Probability and Statistics

Primary data are data collected first-hand by the investigator for the specific purpose of the study (e.g. through questionnaires, interviews, observation or experiments). Secondary data are data that were already collected by someone else for another purpose and are now used by the investigator (e.g. census reports, government publications, journals, websites).

Here $N = 100$ workers, with classes 0-20, 20-40, 40-60, 60-80, 80-100 (income in thousands). Form the cumulative frequency (cf) column from the frequencies given in the paper and use

$P_k = L + \frac{\dfrac{kN}{100} - cf}{f}\times h, \qquad h = 20$

where $L$ is the lower limit of the percentile class, $cf$ the cumulative frequency before that class and $f$ its frequency.

(i) Income limits of the middle 30% of workers. Of the remaining 70%, 35% lie below and 35% above, so the middle 30% lie between $P_{35}$ and $P_{65}$.

$P_{35} = L + \frac{35 - cf}{f}\times 20, \qquad P_{65} = L' + \frac{65 - cf'}{f'}\times 20$

The class of $P_{35}$ is the first class whose cumulative frequency is at least 35, and the class of $P_{65}$ is the first class whose cumulative frequency is at least 65. The middle 30% of workers earn between $P_{35}$ and $P_{65}$ thousand rupees.

(ii) Highest income of the poorest 25% of the workers. This is the first quartile:

$Q_1 = P_{25} = L + \frac{25 - cf}{f}\times 20$

where the class is the first one whose cumulative frequency is at least 25. The poorest 25% have incomes up to $Q_1$ thousand rupees.

Here $N = 100$ workers, with classes 0-20, 20-40, 40-60, 60-80, 80-100 (income in thousands). Form the cumulative frequency (cf) column from the frequencies given in the paper, and use the percentile formula

$P_k = L + \frac{\dfrac{kN}{100} - cf}{f}\times h$

where $L$ is the lower limit of the percentile class, $cf$ the cumulative frequency before it, $f$ its frequency and $h = 20$.

(i) Income limits of the middle 50% of workers. The middle 50% lie between the first quartile $Q_1 = P_{25}$ and the third quartile $Q_3 = P_{75}$.

$Q_1 = L_1 + \frac{25 - cf_1}{f_1}\times 20, \qquad Q_3 = L_3 + \frac{75 - cf_3}{f_3}\times 20$

where the class of $Q_1$ is the first class whose cf is at least 25, and the class of $Q_3$ is the first class whose cf is at least 75. The middle 50% of the workers have incomes between $Q_1$ and $Q_3$ thousand rupees.

(ii) Highest income of the poorest 20%. This is the 20th percentile:

$P_{20} = L + \frac{20 - cf}{f}\times 20$

where the class is the first one whose cf is at least 20. Workers with income up to $P_{20}$ thousand form the poorest 20%.

Ordered data (n = 15): 1, 2, 2, 3, 5, 6, 8, 9, 9, 10, 10, 10, 13, 15, 17

Five-number summary

• Minimum $= 1$, maximum $= 17$
• Median $Q_2$ = 8th value $= 9$
• Lower quartile $Q_1$ = median of the lower seven values (1, 2, 2, 3, 5, 6, 8) $= 3$
• Upper quartile $Q_3$ = median of the upper seven values (9, 10, 10, 10, 13, 15, 17) $= 10$
• Interquartile range $IQR = 10 - 3 = 7$

Outlier check (fences):

$Q_1 - 1.5\,IQR = 3 - 10.5 = -7.5, \qquad Q_3 + 1.5\,IQR = 10 + 10.5 = 20.5$

All values lie between $-7.5$ and $20.5$, so there are no outliers.

Drawing the box plot. On a horizontal scale from 0 to 18 minutes:

1. Draw a box from $Q_1 = 3$ to $Q_3 = 10$.
2. Draw a vertical line inside the box at the median $= 9$.
3. Draw whiskers from the box to the minimum $1$ (left) and to the maximum $17$ (right).

Interpretation

• The middle 50% of waiting times lie between 3 and 10 minutes; the typical wait (median) is 9 minutes.
• The median is close to $Q_3$ (distance 1) and far from $Q_1$ (distance 6), and the left whisker/box part is longer than the upper box part. This shows the central data are bunched at the higher values with a longer tail towards short waits, i.e. the distribution is slightly negatively (left) skewed.
• The waits range from 1 to 17 minutes (range 16), so the waiting time is quite variable, but there are no extreme values.

Let the chances of solving the problem be $P(A) = p_A$, $P(B) = p_B$, $P(C) = p_C$ as given in the paper. The students work independently, so the probabilities of failing are $q_A = 1-p_A$, $q_B = 1-p_B$, $q_C = 1-p_C$.

(i) The problem will be solved (at least one solves it):

$P(\text{solved}) = 1 - P(\text{none solves}) = 1 - q_A q_B q_C$

(ii) Only one of them solves it (exactly one succeeds, the others fail):

$P = p_A q_B q_C + q_A p_B q_C + q_A q_B p_C$

(iii) None of them solves it:

$P = q_A q_B q_C = (1-p_A)(1-p_B)(1-p_C)$

Worked case (chances $\tfrac12, \tfrac13, \tfrac14$): $q_A = \tfrac12,\ q_B = \tfrac23,\ q_C = \tfrac34$.

(iii) $P = \frac12\cdot\frac23\cdot\frac34 = \frac14 = 0.25$

(i) $P = 1 - \frac14 = \frac34 = 0.75$

(ii) $P = \frac12\cdot\frac23\cdot\frac34 + \frac12\cdot\frac13\cdot\frac34 + \frac12\cdot\frac23\cdot\frac14 = \frac{6}{24}+\frac{3}{24}+\frac{2}{24} = \frac{11}{24} = 0.4583$

Substitute the chances printed in the paper in the same formulas to get the required values.

All classes have the same width $h = 10$, so the bar heights are simply the frequencies.

Histogram. Take marks on the X-axis (20 to 90) and number of students on the Y-axis (0 to 80). Draw adjacent rectangles on each class with heights 20, 30, 50, 60, 80, 35, 25. The tallest bar is the class 60-70.

Frequency polygon. Plot the points (25, 20), (35, 30), (45, 50), (55, 60), (65, 80), (75, 35), (85, 25), i.e. the mid-value against the frequency, and join them by straight lines. Close the polygon by adding the points (15, 0) and (95, 0) on the X-axis.

Frequency curve. Draw a smooth free-hand curve through the polygon points, starting at (15, 0) and ending at (95, 0). The curve rises to a peak near the marks 60-70 and then falls, showing that the distribution is unimodal and slightly skewed to the left (a longer tail towards lower marks).

Let $A, B, C$ denote production from units A, B, C and $D$ a defective item.

$P(A) = 0.25,\ P(B) = 0.25,\ P(C) = 0.50; \quad P(D|A) = 0.01,\ P(D|B) = 0.02,\ P(D|C) = 0.03$

Total probability of a defective item:

$P(D) = 0.25(0.01) + 0.25(0.02) + 0.50(0.03) = 0.0025 + 0.005 + 0.015 = 0.0225$

By Bayes' theorem:

$P(C|D) = \frac{P(C)P(D|C)}{P(D)} = \frac{0.015}{0.0225} = \frac23 \approx 0.667$

The probability that the defective item was produced by unit C is 2/3 (about 66.7%).

Let $d =$ after $-$ before:

$n = 10$, $\sum d = 10$, $\sum d^2 = 1+25+49+1+4+64+4+4+81+25 = 258$.

$\bar d = 1, \qquad s_d = \sqrt{\frac{258 - 10(1)^2}{9}} = \sqrt{27.56} = 5.25$

Hypotheses. $H_0: \mu_d = 0$ (training has not benefited); $H_1: \mu_d > 0$ (training has benefited). One-tailed.

Test statistic.

$t = \frac{\bar d}{s_d/\sqrt n} = \frac{1}{5.25/\sqrt{10}} = \frac{1}{1.66} = 0.60$

Critical value. $\alpha = 0.05$, d.f. $= 9$: $t_{0.05} = 1.833$.

Decision. $0.60 < 1.833$, so $H_0$ is not rejected.

Conclusion. At the 5% level the data do not show that the training has significantly improved the IQ scores, so it cannot be concluded that the students benefited from the training.

Given $N = 540$, $n = 60$, $\bar x = 6.2$, $s = 1.368$. The sample is large and taken without replacement from a finite population, so the finite population correction (FPC) is used since $n/N = 0.11 > 0.05$.

(i) Estimated standard error of the mean:

$SE(\bar x) = \frac{s}{\sqrt n}\sqrt{\frac{N-n}{N-1}} = \frac{1.368}{\sqrt{60}}\sqrt{\frac{480}{539}} = 0.1766\times0.9437 = 0.1667$

(ii) 95% confidence interval for the mean ($z = 1.96$):

$\bar x \pm 1.96\,SE = 6.2 \pm 1.96(0.1667) = 6.2 \pm 0.327$

$5.873 < \mu < 6.527$

We are 95% confident that the population mean lies between 5.87 and 6.53.

Given $N = 500$, $n = 50$, $\bar x = 15$, $s = 10$. Since $n/N = 0.10 > 0.05$, the finite population correction is applied.

(i) Estimated standard error of the mean:

$SE(\bar x) = \frac{s}{\sqrt n}\sqrt{\frac{N-n}{N-1}} = \frac{10}{\sqrt{50}}\sqrt{\frac{450}{499}} = 1.4142\times0.9496 = 1.343$

(ii) 95% confidence interval for the mean ($z = 1.96$):

$\bar x \pm 1.96\,SE = 15 \pm 1.96(1.343) = 15 \pm 2.63$

$12.37 < \mu < 17.63$

The 95% confidence interval for the population mean is (12.37, 17.63).

Step 1: Prepare the table. The classes are 300-310, 310-320, 320-330, 330-350, 350-370 and 370-410, so the widths are unequal (10, 10, 10, 20, 20, 40). For a histogram with unequal class widths the bar area must be proportional to frequency, so we plot the adjusted frequency (frequency density) taking the smallest width $h_0 = 10$ as the standard:

$\text{adjusted frequency} = f \times \frac{h_0}{h} = f \times \frac{10}{\text{class width}}$

with $f_1 + \dots + f_6 = 84$ (use the frequencies printed in the paper).

(a) Histogram.

• Draw the weekly salary (Rs) on the X-axis with a continuous scale from 300 to 410, and the adjusted frequency on the Y-axis.
• Draw adjacent rectangles with no gaps; the base of each is its class interval and the height is its adjusted frequency.
• Label the axes and give the title "Weekly salaries of 84 workers".

(b) Frequency polygon.

• Mark a point above each class mid-value at the height of the corresponding bar.
• Join the points by straight lines.
• Close the polygon by adding one empty class at each end (mid-values 295 and 400 with height 0) and joining to them on the X-axis.

(c) Frequency curve.

• Smooth the frequency polygon with a free-hand curve through (or very near) the same points, starting and ending on the X-axis. The curve removes the angularity of the polygon and shows the general shape of the distribution, which is roughly a single-peaked, right-tailed curve for salary data.

The area under the histogram, the polygon and the curve are all approximately equal to the total frequency, 84.

Total students $N = 10+30+36+30+14 = 120$.

(i) Limits of the middle 40% of students. The middle 40% lie between $P_{30}$ and $P_{70}$ (30% below and 30% above).

$P_{30}$: position $= \frac{30 \times 120}{100} = 36$, which falls in the class 20-40 (cf before = 10, $f = 30$).

$P_{30} = 20 + \frac{36-10}{30}\times 20 = 20 + 17.33 = 37.33$

$P_{70}$: position $= \frac{70\times120}{100} = 84$, which falls in the class 60-80 (cf before = 76, $f = 30$).

$P_{70} = 60 + \frac{84-76}{30}\times 20 = 60 + 5.33 = 65.33$

So the middle 40% of students got marks between 37.33 and 65.33.

(ii) Percentage getting more than 75 marks. Marks 75 lie in the class 60-80. Students above 75 in this class $= \frac{80-75}{20}\times 30 = 7.5$. Adding the class 80-100:

$7.5 + 14 = 21.5 \text{ students}, \qquad \frac{21.5}{120}\times 100 = 17.92\%$

(iii) Number who fail (below 35). The class 20-40 contains marks 35. Students from 20 to 35 $= \frac{35-20}{20}\times 30 = 22.5$. Adding the class 0-20:

$10 + 22.5 = 32.5 \approx 33 \text{ students}$

Let $A, B, C$ be the events that the item is from unit A, B, C, and $D$ the event that it is defective.

$P(A) = 0.25,\; P(B) = 0.25,\; P(C) = 0.50$ $P(D|A) = 0.01,\; P(D|B) = 0.02,\; P(D|C) = 0.03$

Total probability of a defective item:

$P(D) = 0.25(0.01) + 0.25(0.02) + 0.50(0.03) = 0.0025 + 0.005 + 0.015 = 0.0225$

By Bayes' theorem, $P(X|D) = \dfrac{P(X)P(D|X)}{P(D)}$:

$P(A|D) = \frac{0.0025}{0.0225} = \frac{1}{9} = 0.1111$

$P(B|D) = \frac{0.005}{0.0225} = \frac{2}{9} = 0.2222$

$P(C|D) = \frac{0.015}{0.0225} = \frac{6}{9} = 0.6667$

Check: $0.1111 + 0.2222 + 0.6667 = 1$. A defective item is most likely to have come from unit C.

Statement. Let $E_1, E_2, \dots, E_n$ be mutually exclusive and exhaustive events (a partition of the sample space $S$) with $P(E_i) > 0$, and let $A$ be any event with $P(A) > 0$ that can occur only with one of the $E_i$. Then the probability that $A$ occurred through the cause $E_i$ is

$P(E_i|A) = \frac{P(E_i)\,P(A|E_i)}{\sum_{j=1}^{n} P(E_j)\,P(A|E_j)}, \qquad i = 1, 2, \dots, n$

Proof. Since $E_1, \dots, E_n$ are exhaustive, $A = (A\cap E_1) \cup (A\cap E_2) \cup \dots \cup (A\cap E_n)$. These parts are mutually exclusive because the $E_i$ are, so by the addition law

$P(A) = \sum_{j=1}^{n} P(A\cap E_j)$

By the multiplication law, $P(A\cap E_j) = P(E_j)\,P(A|E_j)$, hence

$P(A) = \sum_{j=1}^{n} P(E_j)\,P(A|E_j) \quad \text{(theorem of total probability)}$

By the definition of conditional probability,

$P(E_i|A) = \frac{P(E_i\cap A)}{P(A)} = \frac{P(E_i)\,P(A|E_i)}{P(A)}$

Substituting the value of $P(A)$ gives

$P(E_i|A) = \frac{P(E_i)\,P(A|E_i)}{\sum_{j=1}^{n} P(E_j)\,P(A|E_j)}$

which proves the theorem. Here $P(E_i)$ are the prior probabilities, $P(A|E_i)$ the likelihoods and $P(E_i|A)$ the posterior probabilities.

For $n = 6$: $P(X=r) = \binom{6}{r}p^r q^{6-r}$, with $q = 1-p$.

Given $P(3) = 0.2457$ and $P(4) = 0.0819$. Dividing,

$\frac{P(4)}{P(3)} = \frac{\binom{6}{4}p^4q^2}{\binom{6}{3}p^3q^3} = \frac{15}{20}\cdot\frac{p}{q} = \frac34\cdot\frac pq$

$\frac{0.0819}{0.2457} = \frac13 \;\Rightarrow\; \frac34\cdot\frac pq = \frac13 \;\Rightarrow\; \frac pq = \frac49$

Since $p + q = 1$:

$p = \frac{4}{4+9} = \frac{4}{13} \approx 0.308, \qquad q = \frac{9}{13} \approx 0.692$

So the parameters are $p = 4/13 \approx 0.31$ and $q = 9/13 \approx 0.69$.

Given $\bar x = 20$, $s = 3$, $n = 64$ (a large sample), $\alpha = 0.05$.

Standard error:

$SE = \frac{s}{\sqrt n} = \frac{3}{\sqrt{64}} = \frac38 = 0.375$

At the 5% level, $z_{\alpha/2} = 1.96$. The 95% confidence interval for the population mean is

$\bar x \pm z_{\alpha/2}\,SE = 20 \pm 1.96(0.375) = 20 \pm 0.735$

$19.265 < \mu < 20.735$

We are 95% confident that the population mean lies between 19.27 and 20.74.

Marks $X \sim N(\mu = 75,\ \sigma = 5)$.

Lowest mark for grade A (top 5%). The area to the right of the cut-off is $0.05$, so the area to the left is $0.95$ and $z = 1.645$.

$X_A = \mu + z\sigma = 75 + 1.645(5) = 75 + 8.225 = 83.2$

The lowest A is about 83.2 marks (i.e. 83 or 84 marks).

Highest mark for grade F (bottom 25%). The area to the left of the cut-off is $0.25$, so $z = -0.674$.

$X_F = 75 - 0.674(5) = 75 - 3.37 = 71.63$

The highest F is about 71.6 marks (i.e. 71 or 72 marks).

Sample proportion $\hat p = \frac{5}{100} = 0.05$, $\hat q = 0.95$, $n = 100$.

Standard error:

$SE = \sqrt{\frac{\hat p\hat q}{n}} = \sqrt{\frac{0.05\times0.95}{100}} = \sqrt{0.000475} = 0.0218$

95% limits ($z = 1.96$):

$0.05 \pm 1.96(0.0218) = 0.05 \pm 0.0427 \;\Rightarrow\; (0.0073,\ 0.0927)$

99% limits ($z = 2.576$):

$0.05 \pm 2.576(0.0218) = 0.05 \pm 0.0561 \;\Rightarrow\; (-0.0061,\ 0.1061)$

Since a proportion cannot be negative, the 99% interval is $(0,\ 0.1061)$.

Hence the proportion of defective items in the batch lies between about 0.7% and 9.3% (95% confidence) and between 0% and 10.6% (99% confidence).

Probability (random) sampling is a method in which every unit of the population has a known, non-zero chance of being selected, and the selection is made by chance. Non-probability sampling is a method in which units are selected by the judgement, convenience or choice of the investigator, so the chance of selection is unknown.

Example: drawing 50 students by lottery from a list of 500 is probability sampling; interviewing the first 50 students met at the college gate is non-probability (convenience) sampling.

Given $\mu_0 = 44$, $n = 40$, $\bar x = 41$, $s = 9$, $\alpha = 0.05$. The sample is large ($n \ge 30$), so the $z$-test is used.

Step 1: Hypotheses. $H_0: \mu = 44$ (the claim is true) $H_1: \mu \ne 44$ (two-tailed)

Step 2: Level of significance. $\alpha = 0.05$; critical values $z = \pm 1.96$.

Step 3: Test statistic.

$z = \frac{\bar x - \mu_0}{s/\sqrt n} = \frac{41-44}{9/\sqrt{40}} = \frac{-3}{1.423} = -2.11$

Step 4: Decision. $|z| = 2.11 > 1.96$, so $H_0$ is rejected at the 5% level.

Conclusion. The sample provides significant evidence that the mean life of the batteries is not 44 months; the manufacturer's claim is not supported (the sample suggests the mean life is lower). (Using the $t$-distribution with 39 d.f., the critical value is about 2.02 and the conclusion is the same.)

Let $d =$ before $-$ after (reduction in worker-hours lost).

$n = 10$, $\sum d = 52$, $\sum d^2 = 81+169+4+25+4+36+36+25+4+36 = 420$.

$\bar d = \frac{52}{10} = 5.2, \qquad s_d = \sqrt{\frac{420 - 10(5.2)^2}{9}} = \sqrt{\frac{149.6}{9}} = 4.077$

Hypotheses. $H_0: \mu_d = 0$ (the programme is not effective); $H_1: \mu_d > 0$ (it reduces the losses). One-tailed.

Test statistic.

$t = \frac{\bar d}{s_d/\sqrt n} = \frac{5.2}{4.077/\sqrt{10}} = \frac{5.2}{1.289} = 4.03$

Critical value. $t_{0.05}$ with $9$ d.f. $= 1.833$.

Decision. $4.03 > 1.833$, so $H_0$ is rejected.

Conclusion. At the 0.05 level the safety programme is effective in reducing the loss of worker-hours.

Arrange the observed counts in a $2\times2$ table:

(Use the counts printed in the paper for $a, b, c, d$.)

Hypotheses. $H_0$: eye colour and handedness are independent (no association). $H_1$: they are associated.

Expected frequencies. For each cell, $E = \dfrac{\text{row total}\times\text{column total}}{N}$, e.g. $E_a = \dfrac{(a+b)(a+c)}{N}$.

Test statistic.

$\chi^2 = \sum\frac{(O-E)^2}{E} = \frac{N(ad-bc)^2}{(a+b)(c+d)(a+c)(b+d)} \quad (\text{d.f.} = 1)$

Critical value. d.f. $= (2-1)(2-1) = 1$, so $\chi^2_{0.05} = 3.841$.

Decision. If the calculated $\chi^2 > 3.841$, reject $H_0$: there is evidence of an association between eye colour and handedness at the 5% level. If $\chi^2 \le 3.841$, accept $H_0$: there is no evidence of association, i.e. the two attributes are independent.

Method. Spearman's rank correlation coefficient measures the strength of the relation between two variables from their ranks.

Steps.

1. Rank the values of $X$ ($R_x$) and of $Y$ ($R_y$), giving rank 1 to the largest (or smallest) value in both.
2. Find $d = R_x - R_y$ for each pair and then $d^2$ and $\sum d^2$.
3. Apply the formula, with $n$ the number of pairs:

$r_s = 1 - \frac{6\sum d^2}{n(n^2-1)}$

4. If some values are tied, give each the average of the ranks they occupy and add a correction $\dfrac{m(m^2-1)}{12}$ for each group of $m$ tied values to $\sum d^2$:

$r_s = 1 - \frac{6\left[\sum d^2 + \sum \frac{m(m^2-1)}{12}\right]}{n(n^2-1)}$

Interpretation. $r_s$ lies between $-1$ and $+1$. A value near $+1$ means a strong positive relation (the ranks move together), near $-1$ a strong negative relation, and near $0$ almost no relation.

Substituting $n$ and $\sum d^2$ obtained from the ranks of the data in the paper gives the required coefficient.

Let $X$ = experience (years) (independent) and $Y$ = performance rating (dependent), $n = 8$.

Step 1: Compute $\sum X,\ \sum Y,\ \sum XY,\ \sum X^2,\ \sum Y^2$ from the table in the paper.

Step 2: Regression coefficient of $Y$ on $X$.

$b_{yx} = \frac{n\sum XY - \sum X\sum Y}{n\sum X^2 - (\sum X)^2}, \qquad a = \bar Y - b_{yx}\bar X$

Step 3: Regression equation.

$\hat Y = a + b_{yx}X$

Step 4: Estimate for 8 years' experience. Put $X = 8$:

$\hat Y_{8} = a + 8\,b_{yx}$

This is the probable performance rating (good parts per 100 pieces) of an operator with 8 years' experience.

Step 5: Coefficient of determination.

$r = \frac{n\sum XY - \sum X\sum Y}{\sqrt{[n\sum X^2-(\sum X)^2][n\sum Y^2-(\sum Y)^2]}}, \qquad r^2 = b_{yx}\,b_{xy}$

Interpretation. $r^2$ is the proportion of the variation in performance that is explained by experience. For example, $r^2 = 0.81$ would mean 81% of the variation in performance ratings is explained by the operators' experience, and the remaining 19% is due to other factors. A positive $b_{yx}$ means the rating rises on average by $b_{yx}$ good parts per 100 for each additional year of experience.

A random variable is a real-valued function defined on the sample space of a random experiment; it assigns a numerical value to each outcome. It is denoted by capital letters such as $X$, with particular values by $x$.

Example: in tossing two coins, let $X$ be the number of heads. Then $X$ takes the values $0, 1, 2$ with probabilities $\tfrac14, \tfrac12, \tfrac14$.

Types

• Discrete random variable: takes a finite or countably infinite set of values (number of defective items, number of heads). Its probability distribution is given by a probability mass function $p(x)$ with $p(x)\ge 0$ and $\sum p(x) = 1$.
• Continuous random variable: takes any value in an interval (height, weight, time). Its distribution is given by a probability density function $f(x)$ with $f(x)\ge 0$ and $\int_{-\infty}^{\infty} f(x)\,dx = 1$, and $P(a\le X\le b)=\int_a^b f(x)\,dx$.

Expected value and variance: $E(X)=\sum x\,p(x)$ (or $\int x f(x)\,dx$) and $V(X)=E(X^2)-[E(X)]^2$.

An estimator is a sample statistic used to estimate an unknown population parameter. A good estimator should satisfy the following criteria.

1. Unbiasedness: an estimator $T$ of $\theta$ is unbiased if $E(T) = \theta$. On average it neither over- nor under-estimates the parameter. Example: $E(\bar x) = \mu$.
2. Consistency: $T$ is consistent if it approaches the true value as the sample size increases, i.e. $T \to \theta$ in probability as $n\to\infty$ (its bias and variance both tend to zero).
3. Efficiency: among unbiased estimators, the one with the smallest variance is the most efficient. Example: for a normal population the sample mean has smaller variance than the sample median, so it is more efficient.
4. Sufficiency: an estimator is sufficient if it uses all the information about the parameter that is contained in the sample, so no other statistic can add further information about $\theta$.

An estimator having these properties gives estimates that are close to the true value, with small and predictable error.

Conditional probability. The conditional probability of an event $A$ given that event $B$ has already occurred is

$P(A|B) = \frac{P(A\cap B)}{P(B)}, \qquad P(B) > 0$

Chip problem. Let $E$ = chip has an etching defect and $C$ = chip has a crack defect.

$P(E) = 0.32,\quad P(C) = 0.19,\quad P(E\cap C) = 0.05$

By the addition law:

$P(E\cup C) = 0.32 + 0.19 - 0.05 = 0.46$

The probability of neither defect is

$P(\bar E\cap\bar C) = 1 - P(E\cup C) = 1 - 0.46 = 0.54$

So a newly manufactured chip has neither defect with probability 0.54.

Total production per hour $= 2000+3000+5000 = 10000$.

$P(A) = 0.2,\ P(B) = 0.3,\ P(C) = 0.5; \quad P(D|A) = 0.03,\ P(D|B) = 0.02,\ P(D|C) = 0.01$

Probability that the selected article is defective:

$P(D) = 0.2(0.03) + 0.3(0.02) + 0.5(0.01) = 0.006 + 0.006 + 0.005 = 0.017$

By Bayes' theorem:

(a) Machine A:

$P(A|D) = \frac{0.006}{0.017} = \frac{6}{17} \approx 0.353$

(b) Machine B:

$P(B|D) = \frac{0.006}{0.017} = \frac{6}{17} \approx 0.353$

(For completeness, $P(C|D) = 5/17 \approx 0.294$.)

Standard normal variate. If $X$ is a normal variable with mean $\mu$ and standard deviation $\sigma$, then

$Z = \frac{X-\mu}{\sigma}$

is the standard normal variate. It has mean $0$ and variance $1$, i.e. $Z\sim N(0,1)$, and its probabilities are read from the standard normal table.

Problem. Let $X\sim N(\mu,\sigma^2)$.

31% of items are under 45: $P(X<45) = 0.31$. The area to the left is below 0.5, so $z$ is negative: $z_1 = -0.496$.

$\frac{45-\mu}{\sigma} = -0.496 \quad\Rightarrow\quad 45 = \mu - 0.496\sigma \qquad (1)$

8% of items are over 64: $P(X>64) = 0.08$, so the area to the left is $0.92$ and $z_2 = 1.405$.

$\frac{64-\mu}{\sigma} = 1.405 \quad\Rightarrow\quad 64 = \mu + 1.405\sigma \qquad (2)$

Subtracting (1) from (2):

$19 = 1.901\,\sigma \;\Rightarrow\; \sigma = 9.99 \approx 10$

From (2): $\mu = 64 - 1.405(9.99) \approx 49.96 \approx 50$.

Hence the mean is about 50 and the standard deviation about 10.

Scores: 70, 120, 110, 101, 83, 88, 95, 98, 107, 100; $n = 10$ (small sample, so the $t$-distribution is used).

$\sum x = 972, \qquad \bar x = \frac{972}{10} = 97.2$

Deviations from the mean: $-27.2,\ 22.8,\ 12.8,\ 3.8,\ -14.2,\ -9.2,\ -2.2,\ 0.8,\ 9.8,\ 2.8$

$\sum(x-\bar x)^2 = 739.84+519.84+163.84+14.44+201.64+84.64+4.84+0.64+96.04+7.84 = 1833.6$

$s = \sqrt{\frac{1833.6}{9}} = \sqrt{203.73} = 14.27, \qquad SE = \frac{s}{\sqrt n} = \frac{14.27}{\sqrt{10}} = 4.51$

For $95\%$ confidence with $9$ d.f., $t_{0.025} = 2.262$.

$\bar x \pm t\,SE = 97.2 \pm 2.262(4.51) = 97.2 \pm 10.21$

$86.99 < \mu < 107.41$

The 95% confidence interval for the true mean IQ is (86.99, 107.41), approximately 87 to 107.

Group 1 (athletics): $n_1 = 50$, $\bar x_1 = 68.2$, $s_1 = 2.5$. Group 2 (not interested): $n_2 = 50$, $\bar x_2 = 67.5$, $s_2 = 2.8$.

Hypotheses. $H_0: \mu_1 = \mu_2$ $H_1: \mu_1 > \mu_2$ (right-tailed; athletes are taller)

Level of significance. $\alpha = 0.05$; critical value $z_{0.05} = 1.645$ (large samples, so a $z$-test).

Test statistic.

$z = \frac{\bar x_1-\bar x_2}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}} = \frac{0.7}{\sqrt{\dfrac{6.25}{50}+\dfrac{7.84}{50}}} = \frac{0.7}{\sqrt{0.2818}} = \frac{0.7}{0.531} = 1.32$

Decision. $1.32 < 1.645$, so $H_0$ is not rejected.

Conclusion. At the 5% level there is not enough evidence to say that male students who take part in college athletics are taller than the others; the observed difference of 0.7 inch may be due to chance.

Horse A: 28, 30, 32, 33, 33, 29, 34 ($n_1 = 7$). Horse B: 29, 30, 30, 24, 27, 29 ($n_2 = 6$). Small samples, so the two-sample $t$-test is used (assuming equal variances).

Means. $\bar x_1 = \frac{219}{7} = 31.29$, $\bar x_2 = \frac{169}{6} = 28.17$.

Sums of squared deviations. $\sum(x_1-\bar x_1)^2 = 31.43$ (so $s_1^2 = 31.43/6 = 5.24$) $\sum(x_2-\bar x_2)^2 = 26.83$ (so $s_2^2 = 26.83/5 = 5.37$)

Pooled variance:

$s_p^2 = \frac{31.43+26.83}{7+6-2} = \frac{58.26}{11} = 5.297, \qquad s_p = 2.30$

Hypotheses. $H_0: \mu_A = \mu_B$ (same running capacity); $H_1: \mu_A \ne \mu_B$ (two-tailed).

Test statistic.

$t = \frac{\bar x_1-\bar x_2}{s_p\sqrt{\frac1{n_1}+\frac1{n_2}}} = \frac{3.12}{2.30\sqrt{\frac17+\frac16}} = \frac{3.12}{2.30\times0.5594} = \frac{3.12}{1.287} = 2.44$

Critical value. $\alpha = 0.05$, d.f. $= 11$: $t_{0.025} = 2.201$.

Decision. $2.44 > 2.201$, so $H_0$ is rejected.

Conclusion. At the 5% level the two horses do not have the same running capacity; the mean time of horse B is significantly lower, so horse B is faster.

Let $X$ = government deficit (Rs lakhs) and $Y$ = percentage change in GNP, $n = 6$. Compute $\sum X,\ \sum Y,\ \sum XY,\ \sum X^2,\ \sum Y^2$ from the table in the paper.

(i) Correlation coefficient.

$r = \frac{n\sum XY - \sum X\sum Y}{\sqrt{\left[n\sum X^2 - (\sum X)^2\right]\left[n\sum Y^2 - (\sum Y)^2\right]}}$

A positive $r$ near $+1$ shows a strong positive linear relation between the deficit and the change in GNP; a negative $r$ shows an inverse relation.

(ii) Coefficient of determination.

$r^2 = (r)^2$

Interpretation: $100\,r^2\%$ of the variation in the percentage change in GNP is explained by the variation in the government deficit; the remaining $100(1-r^2)\%$ is due to other factors.

(iii) Expected change in GNP when the deficit is Rs 110 lakhs. Fit the regression line of $Y$ on $X$:

$\hat Y = a + bX, \qquad b = \frac{n\sum XY - \sum X\sum Y}{n\sum X^2-(\sum X)^2}, \quad a = \bar Y - b\bar X$

Substituting $X = 110$:

$\hat Y = a + 110\,b$

This is the percentage change in GNP expected in a year when the government deficit is Rs 110 lakhs.

The fundamental elements of a statistical analysis are the basic components that must be identified in any statistical study.

1. Population: the complete collection of all individuals or objects of interest (e.g. all BCA students of a university).
2. Sample: a subset of the population actually selected and studied, chosen so as to represent the population.
3. Variable: a characteristic that varies from one unit to another, e.g. marks, height (quantitative) or gender (qualitative).
4. Experimental/observational unit: the object on which the data are measured (a student, a plant, a bank).
5. Parameter and statistic: a parameter is a numerical measure of a population (e.g. $\mu$, $\sigma$); a statistic is a numerical measure of a sample (e.g. $\bar x$, $s$).
6. Statistical inference: drawing conclusions about the population from the sample, together with a measure of reliability of the inference (confidence level or significance level).

A statistical analysis therefore starts by defining the problem and the population, selecting a sample, collecting and summarising the data, and making an inference with a stated degree of reliability.

Data can be obtained from two main sources.

1. Primary sources. Data collected first-hand by the investigator for a specific purpose, using

• direct personal investigation/interview,
• questionnaires (mailed or online),
• observation,
• experiments and surveys, including telephone interviews.

2. Secondary sources. Data already collected by others, found in

• Published sources: government reports and census, Nepal Rastra Bank and Central Bureau of Statistics publications, statistical abstracts, journals, newspapers, books and international agency reports (UN, World Bank).
• Unpublished sources: records and files of organisations, research reports and theses.
• Websites and databases.

Primary data are original and more reliable but costly and slow to collect; secondary data are cheap and quick but may be outdated or not suited to the present need, so they should be checked for reliability and suitability before use.

(i) Two pieces of information from a stem-and-leaf display.

1. The shape of the distribution: the length of each row (number of leaves) shows where the data are concentrated, how spread out they are and whether the data are symmetric or skewed.
2. The actual data values: unlike a histogram, every original observation can be read back (stem followed by leaf), so the smallest and largest fees, the most common fee and the median can be found.

(ii) Percentage of items before 30. Count all the leaves in the rows with stems below 3 (fees less than 30), call this $k$. With $n = 25$ banks,

$\text{Percentage} = \frac{k}{25}\times 100\%$

(for example, if 10 banks charge less than 30, the percentage is 40%).

(iii) Relative percentage frequency distribution. Take the stem rows (or fee classes such as 10-under 20, 20-under 30, ...) as classes, count the leaves in each row to get the frequency $f$, and compute

$\text{relative frequency} = \frac{f}{25},\qquad \text{percentage frequency} = \frac{f}{25}\times 100$

The percentages add up to 100.