Probability and Statistics

Statistics is the science of collecting, organising, presenting, analysing and interpreting numerical data to support decision making.

Scope of statistics

• Planning and economics: government planning, national income, price indices, population and census studies depend on statistics.
• Business and industry: demand forecasting, market research, quality control, inventory and production planning, and risk assessment.
• Banking and insurance: premium calculation, credit risk and investment analysis use probability and statistics.
• Science and medicine: design of experiments, testing the effectiveness of drugs, and analysis of research data.
• Education and social sciences: examination results, surveys, opinion polls and behavioural studies.
• Information technology: data mining, machine learning, big data analysis and performance evaluation of systems.
• Agriculture and weather: crop yield estimation and weather forecasting.

Limitations of statistics

• Deals with groups, not individuals: it does not give information about a single item; only aggregates and averages are studied.
• Only quantitative data: qualitative features such as honesty or intelligence cannot be studied unless expressed numerically.
• Results are true only on average: conclusions are not exact for every individual case.
• Can be misused: data can be selectively presented or wrongly interpreted, so it needs to be handled by experts and honest persons.
• Requires homogeneous, reliable data: poor or biased data lead to incorrect conclusions.
• Does not explain causes: it shows relationships but does not prove cause and effect.
• Sampling errors: conclusions from samples may differ from population values.

Let $X \sim N(\mu = 200, \sigma = 20)$ and use $Z = \dfrac{X - 200}{20}$.

(i) $P(X > 180)$

$Z = \frac{180-200}{20} = -1$

$P(X>180) = P(Z>-1) = 1 - 0.1587 = 0.8413$

(ii) $P(X < 220)$

$Z = \frac{220-200}{20} = 1$

$P(X<220) = P(Z<1) = 0.8413$

(iii) $P(160 < X < 240)$

$Z_1 = \frac{160-200}{20} = -2,\qquad Z_2 = \frac{240-200}{20} = 2$

$P(-2<Z<2) = 0.9772 - 0.0228 = 0.9544$

(iv) $P(X > 220)$

$P(Z>1) = 1 - 0.8413 = 0.1587$

(v) 10% of the values are less than what value of $X$?

We need $x$ with $P(X < x) = 0.10$. From the normal table, $P(Z < z) = 0.10$ gives $z = -1.28$.

$x = \mu + z\sigma = 200 + (-1.28)(20) = 200 - 25.6 = 174.4$

So 10% of the values are less than $X \approx 174.4$.

Statistics is the science of collecting, organising, analysing and interpreting data. In computer applications it plays the following roles.

1. Data mining and machine learning: regression, classification, clustering and probability models (such as Naive Bayes) are statistical methods used to find patterns and make predictions from large data sets.
2. Database and big data analytics: summaries such as mean, median, variance and correlation are used for reporting, business intelligence and decision support.
3. Performance evaluation and simulation: response time, throughput and load of systems are modelled with distributions (Poisson, exponential) and queuing theory; simulations use random number generation.
4. Software quality and reliability: defect counts, failure rates and mean time between failures are analysed with probability distributions and control charts; testing results are compared using hypothesis tests.
5. Image, speech and signal processing: noise is handled with probabilistic models, and pattern recognition uses statistical decision theory.
6. Networking and security: traffic modelling, packet-loss estimation, anomaly and intrusion detection use statistical tests and probability.
7. Information retrieval and AI: ranking of search results, spam filtering, recommendation systems and natural language processing rely on probabilistic and statistical models.
8. Sampling and testing: sampling techniques and confidence intervals allow conclusions about a large system or user population from a small sample, for example in A/B testing and usability surveys.

Thus statistics gives computer applications the tools for handling uncertainty, validating results and making data-driven decisions.

Sampling: the process of selecting a part (sample) of the units from a population so that conclusions about the whole population can be drawn from the study of that part.

Population: $1, 3, 4, 8$, with $N = 4$.

Population mean:

$\mu = \frac{1+3+4+8}{4} = \frac{16}{4} = 4$

(i) All samples of size 2 without replacement (there are $\binom{4}{2} = 6$):

(ii) Unbiasedness of the sample mean

Mean of the sample means:

$E(\bar x) = \frac{2+2.5+4.5+3.5+5.5+6}{6} = \frac{24}{6} = 4$

Since $E(\bar x) = 4 = \mu$, the sample mean is an unbiased estimate of the population mean.

Let $X$ be the lifetime, $X \sim N(\mu = 5000,\ \sigma = 100)$, and $Z = \dfrac{X-5000}{100}$.

(a) $P(X < 3012)$

$z = \frac{3012-5000}{100} = -19.88$

$P(Z < -19.88) \approx 0$

(b) $P(4000 < X < 6000)$

$z_1 = \frac{4000-5000}{100} = -10,\qquad z_2 = \frac{6000-5000}{100} = 10$

$P(-10 < Z < 10) \approx 1$

(c) $P(X < 4500)$

$z = \frac{4500-5000}{100} = -5$

$P(Z < -5) \approx 0.0000003 \approx 0$

(d) $P(X > 7000)$

$z = \frac{7000-5000}{100} = 20$

$P(Z > 20) \approx 0$

The standard deviation is small compared with these limits, so almost all components last between about $4700$ and $5300$ hours ($\mu \pm 3\sigma$). Hence the required probabilities are approximately $0$, $1$, $0$ and $0$ respectively; the standard normal table ends at $z = 3.49$ (area $0.9998$), so values beyond it are taken as $0$ or $1$.

Normal distribution: a continuous distribution of a variable $X$ with mean $\mu$ and standard deviation $\sigma$, having the bell-shaped, symmetric density

$f(x) = \frac{1}{\sigma\sqrt{2\pi}}\,e^{-\frac{(x-\mu)^2}{2\sigma^2}},\qquad -\infty < x < \infty$

Standard normal distribution: the normal distribution with $\mu = 0$ and $\sigma = 1$, obtained by the transformation $Z = \dfrac{X-\mu}{\sigma}$; its density is $\phi(z) = \dfrac{1}{\sqrt{2\pi}}e^{-z^2/2}$ and its areas are tabulated.

Here $X \sim N(5000,\ 1000)$.

(a) $P(X < 5012)$

$z = \frac{5012-5000}{1000} = 0.012 \approx 0.01$

$P(Z < 0.01) = 0.5040 \quad(\text{exactly } 0.5048)$

(b) $P(4000 < X < 6000)$

$z_1 = \frac{4000-5000}{1000} = -1,\qquad z_2 = \frac{6000-5000}{1000} = 1$

$P(-1<Z<1) = 0.8413 - 0.1587 = 0.6826$

(c) $P(X > 7000)$

$z = \frac{7000-5000}{1000} = 2$

$P(Z > 2) = 1 - 0.9772 = 0.0228$

The sample size is the number of units selected from the population for study. It must be large enough for reliable results but small enough to limit cost and time.

Factors that determine the sample size

• Population variability ($\sigma$): the more variable the population, the larger the sample required.
• Desired precision (margin of error $E$): smaller permissible error needs a larger sample.
• Confidence level: a higher confidence level (e.g. 99% instead of 95%) needs a larger sample, through a larger $Z$ value.
• Population size: matters mainly for small populations; a finite population correction can reduce the sample.
• Resources: available time, money and manpower.
• Sampling method and nature of analysis: complex designs or subgroup analysis need larger samples.

Formulas

For estimating a population mean (large or infinite population):

$n = \left(\frac{Z_{\alpha/2}\,\sigma}{E}\right)^2$

For estimating a proportion:

$n = \frac{Z_{\alpha/2}^2\,p\,(1-p)}{E^2}$

If $p$ is unknown, $p = 0.5$ is used, as it gives the maximum $n$. For a finite population of size $N$, the adjusted sample is

$n' = \frac{n}{1 + \dfrac{n-1}{N}}$

Example: for 95% confidence ($Z = 1.96$), $\sigma = 10$ and $E = 2$: $n = (1.96 \times 10 / 2)^2 = 96.04 \approx 97$.

Simple Random Sampling (SRS) is a probability sampling method in which every unit of the population has an equal chance of being selected, and every possible sample of a given size has the same chance of being chosen.

Types

• SRS with replacement (SRSWR): a selected unit is returned to the population before the next draw, so it can be selected again. Number of possible samples is $N^n$.
• SRS without replacement (SRSWOR): a selected unit is not returned. Number of possible samples is $\binom{N}{n}$.

Methods of selection

• Lottery method: each unit is given a number on a slip, the slips are mixed well and $n$ are drawn.
• Random number table or computer generated random numbers: units are numbered from 1 to $N$ and the selected numbers decide the sample.

Merits

• Simple and easy to understand and apply.
• Free from personal bias and selection is unbiased.
• Sample mean is an unbiased estimator of the population mean; sampling error can be measured.
• Basis for other sampling methods.

Demerits

• Needs a complete list (sampling frame) of the population.
• Selected units may be widely scattered, making the survey costly.
• Not suitable for a heterogeneous population, where stratified sampling is better.
• Sample may not represent small subgroups.

Use: best when the population is homogeneous and of moderate size.

(b) 2

On the basis of source of collection, data are of two types: primary data (collected first hand by the investigator) and secondary data (already collected by others).

(a) Arithmetic mean

Profit is a quantitative, additive measure, and the arithmetic mean uses all values and gives the average profit per unit or period.

(c) -1 to 1

The correlation coefficient satisfies $-1 \le r \le +1$, where $-1$ is perfect negative and $+1$ is perfect positive correlation.

(d) 4% of total variation in dependent variable has been explained by independent variable.

The coefficient of determination is $r^2 = (0.2)^2 = 0.04$, which is 4%.

(c) 0

Probability always lies between 0 and 1, so the minimum value is 0 (an impossible event).

(b) Mean = Median

The normal distribution is symmetric about its centre, so the mean, median and mode coincide.

(d) $(\bar{X}, \bar{Y})$

Both regression lines pass through the point of the two sample means, so they intersect at $(\bar{X}, \bar{Y})$.

(b) sample mean is unbiased estimator population mean.

In systematic sampling, each unit has an equal chance of being included, so the expected value of the sample mean equals the population mean (when $N$ is a multiple of the sampling interval $k$).

(d) n

For a chi-square distribution with $n$ degrees of freedom, the mean is $n$ and the variance is $2n$.

(d) (k-1, n-k)

For $k$ groups and $n$ total observations, the between-group degrees of freedom are $k-1$ and the within-group degrees of freedom are $n-k$, so $F$ has $(k-1,\; n-k)$ degrees of freedom.

Primary data are data collected first hand by the investigator for a specific purpose, directly from the original source. The process of collecting them involves the following steps.

1. Define the objective: state clearly what problem is to be studied and what information is needed.
2. Decide the scope and population: determine the units of study, the area covered and the time period.
3. Choose the method of collection:
   • Direct personal investigation (interview): the investigator meets respondents face to face; gives accurate and detailed data but is costly and time consuming.
   • Indirect oral investigation: data are obtained from persons who know about the subject.
   • Questionnaire / schedule: a set of written questions sent by mail or online (questionnaire) or filled by an enumerator (schedule).
   • Observation: recording behaviour or events directly.
   • Experiments: data produced under controlled conditions.
   • Telephone or online surveys: quick and cheaper for wide coverage.
4. Prepare the instrument: design a clear, short, unambiguous questionnaire and test it in a pilot survey, then correct defects.
5. Select the sample or plan a census: use a suitable sampling method if the whole population cannot be covered.
6. Train enumerators and collect data: field workers are trained and the actual collection is carried out with supervision.
7. Check and edit the data: verify completeness, consistency and accuracy before tabulation and analysis.

Primary data are reliable and suited to the purpose of the study, but collecting them costs more time, money and effort than using secondary data.

Sampling is the process of selecting a part (sample) of the units from a population, studying it, and using the results to draw conclusions about the whole population. It saves time and cost and is used when studying every unit (census) is impractical.

Sample size for estimating a mean

Given: $\sigma = 0.5$, error $E = 0.01$, confidence level 95% so $Z_{\alpha/2} = 1.96$.

$n = \left(\frac{Z_{\alpha/2}\,\sigma}{E}\right)^2 = \left(\frac{1.96 \times 0.5}{0.01}\right)^2 = \left(\frac{0.98}{0.01}\right)^2 = (98)^2 = 9604$

A sample of at least 9604 students must be taken to be 95% confident that the error of the estimate does not exceed 0.01.

Total items $= 50$, defective $= 20$, so non-defective $= 50-20 = 30$.

$P(\text{non-defective}) = \frac{\text{non-defective items}}{\text{total items}} = \frac{30}{50} = \frac{3}{5} = 0.6$

The probability that the selected item is non-defective is 0.6. (Check: $P(\text{defective}) = 20/50 = 0.4$, and $0.4 + 0.6 = 1$.)

Simple random sampling (SRS) is a probability sampling method in which every unit of the population has an equal chance of being selected, and every possible sample of size $n$ has the same chance of being chosen. Selection of one unit does not influence the selection of another.

Procedure

1. Prepare a complete list (sampling frame) of the $N$ units of the population and number them $1$ to $N$.
2. Decide the sample size $n$.
3. Select $n$ numbers using a lottery method or a table of random numbers / a random number generator.
4. The units carrying the selected numbers form the sample.

Types

• SRS without replacement (SRSWOR): a selected unit is not returned to the population; the number of possible samples is $\binom{N}{n}$ and each has probability $1/\binom{N}{n}$.
• SRS with replacement (SRSWR): a selected unit is returned before the next draw; the number of possible samples is $N^n$.

Example: A college has $N = 500$ BCA students and the principal wants the opinion of $n = 50$ of them. Each student is given a roll number from 1 to 500. Fifty distinct numbers are drawn using a random number generator (or 500 slips are mixed in a box and 50 are drawn one by one without replacement). The 50 students so chosen form a simple random sample; each student had the probability $50/500 = 0.1$ of being selected.

Merits: it is free from personal bias, simple to understand, and the sample is representative on average, so statistical theory (standard error, confidence intervals) applies directly.

Demerits: it needs a complete list of the population, it is costly if the population is widely scattered, and for a very heterogeneous population it may fail to represent all groups unless the sample is large.

Total items $= 30$, defective $= 10$, non-defective $= 20$. Two items are drawn without replacement.

(a) Both defective

$P = \frac{10}{30}\times\frac{9}{29} = \frac{90}{870} = \frac{3}{29} \approx 0.1034$

(equivalently $\binom{10}{2}/\binom{30}{2} = 45/435$).

(b) Both non-defective

$P = \frac{20}{30}\times\frac{19}{29} = \frac{380}{870} = \frac{38}{87} \approx 0.4368$

(equivalently $\binom{20}{2}/\binom{30}{2} = 190/435$).

Example of convenient sampling: to know students' opinion on the canteen, a researcher interviews the first 30 students he meets at the college gate.

Example of systematic sampling: from a list of 1000 students, a sample of $n = 50$ is needed. Then $k = 1000/50 = 20$; a random number between 1 and 20, say 7, is chosen, and the students numbered $7, 27, 47, 67, \dots$ are selected.

Here $n = 300$ is large and $p = 0.02$ is small, so the Poisson distribution is a suitable approximation to the binomial with

$\lambda = np = 300 \times 0.02 = 6,\qquad P(X = x) = \frac{e^{-\lambda}\lambda^{x}}{x!}$

Using $e^{-6} = 0.002479$:

(i) Exactly two defective articles

$P(X=2) = \frac{e^{-6}\,6^2}{2!} = \frac{0.002479\times 36}{2} = 0.0446$

(ii) At least two defective articles

$P(X\ge 2) = 1 - P(X=0) - P(X=1) = 1 - e^{-6}(1 + 6) = 1 - 7(0.002479) = 1 - 0.01735 = 0.9827$

Let $X \sim N(\mu = 53,\ \sigma = 3)$ and $Z = \dfrac{X-53}{3}$.

(i) $P(X < 57)$

$z = \frac{57-53}{3} = 1.33$

$P(Z < 1.33) = 0.9082$

(ii) $P(X > 64)$

$z = \frac{64-53}{3} = 3.67$

$P(Z > 3.67) = 1 - 0.9999 \approx 0.0001$

(iii) $P(50 < X < 56)$

$z_1 = \frac{50-53}{3} = -1,\qquad z_2 = \frac{56-53}{3} = 1$

$P(-1 < Z < 1) = 0.8413 - 0.1587 = 0.6826$

Binomial distribution: a discrete distribution for the number of successes $X$ in $n$ independent trials, each having only two outcomes (success/failure) with a constant probability of success $p$. Its probability function is

$P(X = x) = \binom{n}{x}p^{x}q^{\,n-x},\quad x = 0,1,\dots,n,\quad q = 1-p$

with mean $np$ and variance $npq$.

Here $n = 15$, $p = 0.7$, $q = 0.3$.

(i) At least 12 thick coatings

$P(X\ge 12) = P(12)+P(13)+P(14)+P(15)$

$P(12) = \binom{15}{12}(0.7)^{12}(0.3)^{3} = 455(0.013841)(0.027) = 0.1701$

$P(13) = \binom{15}{13}(0.7)^{13}(0.3)^{2} = 105(0.0096889)(0.09) = 0.0916$

$P(14) = \binom{15}{14}(0.7)^{14}(0.3) = 15(0.0067823)(0.3) = 0.0305$

$P(15) = (0.7)^{15} = 0.0047$

$P(X\ge 12) = 0.1701+0.0916+0.0305+0.0047 = 0.2969$

(ii) Exactly 10 thick coatings

$P(10) = \binom{15}{10}(0.7)^{10}(0.3)^{5} = 3003(0.0282475)(0.00243) = 0.2061$

For estimating a mean with known $\sigma$, the maximum error at confidence level $1-\alpha$ is $E = z_{\alpha/2}\dfrac{\sigma}{\sqrt n}$, so

$n = \left(\frac{z_{\alpha/2}\,\sigma}{E}\right)^{2}$

Given $\sigma = 0.95$, $E = 0.01$, and for 95% confidence $z_{0.025} = 1.96$:

$n = \left(\frac{1.96\times 0.95}{0.01}\right)^2 = (186.2)^2 = 34670.44$

Rounding up, the psychologist must take a sample of at least $n = 34{,}671$ measurements.

Let $X \sim N(\mu = 266,\ \sigma = 16)$ and $Z = \dfrac{X-266}{16}$.

(i) $P(240 < X < 270)$

$z_1 = \frac{240-266}{16} = -1.625,\qquad z_2 = \frac{270-266}{16} = 0.25$

$P(-1.625 < Z < 0.25) = P(Z<0.25) - P(Z<-1.625) = 0.5987 - 0.0521 = 0.5466$

(With table values rounded to $z=-1.63$, the area is $0.0516$, giving $\approx 0.547$.)

(ii) $P(X > 275)$

$z = \frac{275-266}{16} = 0.5625 \approx 0.56$

$P(Z > 0.56) = 1 - 0.7123 = 0.2877 \approx 0.287$

(iii) $P(X < 272)$

$z = \frac{272-266}{16} = 0.375 \approx 0.38$

$P(Z < 0.38) = 0.6480 \approx 0.646$

So about 54.7% of pregnancies last 240 to 270 days, about 28.7% last more than 275 days, and about 64.6% last less than 272 days (small differences are due to table rounding).

Poisson distribution: a discrete distribution for the number of occurrences $X$ of a rare event in a fixed interval, obtained as a limit of the binomial when $n$ is large, $p$ is small and $np = \lambda$ is finite:

$P(X=x) = \frac{e^{-\lambda}\lambda^{x}}{x!},\qquad x = 0,1,2,\dots$

Its mean and variance are both $\lambda$.

Here $p = 1/500$ and $n = 10$, so $\lambda = np = 10/500 = 0.02$ and $e^{-0.02} = 0.9802$.

(i) No defective lens

$P(X=0) = e^{-0.02} = 0.9802$

(ii) At least one defective lens

$P(X\ge 1) = 1 - P(X=0) = 1 - 0.9802 = 0.0198$

(iii) At most two defective lenses

$P(X\le 2) = e^{-0.02}\left(1 + 0.02 + \frac{0.02^2}{2}\right) = 0.9802(1.0202) = 0.99998 \approx 1.0000$

The sample size needed to estimate a mean with maximum error $E$ at 95% confidence, when $\sigma$ is known, is

$n = \left(\frac{z_{\alpha/2}\,\sigma}{E}\right)^2$

Given $\sigma = 1.40$, $E = 0.25$, $z_{0.025} = 1.96$:

$n = \left(\frac{1.96\times 1.40}{0.25}\right)^2 = \left(\frac{2.744}{0.25}\right)^2 = (10.976)^2 = 120.47$

Rounding up, the dean must take a sample of at least $n = 121$ students.

Statistics is used wherever computer applications must collect, summarise or draw conclusions from data. Its main applications are:

1. Machine learning and data mining: regression, classification, clustering, Bayesian methods and hypothesis testing underlie prediction and pattern discovery.
2. Database management and big data analytics: descriptive measures (mean, variance, correlation) support reports, dashboards and decision making.
3. Software engineering: defect and failure data are analysed with probability distributions to estimate reliability; sampling is used in testing; control charts monitor quality.
4. Performance modelling and simulation: queuing models, Poisson arrivals and random number generation are used to study server load, response time and network traffic.
5. Networks and security: anomaly and intrusion detection, spam filtering and error estimation are based on probabilistic models.
6. Image and speech processing, AI and NLP: noise removal, pattern recognition, speech recognition, search ranking and recommendation use statistical models.
7. Human-computer interaction and e-commerce: surveys, A/B tests and confidence intervals evaluate usability and user behaviour.
8. Data compression and cryptography: the frequency distribution of symbols (entropy) is used in coding, and randomness tests are used to check keys.

Hence statistics is a foundation for designing, evaluating and improving computer systems and for handling uncertainty in data.

Probability: a numerical measure of the chance that an event occurs. For an experiment with equally likely outcomes,

$P(E) = \frac{\text{number of outcomes favourable to } E}{\text{total number of outcomes}},\qquad 0\le P(E)\le 1$

Probability of A solving. Odds against A are $8:6$, so

$P(A) = \frac{6}{8+6} = \frac{3}{7},\qquad P(\bar A) = \frac{4}{7}$

Probability of B solving. Odds in favour of B are $14:10$, so

$P(B) = \frac{14}{14+10} = \frac{7}{12},\qquad P(\bar B) = \frac{5}{12}$

A and B work independently.

(i) Both solve it

$P(A\cap B) = \frac{3}{7}\times\frac{7}{12} = \frac{1}{4} = 0.25$

(ii) A solves it but B fails

$P(A\cap \bar B) = \frac{3}{7}\times\frac{5}{12} = \frac{5}{28} \approx 0.1786$

Example: the average height of all 5000 students of a university is a parameter $\mu$; the average height $\bar x$ of 100 students selected from them is a statistic used to estimate $\mu$.

Sampling distribution: the probability distribution of a statistic (such as the sample mean $\bar x$) calculated from all possible samples of the same size $n$ drawn from a population. Its mean, standard deviation (called the standard error) and shape are used for estimation and testing.

Population: $2, 4, 6, 8$, $N = 4$, $\mu = \dfrac{2+4+6+8}{4} = 5$.

(i) All samples of size 2 without replacement ($\binom42 = 6$) and their means:

Frequency distribution of the sample mean

(ii) Unbiasedness

$E(\bar x) = \frac{\sum f\bar x}{\sum f} = \frac{3+4+10+6+7}{6} = \frac{30}{6} = 5$

Since $E(\bar x) = 5 = \mu$, the sample mean is an unbiased estimate of the population mean.

Let $X \sim N(\mu = 68.22,\ \sigma^2 = 10.8)$, so $\sigma = \sqrt{10.8} = 3.286$. Batch size $= 1000$.

(a) Over 72 grams

$z = \frac{72-68.22}{3.286} = 1.15$

$P(X>72) = 1 - 0.8749 = 0.1251$

Expected number $= 1000\times 0.1251 \approx$ 125 products.

(b) Between 70 and 72 grams

$z_1 = \frac{70-68.22}{3.286} = 0.54,\qquad z_2 = 1.15$

$P(70<X<72) = 0.8749 - 0.7054 = 0.1695$

Expected number $\approx$ 169 products (about 170).

(c) Below 65 grams

$z = \frac{65-68.22}{3.286} = -0.98$

$P(X<65) = 0.1635$

Expected number $\approx$ 164 products.

Hypotheses

$H_0:\ \mu_1=\mu_2=\mu_3$ (mean breaking force is the same at the three positions)

$H_1:$ at least one mean differs. Level of significance $\alpha = 0.05$.

Data (taking the values as: Position 1: 90, 82, 79, 98, 83, 91; Position 2: 95, 83, 75, 102, 59, 90, 92; Position 3: 83, 89, 80, 94)

$N = 17$, grand total $T = 523+596+346 = 1465$, $\sum x^2 = 127873$.

Sums of squares

$CF = \frac{T^2}{N} = \frac{1465^2}{17} = 126248.53$

$SST = 127873 - 126248.53 = 1624.47$

$SSB = \sum\frac{T_i^2}{n_i} - CF = 126262.31 - 126248.53 = 13.78$

$SSW = SST - SSB = 1610.69$

ANOVA table

Decision: the tabulated value is $F_{0.05}(2,14) = 3.74$. Since the calculated $F = 0.06 < 3.74$, $H_0$ is not rejected. The differences among the sample means at the three positions are not significant at the 0.05 level.