pm 0.001645 = (10.67836,\; 10.68165) \text{ inches.}$

The 90% interval is narrower than the 95% interval (lower confidence gives a shorter interval). Neither interval contains the expected length of 11 inches, which suggests that the process mean has shifted from the specification.

Kruskal-Wallis H test. $H_0$: the three catalysts give the same distribution of alcohol yield (no significant difference). $H_1$: at least one differs. $\alpha = 0.05$.

Rank all $N = 11$ observations together (ties get the average rank): 250 (1), 270, 270 (2.5 each), 290 (4), 350 (5), 380, 380 (6.5 each), 400 (8), 410 (9), 430 (10), 450 (11).

Check: $25.5 + 15 + 25.5 = 66 = 11\times12/2$.

$H = \frac{12}{N(N+1)}\sum\frac{R_i^2}{n_i} - 3(N+1) = \frac{12}{11\times12}\left(\frac{25.5^2}{3} + \frac{15^2}{5} + \frac{25.5^2}{3}\right) - 36$ $= \frac{1}{11}(216.75 + 45 + 216.75) - 36 = 43.5 - 36 = 7.50 .$

With the correction for ties (two tied pairs): $1 - \dfrac{6+6}{11^3-11} = 0.9909$, so $H = 7.50/0.9909 = 7.57$.

Degrees of freedom $= 3 - 1 = 2$; $\chi^2_{0.05}(2) = 5.991$.

Since $7.5\ (\text{or } 7.57) > 5.991$, $H_0$ is rejected. There is a significant difference among the three catalysts in the amount of alcohol distilled at the 5% level.

Completing the table. Use $MS = SS/df$ and $F = MS/MSE$.

• Error: $MSE = 12$, $df = 6$, so $SSE = 12\times6 = 72$.
• Column: $F = 2 = MS_{col}/12$, so $MS_{col} = 24$; $df = 72/24 = 3$.
• Rows: $MS = 36$ and $F = 36/12 = 3$. The design has the same number of rows, columns and treatments, so $df = 3$ and $SS_{rows} = 36\times3 = 108$.
• Treatments: $df = 3$, $MS = 180/3 = 60$, $F = 60/12 = 5$.
• Total: $SS = 72 + 108 + 180 + 72 = 432$; $df = 3+3+3+6 = 15$.

Answers

• The design employed is a Latin Square Design: two blocking factors (rows and columns), the treatments, and error d.f. $= (t-1)(t-2) = 3\times2 = 6$, with $t^2 = 16$ observations ($15$ total d.f.).
• Four treatments were compared (treatment d.f. $= 3$, so $t = 4$).

(For information: $F_{0.05}(3,6) = 4.76$, so the treatment effect, $F = 5$, is significant at 5%, whereas the row and column effects are not.)

A queueing system is one in which customers (jobs, calls, vehicles, patients) arrive, wait if necessary for a service facility, are served and then leave. Its main components are:

1. Input source (calling population). The population from which customers come. It may be finite or infinite; customers may arrive singly or in batches.
2. Arrival process. Describes how customers arrive, usually through the distribution of inter-arrival times. Most models assume Poisson arrivals with mean rate $\lambda$ (equivalently, exponential inter-arrival times). Customers' behaviour may include balking, reneging or jockeying.
3. Queue (waiting line). The line in which customers wait for service. Its capacity may be finite or infinite.
4. Queue discipline. The rule by which the next customer is chosen for service: FIFO (first in, first out), LIFO, service in random order, or priority-based.
5. Service mechanism. The number of servers (single or multiple, in series or in parallel) and the service-time distribution, usually exponential with mean service rate $\mu$ per server.
6. Departure (exit). After service, customers leave the system (or may rejoin it).

Together these are summarised in Kendall's notation $(a/b/c):(d/e)$, for example M/M/1.

Arrival rate $\lambda = 4$ jobs per minute.

a) Frame size. In a binomial (Bernoulli) arrival process, a job arrives in a frame of length $\Delta$ with probability $p = \lambda\Delta$. For $p = 0.1$:

$\Delta = \frac{p}{\lambda} = \frac{0.1}{4} = 0.025 \text{ minute (1.5 seconds)} .$

b) More than 4 jobs in one minute. One minute has $n = 1/0.025 = 40$ frames, so the number of jobs $X \sim \text{Binomial}(n = 40,\ p = 0.1)$, with mean $np = 4$.

$P(X = k) = \binom{40}{k}(0.1)^k(0.9)^{40-k}$

$P(0) = 0.01478$, $P(1) = 0.06569$, $P(2) = 0.14233$, $P(3) = 0.20032$, $P(4) = 0.20589$.

$P(X \le 4) = 0.62902, \qquad P(X > 4) = 1 - 0.62902 = 0.371 .$

The probability of more than 4 jobs in one minute is about 0.371.

c) Mean and variance of inter-arrival time. The number of frames until the next arrival is geometric with $p = 0.1$:

$E(N) = \frac1p = 10 \text{ frames}, \qquad Var(N) = \frac{1-p}{p^2} = \frac{0.9}{0.01} = 90 \text{ frames}^2 .$

Converting with the frame length 0.025 minute:

$\text{Mean} = 10\times0.025 = 0.25 \text{ minute (15 seconds)}$ $\text{Variance} = 90\times(0.025)^2 = 0.05625 \text{ minute}^2 .$

(The mean agrees with $1/\lambda = 0.25$ minute; the variance is close to the exponential value $1/\lambda^2 = 0.0625$ of the continuous Poisson model.)

Need of non-parametric statistical methods

Parametric tests (t, Z, F) assume that the data come from a specific distribution, usually normal, with parameters such as the mean and variance, and that the measurements are on an interval or ratio scale. Non-parametric (distribution-free) methods are needed because:

• The population distribution is unknown or clearly non-normal (skewed, heavy-tailed), or the sample is too small to rely on the central limit theorem.
• The data are ordinal (ranks, ratings, grades) or nominal (counts, categories), where means and variances are meaningless.
• The data contain outliers, which affect parametric tests strongly but ranks only slightly.
• They are simple, quick and need few assumptions, so they remain valid under many conditions (examples: sign test, Wilcoxon, Mann-Whitney U, Kruskal-Wallis, chi-square, run test).

Their drawbacks are lower power than parametric tests when the parametric assumptions do hold, and loss of information when actual values are replaced by ranks.

Efficiency of Randomized Block Design (RBD) relative to Completely Randomized Design (CRD)

RBD removes the variation between blocks from the error, so it is more precise than CRD whenever the units are heterogeneous. The relative efficiency of RBD over CRD, with $b$ blocks and $t$ treatments, is

$RE = \frac{\hat\sigma^2_{CRD}}{\hat\sigma^2_{RBD}} = \frac{(b-1)\,MS_B + b(t-1)\,MS_E}{(bt-1)\,MS_E},$

where $MS_B$ is the block mean square and $MS_E$ the error mean square of the RBD analysis. If $RE > 1$, RBD is more efficient than CRD: $RE = 1.5$ means CRD would need about 50% more replications to give the same precision. $RE$ close to 1 means blocking was not useful (blocks were nearly homogeneous), and the loss of error degrees of freedom was wasted.

Concept. The sampling distribution of the mean is the probability distribution of the sample means $\bar x$ over all possible samples of the same size $n$ drawn from a population. It shows how the sample mean varies from sample to sample and is the basis of estimation and testing.

Population: $X = 20, 21, 22, 23$, $N = 4$, each value with probability $1/4$.

1. Population mean, variance and distribution

$\mu = \frac{20+21+22+23}{4} = 21.5, \qquad \sigma^2 = \frac{(-1.5)^2 + (-0.5)^2 + 0.5^2 + 1.5^2}{4} = \frac{5}{4} = 1.25, \quad \sigma = 1.118 .$

The population distribution is uniform: $P(X = x) = 0.25$ for each of 20, 21, 22, 23.

2. Samples of size $n = 2$ with replacement ($N^n = 16$ samples). Sample variance is $s^2 = \sum(x-\bar x)^2/(n-1) = (a-b)^2/2$ for a pair $(a,b)$.

Sampling distribution of $\bar x$:

Mean of sample means:

$E(\bar x) = \frac{20(1) + 20.5(2) + 21(3) + 21.5(4) + 22(3) + 22.5(2) + 23(1)}{16} = \frac{344}{16} = 21.5 .$

Variance of sample means:

$Var(\bar x) = \frac{1(1.5)^2 + 2(1)^2 + 3(0.5)^2 + 0 + 3(0.5)^2 + 2(1)^2 + 1(1.5)^2}{16} = \frac{10}{16} = 0.625 .$

Distribution of $s^2$: value 0 (4 samples), 0.5 (6), 2 (4), 4.5 (2). Mean of sample variances:

$E(s^2) = \frac{0(4) + 0.5(6) + 2(4) + 4.5(2)}{16} = \frac{20}{16} = 1.25 .$

(If the divisor $n$ were used, the mean would be 0.625, which underestimates $\sigma^2$.)

3. Comparison

4. Standard error of the mean

$SE(\bar x) = \sqrt{Var(\bar x)} = \sqrt{0.625} = 0.791 = \frac{\sigma}{\sqrt n} = \frac{1.118}{\sqrt2} .$

5. Final comments

• The mean of the sampling distribution equals the population mean, so $\bar x$ is an unbiased estimator of $\mu$; likewise $s^2$ (with divisor $n-1$) is unbiased for $\sigma^2$.
• The variance of $\bar x$ is $\sigma^2/n$, half the population variance here, so sample means are less spread out than individual values; a larger $n$ gives a smaller standard error and a more precise estimate.
• Although the population is uniform, the distribution of $\bar x$ is already symmetric and peaked at $\mu$, illustrating the central limit theorem: it tends to normality as $n$ increases.

Dependent variable: the lifetime of the computer ($Y$, years), because it is the outcome that is affected by the other two. The independent variables are play time per day ($X_1$, hours) and RAM ($X_2$, MB).

Model: $\hat Y = b_0 + b_1X_1 + b_2X_2$, with $n = 7$.

Sums: $\sum Y = 28$, $\sum X_1 = 27$, $\sum X_2 = 32$, $\sum X_1^2 = 143$, $\sum X_2^2 = 182$, $\sum X_1X_2 = 91$, $\sum X_1Y = 77$, $\sum X_2Y = 154$, $\sum Y^2 = 140$.

Means: $\bar Y = 4$, $\bar X_1 = 3.857$, $\bar X_2 = 4.571$.

Corrected sums:

$S_{11} = 143 - \frac{27^2}{7} = 38.857, \quad S_{22} = 182 - \frac{32^2}{7} = 35.714, \quad S_{12} = 91 - \frac{27\times32}{7} = -32.429,$ $S_{1y} = 77 - \frac{27\times28}{7} = -31, \quad S_{2y} = 154 - \frac{32\times28}{7} = 26, \quad S_{yy} = 140 - \frac{28^2}{7} = 28 .$

Normal equations:

$38.857\,b_1 - 32.429\,b_2 = -31, \qquad -32.429\,b_1 + 35.714\,b_2 = 26 .$

Determinant $= 38.857\times35.714 - 32.429^2 = 1387.76 - 1051.67 = 336.1$.

$$b_1 = \frac{(-31)(35.714) - (-32.429)(26)}{336.1} = \frac{-264.0}{336.1} = -0.785, \qquad b_2 = \frac{(38.857)(26) - (-32.429)(-31)}{336.1} = \frac{5.0}{336.1} = 0.015 .$$

$b_0 = 4 - (-0.785)(3.857) - (0.015)(4.571) = 6.96 .$

Fitted model:

$\hat Y = 6.96 - 0.785\,X_1 + 0.015\,X_2 .$

Goodness of fit. $SSR = b_1S_{1y} + b_2S_{2y} = 24.33 + 0.39 = 24.73$, $SST = 28$, $SSE = 3.27$.

$R^2 = \frac{24.73}{28} = 0.883, \qquad \bar R^2 = 1 - (1-0.883)\frac{6}{4} = 0.825 .$

$F = \frac{24.73/2}{3.27/4} = 15.14 > F_{0.05}(2,4) = 6.94 ,$

so the model as a whole is significant.

Problem-specific interpretation

• Intercept 6.96: a computer with no daily game play and zero RAM would be expected to last about 7 years (a baseline with no practical meaning for RAM = 0).
• $b_1 = -0.785$: holding RAM constant, each additional hour per day that children spend playing games reduces the expected computer lifetime by about 0.79 year (about 9 months). This supports the report that rough use through gaming shortens a computer's life. Its individual t value is $-2.67$ (standard error 0.295) against the critical $t_{0.025,4} = 2.776$, so it is borderline and not significant at 5% with such a small sample.
• $b_2 = 0.015$: holding play time constant, each extra MB of RAM adds only 0.015 year, a negligible effect ($t = 0.05$, not significant). In these data RAM adds nothing once play time is accounted for, partly because RAM and play time are strongly correlated ($r = -0.87$; computers with more RAM were used for less play time), so the separate effects cannot be distinguished (multicollinearity).
• About 88% of the variation in lifetime is explained jointly by play time and RAM; the main driver is play time.

Fundamental concepts of Latin Square Design (LSD). LSD is an experimental design for $t$ treatments in which the experimental units are classified in two directions, rows and columns, each of $t$ levels, and the treatments are allotted to the $t\times t$ cells so that every treatment occurs once and only once in each row and each column. Rows and columns act as two blocking factors, so the design removes two sources of variation (row and column effects) from the error. The model is

$y_{ijk} = \mu + \alpha_i + \beta_j + \tau_k + \varepsilon_{ijk}, \quad \varepsilon_{ijk}\sim N(0,\sigma^2).$

Necessary conditions

• Number of rows = number of columns = number of treatments $= t$, and total units $= t^2$.
• Each treatment appears exactly once in every row and every column; treatments are randomly allocated subject to this.
• The three factors (rows, columns, treatments) are additive: no interaction among them.
• Errors are independent, normally distributed with constant variance; $t$ is not too small or large (usually 4 to 12).

Analysis of variance ($t = 3$, $N = 9$):

Treatment totals: A = 5 + 15 + 10 = 30, B = 10 + 10 + 20 = 40, C = 15 + 20 + 5 = 40.

$H_0$: the treatment means are equal (and similarly for rows and columns); $\alpha = 0.05$.

$CF = \frac{110^2}{9} = 1344.44, \qquad \sum y^2 = 1600, \qquad TSS = 1600 - 1344.44 = 255.56$ $SSR = \frac{30^2+45^2+35^2}{3} - CF = \frac{4150}{3} - 1344.44 = 38.89$ $SSC = \frac{45^2+30^2+35^2}{3} - CF = \frac{4150}{3} - 1344.44 = 38.89$ $SS_{Tr} = \frac{30^2+40^2+40^2}{3} - CF = \frac{4100}{3} - 1344.44 = 22.22$ $SSE = 255.56 - 38.89 - 38.89 - 22.22 = 155.56$

Critical value $F_{0.05}(2,2) = 19.00$.

Final comments. All calculated F values (0.25, 0.25, 0.14) are far below 19.00, so $H_0$ is accepted in each case. There is no significant difference among treatments A, B and C, and the row and column effects are also insignificant at the 5% level. Most of the variation (about 61%) is in the error term, and with only 2 error degrees of freedom the test has very low power, so a larger Latin square (or replication) would be needed for a more conclusive result.

Hypotheses. The dealer claims $\mu>60000$ hours, so

$H_0:\mu\le 60000\qquad H_1:\mu>60000\ (\text{right-tailed})$

Test statistic. $n=100$ is large and $\sigma=6000$ is known, so use the $Z$ test:

$Z=\frac{\bar x-\mu_0}{\sigma/\sqrt n}=\frac{55000-60000}{6000/\sqrt{100}}=\frac{-5000}{600}=-8.33$

Critical value. At $\alpha=0.05$ (one-tailed), $Z_{0.05}=1.645$. Reject $H_0$ only if $Z>1.645$.

Decision. $Z=-8.33<1.645$, so $H_0$ is not rejected.

Conclusion. The sample mean (55,000 h) is actually well below 60,000 h, so the sample gives no support for the claim. At the 5% level there is no evidence that the average lifetime exceeds 60,000 hours; the dealer's claim is not accepted.

Given: $\bar x_1=10,\ s_1=3,\ n_1=49$; $\bar x_2=15,\ s_2=5,\ n_2=64$. Assume equal population variances (independent pooled t-test).

(a) Test of difference of means

$H_0:\mu_1=\mu_2\qquad H_1:\mu_1\ne\mu_2$

Pooled variance:

$s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}=\frac{48(9)+63(25)}{111}=\frac{2007}{111}=18.08,\quad s_p=4.252$

$SE=s_p\sqrt{\frac1{n_1}+\frac1{n_2}}=4.252\sqrt{0.020408+0.015625}=4.252(0.1898)=0.807$

$t=\frac{\bar x_1-\bar x_2}{SE}=\frac{10-15}{0.807}=-6.19$

With $df=111$, the critical value is $t_{0.025}\approx 1.98$. Since $|t|=6.19>1.98$, reject $H_0$: the two means differ significantly at the 5% level.

(b) 95% confidence interval for $\mu_1-\mu_2$

$(\bar x_1-\bar x_2)\pm t_{0.025}\,SE=-5\pm1.98(0.807)=-5\pm1.60$

$95\%\text{ CI}: (-6.60,\ -3.40)$

(c) Linkage between testing and interval estimation. The CI does not contain 0 (the value of $\mu_1-\mu_2$ under $H_0$), which is exactly why the two-sided test at $\alpha=0.05$ rejects $H_0$. A two-sided test at level $\alpha$ rejects $H_0:\mu_1-\mu_2=0$ if and only if the $(1-\alpha)100\%$ CI excludes 0. The interval also adds information: Group B's mean exceeds Group A's by between about 3.4 and 6.6 units.

Given: the pilot study gives $\hat p=\dfrac{300}{1000}=0.30$, margin of error $E=0.01$, confidence 99% so $Z_{0.005}=2.576$.

Sample size for a proportion:

$n=\frac{Z^2\,\hat p(1-\hat p)}{E^2}=\frac{(2.576)^2(0.30)(0.70)}{(0.01)^2}=\frac{6.6358\times0.21}{0.0001}=\frac{1.3935}{0.0001}=13935.1$

Rounding up, $n\approx 13{,}936$ computer engineers must be surveyed.

(If no prior estimate were available, $\hat p=0.5$ would give the larger, conservative size $n=16{,}590$.)

Sample statistics ($n=6$, population $\sigma$ unknown, normal population, so use the $t$ distribution):

$\bar x=\frac{8.24+8.26+8.20+8.28+8.21+8.23}{6}=\frac{49.42}{6}=8.2367$

$\sum(x-\bar x)^2=0.004533,\qquad s=\sqrt{\frac{0.004533}{5}}=0.0301$

95% CI. $df=5$, $t_{0.025,5}=2.571$:

$\bar x\pm t_{0.025}\frac{s}{\sqrt n}=8.2367\pm2.571\times\frac{0.0301}{\sqrt6}=8.2367\pm2.571(0.01229)=8.2367\pm0.0316$

$\boxed{8.205\ \text{mm}<\mu<8.268\ \text{mm}}$

Interpretation. We are 95% confident that the true mean diameter of rods produced by this machine lies between about 8.205 mm and 8.268 mm. If many such samples were drawn and an interval built from each, about 95% of those intervals would contain the true mean diameter. The narrow width (about 0.063 mm) shows the machine's output is very consistent.

Hypotheses. $H_0$: the satisfaction scores of the two gadgets come from identical populations. $H_1$: they differ (two-tailed), $\alpha=0.05$.

Ranking the 8 pooled values (ties get average ranks): 10, 20, 30, 30, 40, 40, 40, 50 receive ranks 1, 2, 3.5, 3.5, 6, 6, 6, 8.

Check: $R_A+R_B=36=\frac{8\cdot9}{2}$.

U statistics ($n_1=n_2=4$):

$U_A=n_1n_2+\frac{n_1(n_1+1)}2-R_A=16+10-19.5=6.5$

$U_B=16+10-16.5=9.5\qquad (U_A+U_B=16=n_1n_2)$

$U=\min(U_A,U_B)=6.5$.

Critical value. For $n_1=n_2=4$, two-tailed $\alpha=0.05$, the table value is $U_{crit}=0$. Reject $H_0$ only if $U\le0$.

Decision. $6.5>0$, so $H_0$ is not rejected. There is no significant difference in satisfaction between Gadget A and Gadget B at the 5% level.

Definition. A Markov chain is a stochastic process $\{X_n,\ n=0,1,2,\dots\}$ taking values in a set of states $S$ such that the future depends on the past only through the present state:

$P(X_{n+1}=j\mid X_n=i,X_{n-1}=i_{n-1},\dots,X_0=i_0)=P(X_{n+1}=j\mid X_n=i)=p_{ij}$

This is the Markov (memoryless) property.

Characteristics

1. State space. The set of possible states (e.g. sunny/rainy) is finite or countable.
2. Transition probabilities. $p_{ij}$ is the probability of moving from state $i$ to state $j$ in one step. They are non-negative and each row sums to 1: $\sum_j p_{ij}=1$.
3. Transition matrix. $P=[p_{ij}]$ is a square stochastic matrix. The $n$-step matrix is $P^{(n)}=P^n$ (Chapman-Kolmogorov equations).
4. Stationary (time-homogeneous). $p_{ij}$ does not change with time $n$.
5. Initial distribution. The vector $\pi^{(0)}$ gives the starting probabilities, and $\pi^{(n)}=\pi^{(0)}P^n$.
6. Steady state. For a regular chain, $\pi^{(n)}$ converges to a unique vector $\pi$ with $\pi P=\pi$, $\sum\pi_i=1$, independent of the starting state.
7. Classification of states. States may be absorbing, transient, recurrent, or periodic/aperiodic, depending on whether they can be left or revisited.