Introduction to Statistics and Data Analysis

(a) Theoretical (expected) frequencies

The number of typing mistakes per page is usually fitted by a Poisson distribution.

Total pages: [ N=\sum f=122+60+15+2+1=200 ]

Mean number of mistakes per page: [ \bar X=\frac{\sum fX}{N} =\frac{0(122)+1(60)+2(15)+3(2)+4(1)}{200} =\frac{100}{200}=0.50 ]

So take (\lambda=0.50), and (e^{-0.5}=0.6065).

Poisson probability: [ P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!} ]

Expected frequency: [ E_x=N\cdot P(X=x) ]

Hence the theoretical frequencies are approximately:

[ \boxed{121.31,\ 60.65,\ 15.16,\ 2.53,\ 0.32,\ 0.03,\ 0.00} ]

for (X=0,1,2,3,4,5,\ge 6) respectively.

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(b) Paired (t)-test for effectiveness of training

Let [ d=\text{After training IQ}-\text{Before training IQ} ]

[ n=10,\qquad \sum d=29 ]

[ \bar d=\frac{29}{10}=2.9 ]

[ \sum d^2=4^2+5^2+(-1)^2+3^2+3^2+4^2+5^2+4^2+(-1)^2+3^2=127 ]

Sample standard deviation: [ s_d=\sqrt{\frac{\sum d^2-\frac{(\sum d)^2}{n}}{n-1}} ]

[ s_d=\sqrt{\frac{127-\frac{29^2}{10}}{9}} =\sqrt{\frac{127-84.1}{9}} =\sqrt{\frac{42.9}{9}} =2.1833 ]

Test statistic: [ t=\frac{\bar d}{s_d/\sqrt n} =\frac{2.9}{2.1833/\sqrt{10}} =\frac{2.9}{0.6904} =4.20 ]

Degrees of freedom: [ df=n-1=9 ]

At (5%) level of significance, for a one-tailed test: [ t_{0.05,9}=1.833 ]

Since [ t=4.20>1.833 ] we reject (H_0).

[ \boxed{\text{The training program is effective at the 5% level of significance.}} ]

The mean increase in IQ after training is (2.9) points, and this increase is statistically significant.

(a) Normal distribution

Given:

• $N = 5000$
• Mean $\mu = 60$
• Variance $\sigma^2 = 100$
• Standard deviation $\sigma = \sqrt{100} = 10$

Let $X$ be the marks obtained. Then

$Z = \frac{X-\mu}{\sigma} = \frac{X-60}{10} \sim N(0,1)$

i. Highest marks of poorest 20% students

The poorest 20% students are the bottom 20% of the distribution.
So we need the 20th percentile, i.e.

$P(X \le x) = 0.20$

From the standard normal table,

$P(Z \le -0.8416) = 0.20$

Thus,

$x = \mu + z\sigma = 60 + (-0.8416)(10)$

$x = 60 - 8.416 = 51.584$

Therefore, the highest marks of the poorest 20% students is approximately

$\boxed{51.58 \text{ marks} \approx 52 \text{ marks}}$

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ii. Minimum marks of top 10% students

Top 10% students means the upper 10% of the distribution.
So we need the 90th percentile:

$P(X \ge x) = 0.10$

or

$P(X \le x) = 0.90$

From the standard normal table,

$P(Z \le 1.2816) = 0.90$

Thus,

$x = \mu + z\sigma = 60 + (1.2816)(10)$

$x = 60 + 12.816 = 72.816$

Therefore, the minimum marks of the top 10% students is approximately

$\boxed{72.82 \text{ marks} \approx 73 \text{ marks}}$

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(b) Test for difference between two means

Given:

We test:

$H_0: \mu_1 = \mu_2$

$H_1: \mu_1 \ne \mu_2$

Level of significance:

$\alpha = 0.05$

Since both sample sizes are large, we use the $Z$-test for difference of means.

The standard error of the difference is

$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

$SE = \sqrt{\frac{36^2}{50} + \frac{40^2}{50}}$

$SE = \sqrt{\frac{1296}{50} + \frac{1600}{50}}$

$SE = \sqrt{25.92 + 32} = \sqrt{57.92} = 7.61$

The test statistic is

$Z = \frac{\bar{X}_1 - \bar{X}_2}{SE}$

$Z = \frac{1234 - 1215}{7.61} = \frac{19}{7.61} = 2.50$

For a two-tailed test at $\alpha = 0.05$, the critical value is

$Z_{0.025} = \pm 1.96$

Since

$|Z| = 2.50 > 1.96$

we reject $H_0$.

Therefore, there is sufficient evidence at the 5% level of significance to conclude that the average lengths of life of the two types of electric bulbs are significantly different.

$\boxed{\text{Reject } H_0: \text{ the two average lives are significantly different.}}$

Given: $r_{12}=0.51,\quad r_{13}=0.42,\quad r_{23}=0.71$

(i) Partial correlation coefficient $r_{12.3}$

Formula: $r_{12.3}=\frac{r_{12}-r_{13}r_{23}}{\sqrt{(1-r_{13}^2)(1-r_{23}^2)}}$

Substituting: $r_{12.3}=\frac{0.51-(0.42)(0.71)}{\sqrt{(1-0.42^2)(1-0.71^2)}}$

$=\frac{0.51-0.2982}{\sqrt{(0.8236)(0.4959)}}$

$=\frac{0.2118}{\sqrt{0.40842324}}$

$=\frac{0.2118}{0.63908}$

$\boxed{r_{12.3}=0.3314}$

So, the partial correlation between variables 1 and 2, keeping variable 3 constant, is 0.3314.

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(ii) Multiple correlation coefficient $R_{1.23}$

Formula: $R_{1.23}=\sqrt{\frac{r_{12}^2+r_{13}^2-2r_{12}r_{13}r_{23}}{1-r_{23}^2}}$

Substituting: $R_{1.23}=\sqrt{\frac{(0.51)^2+(0.42)^2-2(0.51)(0.42)(0.71)}{1-(0.71)^2}}$

$=\sqrt{\frac{0.2601+0.1764-0.304164}{0.4959}}$

$=\sqrt{\frac{0.132336}{0.4959}}$

$=\sqrt{0.266848}$

$\boxed{R_{1.23}=0.5166}$

So, the multiple correlation coefficient of variable 1 on variables 2 and 3 is 0.5166.

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Final Answers: $\boxed{r_{12.3}=0.3314,\qquad R_{1.23}=0.5166}$

Let $X$ be the number of heads obtained when 10 fair coins are tossed. Then

$X \sim \text{Binomial}\left(n=10,\ p=\frac12\right)$

Total number of possible outcomes is

$2^{10}=1024$

The probability of getting exactly $r$ heads is

$P(X=r)=\binom{10}{r}\left(\frac12\right)^{10}=\frac{\binom{10}{r}}{1024}$

(a) No head

$P(X=0)=\frac{\binom{10}{0}}{1024}=\frac{1}{1024}$

(b) One head

$P(X=1)=\frac{\binom{10}{1}}{1024}=\frac{10}{1024}=\frac{5}{512}$

(c) Exactly two heads

$P(X=2)=\frac{\binom{10}{2}}{1024}=\frac{45}{1024}$

(d) At least one head

$P(X\ge 1)=1-P(X=0)=1-\frac{1}{1024}=\frac{1023}{1024}$

Answers:

$(a)\ \frac{1}{1024},\quad (b)\ \frac{5}{512},\quad (c)\ \frac{45}{1024},\quad (d)\ \frac{1023}{1024}$

Total ways: $\binom{30}{3}=\frac{30\cdot29\cdot28}{6}=4060$

(i) 2 girls and 1 boy:

$\binom{18}{2}\binom{12}{1}=153\times12=1836$

$P=\frac{1836}{4060}=\frac{459}{1015}\approx0.452$

(ii) 2 boys and 1 girl:

$\binom{12}{2}\binom{18}{1}=66\times18=1188$

$P=\frac{1188}{4060}=\frac{297}{1015}\approx0.293$

Check: with 3 girls ($816$ ways) and 3 boys ($220$ ways), $1836+1188+816+220=4060$.

Given:

• Margin of error, $E = 10\% = 0.10$
• Confidence level: assumed $95\%$, so $Z_{\alpha/2} = 1.96$

For estimating a population proportion, the required sample size is:

$n = \frac{Z^2 p q}{E^2}$

where $q = 1-p$.

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(a) Previous estimate available: $p = 0.30$

Here, $p = 0.30,\quad q = 0.70,\quad E = 0.10,\quad Z = 1.96$

$n = \frac{(1.96)^2 \times 0.30 \times 0.70}{(0.10)^2}$

$n = \frac{3.8416 \times 0.21}{0.01}$

$n = \frac{0.806736}{0.01} = 80.6736 \approx 81$

Required sample size = 81 students.

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(b) No previous estimate available

When no prior estimate of $p$ is available, we take $p = 0.50$ and $q = 0.50$, because this gives the maximum required sample size.

$n = \frac{(1.96)^2 \times 0.50 \times 0.50}{(0.10)^2}$

$n = \frac{3.8416 \times 0.25}{0.01}$

$n = \frac{0.9604}{0.01} = 96.04 \approx 97$

Required sample size = 97 students.

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Answer:

• With previous estimate $p = 0.30$: 81 students
• With no previous estimate: 97 students

(Assumed 95% confidence level. If another confidence level is specified, $Z$ changes accordingly.)

Given population: (1, 3, 5, 7, 9) with (N = 5), sample size (n = 2), sampling without replacement.

(i) Enumeration of all possible samples of size 2

Number of possible samples (= \binom{5}{2} = 10)

(ii) Verification that the mean of the sampling distribution equals the population mean

Population mean:

$\mu = \frac{1+3+5+7+9}{5} = \frac{25}{5} = 5$

Mean of the sampling distribution of sample means:

$\bar{\bar{x}} = \frac{\sum \bar{x}}{C(N,n)} = \frac{2+3+4+5+4+5+6+6+7+8}{10}$

$\bar{\bar{x}} = \frac{50}{10} = 5$

Conclusion:

$\bar{\bar{x}} = \mu = 5$

Hence, the mean of the sampling distribution of sample means is equal to the population mean. (\blacksquare)

Short Notes (any two)

(a) Non-probability Sampling

Meaning. Non-probability sampling is a sampling technique in which sample units are selected on the basis of the personal judgment, convenience or purpose of the investigator rather than by chance. The probability of selection of each unit is not known, and some units of the population may have no chance of being selected at all. It is therefore also called non-random, purposive or judgment sampling.

Main types

Merits

1. Simple, quick and cheap; no need of a complete sampling frame.
2. Useful in exploratory, pilot and qualitative studies, and where the population is not clearly defined.
3. Gives good results when the investigator has expert knowledge of the population.

Demerits

1. Selection bias — the sample may not be representative.
2. Sampling error cannot be measured, so reliability of estimates is unknown.
3. Results cannot be safely generalised to the population, and tests of significance are not strictly valid.

Use. Market research, opinion polls, pilot surveys and studies where time, cost and accessibility are the main constraints.

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(b) Type I and Type II Errors

In testing a hypothesis we set a null hypothesis $H_0$ against an alternative $H_1$. Since the decision is based on sample evidence, it may be wrong in two ways. The four possible situations are:

Type I error. Rejecting $H_0$ when it is actually true. Its probability is $\alpha$, the level of significance (usually 1% or 5%). It is also called the producer's risk or a false positive. Example: a good consignment of goods is rejected by a quality-control test.

Type II error. Accepting $H_0$ when it is actually false. Its probability is $\beta$, and it is called the consumer's risk or a false negative. The quantity $1-\beta$ is the power of the test, i.e. the probability of correctly rejecting a false $H_0$. Example: a defective consignment is accepted as good.

Relationship

1. For a fixed sample size, reducing $\alpha$ (shifting the critical region) increases $\beta$, and vice versa — there is a trade-off between the two errors.
2. Both errors can be reduced simultaneously only by increasing the sample size $n$.
3. Conventionally $\alpha$ is fixed first (as it is the more serious error in most problems), and then the test with the smallest $\beta$ is chosen.

Hypotheses. $H_0: \mu_d = 0$ (the program is not effective) against $H_1: \mu_d > 0$ (weight falls, i.e. the program is effective), one-tailed. $\alpha = 0.01$.

Paired differences $d$ = before $-$ after:

$\sum d = 28$, $\sum d^2 = 258$, $n = 8$.

$\bar d = \frac{28}{8} = 3.5$

$s_d^2 = \frac{258 - 28^2/8}{7} = \frac{160}{7} = 22.857,\qquad s_d = 4.781$

Test statistic:

$t = \frac{\bar d}{s_d/\sqrt{n}} = \frac{3.5}{4.781/\sqrt 8} = \frac{3.5}{1.690} = 2.071$

Critical value: $t_{0.01}(7\ \text{d.f.}) = 2.998$ (one-tailed).

Decision: Since $2.071 < 2.998$, we fail to reject $H_0$.

Conclusion: At the 1% level the data do not provide sufficient evidence that the prescribed exercise program is effective in reducing weight.

Hypothesis test for difference between two population proportions

Let
$p_1$ = proportion of arthritis victims relieved by the first drug
$p_2$ = proportion of arthritis victims relieved by the second drug

Null hypothesis:
$H_0: p_1 = p_2$ i.e. the effectiveness of the two drugs does not differ significantly.

Alternative hypothesis:
$H_1: p_1 \ne p_2$ (two-tailed test)

Given:

$$n_1=200,\quad x_1=120,\quad n_2=150,\quad x_2=80$$

Sample proportions:

$$\hat p_1=\frac{120}{200}=0.60$$ $$\hat p_2=\frac{80}{150}=0.5333$$

Pooled proportion:

$$\hat p=\frac{x_1+x_2}{n_1+n_2} =\frac{120+80}{200+150} =\frac{200}{350} =0.5714$$ $$\hat q=1-\hat p=0.4286$$

Standard error of difference:

$$SE=\sqrt{\hat p\hat q\left(\frac1{n_1}+\frac1{n_2}\right)}$$ $$SE=\sqrt{0.5714\times0.4286\left(\frac1{200}+\frac1{150}\right)}$$ $$SE=\sqrt{0.2449\times0.011667} =\sqrt{0.002857} =0.05345$$

Test statistic:

$$Z=\frac{\hat p_1-\hat p_2}{SE} =\frac{0.60-0.5333}{0.05345} =\frac{0.06667}{0.05345} =1.247$$

At 5% level of significance for a two-tailed test, the critical value is:

$$Z_{0.025}=1.96$$

Since

$$|Z|=1.247<1.96$$

we fail to reject $H_0$. Also, the p-value is approximately $0.213$, which is greater than $0.05$.

Conclusion:
At the 5% level of significance, there is no significant difference in the effectiveness of the two drugs. Hence, the two drugs are equally effective in providing relief to arthritis sufferers.

Mean of the data:

$m = \frac{\sum fx}{\sum f} = \frac{0(211)+1(90)+2(19)+3(5)+4(0)}{325} = \frac{143}{325} = 0.44$

Poisson probabilities with $m = 0.44$ and $e^{-0.44} = 0.6440$:

$P(x) = \frac{e^{-m}m^x}{x!}$

• $P(0) = 0.6440$
• $P(1) = 0.6440 \times 0.44 = 0.2834$
• $P(2) = 0.6440 \times 0.1936 / 2 = 0.0623$
• $P(3) = 0.6440 \times 0.08518 / 6 = 0.0091$
• $P(4) = 0.6440 \times 0.03748 / 24 = 0.0010$

Expected frequencies $= 325 \times P(x)$:

The fitted Poisson frequencies (about 209, 92, 20, 3, 0) are close to the observed ones, so the Poisson distribution fits the data well.

Let (X) be the length of stay in days.
Given:

[ \mu = 60,\qquad \sigma = 15 ]

Since (X) is approximately normal,

[ Z=\frac{X-\mu}{\sigma}=\frac{X-60}{15} ]

(a) Probability that length of stay is more than 50 days

[ P(X>50)=P\left(Z>\frac{50-60}{15}\right) ]

[ =P(Z>-0.6667)=P(Z<0.6667) ]

Using the standard normal table,

[ P(Z<0.6667)\approx 0.7475 ]

[ \boxed{P(X>50)\approx 0.7475 \text{ or } 74.75%} ]

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(b) Probability that length of stay is less than 30 days

[ P(X<30)=P\left(Z<\frac{30-60}{15}\right) ]

[ =P(Z<-2) ]

Using the standard normal table,

[ P(Z<-2)=0.0228 ]

[ \boxed{P(X<30)\approx 0.0228 \text{ or } 2.28%} ]

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(c) Probability that length of stay is between 30 and 80 days

[ P(30<X<80)=P\left(\frac{30-60}{15}<Z<\frac{80-60}{15}\right) ]

[ =P(-2<Z<1.3333) ]

[ =\Phi(1.3333)-\Phi(-2) ]

From the standard normal table,

[ \Phi(1.3333)\approx 0.9088,\qquad \Phi(-2)=0.0228 ]

[ P(-2<Z<1.3333)=0.9088-0.0228=0.8860 ]

[ \boxed{P(30<X<80)\approx 0.8860 \text{ or } 88.60%} ]

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Final answers:

[ \boxed{P(X>50)\approx 0.7475} ]

[ \boxed{P(X<30)\approx 0.0228} ]

[ \boxed{P(30<X<80)\approx 0.8860} ]

So, about 74.75% of patients stay more than 50 days, about 2.28% stay less than 30 days, and about 88.60% stay between 30 and 80 days.

Hypotheses. $H_0: \mu_1 = \mu_2$ (mean weights of the two processes are equal) against $H_1: \mu_1 \ne \mu_2$ (two-tailed). $\alpha = 0.05$.

Given: $n_1 = 250,\ \bar x_1 = 120,\ s_1 = 10$; $n_2 = 400,\ \bar x_2 = 124,\ s_2 = 12$. The samples are large, so use the z-test.

$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{100}{250} + \frac{144}{400}} = \sqrt{0.76} = 0.8718$

$z = \frac{\bar x_1 - \bar x_2}{SE} = \frac{120 - 124}{0.8718} = -4.59$

Critical value: $z_{0.025} = 1.96$.

Decision: $|z| = 4.59 > 1.96$, so reject $H_0$.

Conclusion: The mean weights of items from the two processes differ significantly at the 5% level.

Given: [ r_{12}=0.59,\qquad r_{13}=0.46,\qquad r_{23}=0.77 ]

(i) Partial correlation coefficient (r_{12.3})

The partial correlation between (X_1) and (X_2), keeping (X_3) constant, is [ r_{12.3}=\frac{r_{12}-r_{13}r_{23}} {\sqrt{(1-r_{13}^2)(1-r_{23}^2)}} ]

Substituting the values: [ r_{12.3}=\frac{0.59-(0.46)(0.77)} {\sqrt{(1-0.46^2)(1-0.77^2)}} ]

[ =\frac{0.59-0.3542} {\sqrt{(0.7884)(0.4071)}} ]

[ =\frac{0.2358}{\sqrt{0.32095764}} ]

[ =\frac{0.2358}{0.566531} ]

[ \boxed{r_{12.3}=0.4162} ]

So the partial correlation coefficient is approximately 0.4162.

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(ii) Multiple correlation coefficient (R_{1.23})

The multiple correlation of (X_1) on (X_2) and (X_3) is [ R_{1.23}=\sqrt{\frac{r_{12}^2+r_{13}^2-2r_{12}r_{13}r_{23}} {1-r_{23}^2}} ]

Substituting the values: [ R_{1.23}=\sqrt{\frac{(0.59)^2+(0.46)^2-2(0.59)(0.46)(0.77)} {1-(0.77)^2}} ]

[ =\sqrt{\frac{0.3481+0.2116-0.417956}{0.4071}} ]

[ =\sqrt{\frac{0.141744}{0.4071}} ]

[ =\sqrt{0.34818} ]

[ \boxed{R_{1.23}=0.5901} ]

So the multiple correlation coefficient is approximately 0.5901.

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Final Answers

[ \boxed{r_{12.3}=0.4162} ] [ \boxed{R_{1.23}=0.5901} ]

Given:

• Number of tosses, (n = 400)
• Number of heads, (X = 240)
• Level of significance, (\alpha = 5% = 0.05)

Hypotheses:

• (H_0): The coin is unbiased, i.e. (P(\text{head}) = p = 0.5)
• (H_1): The coin is biased, i.e. (p \neq 0.5) (two-tailed test)

Test statistic: Under (H_0), the sample proportion is

[ \hat{p} = \frac{X}{n} = \frac{240}{400} = 0.6 ]

The standard error of the proportion is

[ SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5 \times 0.5}{400}} = \sqrt{\frac{0.25}{400}} = \sqrt{0.000625} = 0.025 ]

[ z = \frac{\hat{p} - p}{SE} = \frac{0.6 - 0.5}{0.025} = \frac{0.1}{0.025} = 4 ]

Critical value: At 5% level of significance for a two-tailed test,

[ z_{0.025} = 1.96 ]

Decision: Since (|z| = 4 > 1.96), we reject (H_0).

Conclusion: The calculated value of (z) is much greater than the critical value, so the null hypothesis is rejected at the 5% level of significance. Hence, the coin cannot be regarded as unbiased; the data provide sufficient evidence that the coin is biased.

Given:

• Sample size, $n = 100$
• Sample mean, $\bar{x} = 30{,}214$ km
• Population standard deviation, $\sigma = 860$ km

Confidence limits for the population mean:

$\bar{x} \pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}$

Standard error:

$SE = \frac{\sigma}{\sqrt{n}} = \frac{860}{\sqrt{100}} = \frac{860}{10} = 86 \text{ km}$

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95% confidence limits ($z_{0.025} = 1.96$):

$30{,}214 \pm 1.96(86) = 30{,}214 \pm 168.56$

$= (30{,}214 - 168.56,\; 30{,}214 + 168.56) = (30{,}045.44,\; 30{,}382.56)$

95% confidence limits: 30,045.44 km to 30,382.56 km

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99% confidence limits ($z_{0.005} = 2.58$):

$30{,}214 \pm 2.58(86) = 30{,}214 \pm 221.88$

$= (30{,}214 - 221.88,\; 30{,}214 + 221.88) = (29{,}992.12,\; 30{,}435.88)$

99% confidence limits: 29,992.12 km to 30,435.88 km

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Conclusion: We are 95% confident that the true mean life of the tyres lies between 30,045.44 km and 30,382.56 km, and 99% confident that it lies between 29,992.12 km and 30,435.88 km. The 99% interval is wider than the 95% interval because a higher confidence level requires a larger margin of error.

Hypotheses. $H_0: \sigma_1^2 = \sigma_2^2$ (variability of boys' and girls' marks is the same) against $H_1: \sigma_1^2 \ne \sigma_2^2$ (two-tailed). $\alpha = 0.05$.

Given: Boys $n_1 = 11,\ s_1^2 = 81$; Girls $n_2 = 10,\ s_2^2 = 121$.

Test statistic (larger variance in the numerator):

$F = \frac{121}{81} = 1.494$

Degrees of freedom: $v_1 = 10 - 1 = 9$ (girls, numerator), $v_2 = 11 - 1 = 10$ (boys, denominator).

Critical value: for the two-tailed test at 5%, use the upper 2.5% point: $F_{0.025}(9, 10) \approx 3.78$.

Decision: Since $1.494 < 3.78$, we fail to reject $H_0$.

Conclusion: There is no significant difference in the variability of marks between boys and girls at the 5% level.

Total ways of selecting 3 children from 8:

$N = \binom{8}{3} = 56$

(i) No girl (all 3 are boys)

Favourable ways:

$\binom{5}{3} = 10$

$\therefore P(\text{no girl}) = \frac{10}{56} = \frac{5}{28}$

(ii) 2 boys and 1 girl

Favourable ways:

$\binom{5}{2} \times \binom{3}{1} = 10 \times 3 = 30$

$\therefore P(\text{2 boys and 1 girl}) = \frac{30}{56} = \frac{15}{28}$

Answers:

$(i)\ \frac{5}{28} \qquad (ii)\ \frac{15}{28}$

(a) Systematic Random Sampling

Definition. Systematic random sampling is a probability sampling method in which the first unit is selected at random from the first $k$ units and the remaining units are selected at a fixed interval, called the sampling interval $k$. It is also known as interval sampling.

Procedure.

1. Prepare a list of all $N$ units of the population and number them from $1$ to $N$.
2. Decide the sample size $n$ and compute the sampling interval $k=\frac{N}{n}.$ If $N/n$ is not an integer, a circular systematic sampling or a suitable modified procedure is used; simple rounding down may not give exactly $n$ units or may omit some end units.
3. Select one random number $r$ between $1$ and $k$ by the lottery method or a random number table. This gives the first selected unit.
4. Then select every $k$-th unit after $r$, i.e. $r,\; r+k,\; r+2k,\; \dots,\; r+(n-1)k.$

Example. If $N=100$ and $n=20$, then $k=5$. If the random start is $r=3$, the sample consists of units $3,8,13,18,\dots,98$.

Merits.

• Simple and easy to operate; only one random number is needed.
• Less time, labour and cost than simple random sampling.
• The sample is spread evenly over the whole population, so it is generally representative.
• Useful when units are already arranged in a natural order, such as houses along a street, ledger pages or a queue.
• Convenient for large populations.

Demerits.

• A sampling frame or ordered list is generally required, though a separate list may not be necessary if units occur in a natural sequence.
• If the population is arranged in a periodic pattern matching the interval $k$, the sample becomes biased.
• It is not simple random sampling because all possible samples of size $n$ do not have equal probability. When $N=nk$, only $k$ possible samples exist, although each unit has equal inclusion probability $n/N$.
• Less efficient than simple random sampling when periodicity is present.

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(b) Simple Random Sampling

Definition. Simple random sampling 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 an equal probability of being chosen. In with-replacement sampling, selections are independent; in the usual without-replacement sampling, selections are not independent, but inclusion probabilities remain equal.

Methods.

1. Lottery method — each unit is given a slip, slips are mixed thoroughly, and the required number is drawn at random.
2. Random number table method — units are numbered and selected by reading numbers from a random number table, ignoring repetitions.
3. Computer/calculator generated random numbers.

Properties.

• Sampling is usually done without replacement. The number of possible samples of size $n$ from $N$ units is $\binom{N}{n}=\frac{N!}{n!(N-n)!}.$
• The sample mean $\bar{x}$ is an unbiased estimator of the population mean $\mu$.
• For simple random sampling without replacement, $S.E.(\bar{x})=\frac{\sigma}{\sqrt{n}}\sqrt{\frac{N-n}{N-1}}.$ For simple random sampling with replacement, $S.E.(\bar{x})=\frac{\sigma}{\sqrt{n}}.$

Merits.

• Free from personal bias; selection is purely by chance.
• Simple to understand and easy to apply.
• Sampling error can be measured and the reliability of estimates tested.
• Results can be generalised to the whole population.
• Serves as the theoretical base for other sampling designs.

Demerits.

• A complete sampling frame of all units is essential.
• Not practicable for very large populations.
• Selected units are widely scattered, so more time, cost and administrative difficulty may arise.
• May fail to include rare or minority characteristics of the population.
• Requires units to be numbered and individually identifiable.

(a) Chi-square test for association

Let

• (H_0): There is no association between students’ interest in business administration and ability in statistics.
• (H_1): There is an association between them.
• Level of significance: (\alpha=0.05)

Observed frequencies are given. Expected frequency:

[ E_{ij}=\frac{(\text{Row total})(\text{Column total})}{N} ]

Expected frequencies:

Now,

[ \chi^2=\sum \frac{(O-E)^2}{E} ]

[ \chi^2= \frac{(230-229.5)^2}{229.5} +\frac{(120-130.5)^2}{130.5} +\frac{(100-90)^2}{90} ]

[ +\frac{(150-153)^2}{153} +\frac{(100-87)^2}{87} +\frac{(50-60)^2}{60} ]

[ +\frac{(130-127.5)^2}{127.5} +\frac{(70-72.5)^2}{72.5} +\frac{(50-50)^2}{50} ]

[ \chi^2 \approx 5.76 ]

Degrees of freedom:

[ df=(r-1)(c-1)=(3-1)(3-1)=4 ]

Critical value at 5% level:

[ \chi^2_{0.05,4}=9.488 ]

Since

[ 5.76 < 9.488 ]

we fail to reject (H_0).

Conclusion: At the 5% level of significance, there is no significant association between students’ interest in business administration and their ability in statistics.

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(b) Test for difference between two proportions

Let

• (p_E) = proportion of tea drinkers in Eastern region
• (p_W) = proportion of tea drinkers in Western region

Hypotheses:

[ H_0: p_E \le p_W ]

[ H_1: p_E > p_W ]

This is a right-tailed test.

Sample proportions:

[ \hat p_E=\frac{450}{550}=0.8182 ]

[ \hat p_W=\frac{340}{450}=0.7556 ]

Pooled proportion:

[ \hat p=\frac{450+340}{550+450}=\frac{790}{1000}=0.79 ]

[ \hat q=1-\hat p=0.21 ]

Standard error:

[ SE=\sqrt{\hat p \hat q\left(\frac{1}{n_1}+\frac{1}{n_2}\right)} ]

[ SE=\sqrt{0.79\times0.21\left(\frac{1}{550}+\frac{1}{450}\right)} ]

[ SE\approx 0.02589 ]

Test statistic:

[ z=\frac{\hat p_E-\hat p_W}{SE} ]

[ z=\frac{0.8182-0.7556}{0.02589} ]

[ z\approx 2.42 ]

Critical value at 5% level for a right-tailed test:

[ z_{0.05}=1.645 ]

Since

[ 2.42 > 1.645 ]

we reject (H_0).

Conclusion: At the 5% level of significance, the proportion of tea drinkers is significantly higher in the Eastern Development Region than in the Western Development Region.

(a) Normal distribution

Given:

• $N = 10{,}000$
• Mean $\mu = 55$
• Standard deviation $\sigma = 10$

Let $X$ be the marks obtained. Then

$Z=\frac{X-\mu}{\sigma}=\frac{X-55}{10}$

(i) Lowest marks of top 10% of students

Top 10% means 90% of students are below this mark. So we need the 90th percentile.

$P(Z>z)=0.10 \Rightarrow P(Z<z)=0.90$

From the standard normal table,

$z=1.2816$

Hence,

$X=\mu+z\sigma=55+1.2816(10)$

$X=55+12.816=67.816$

$\boxed{X \approx 67.82 \text{ marks}}$

So, the lowest marks of the top 10% students is about 67.82 marks.

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(ii) Highest marks of lowest 20% of students

Lowest 20% means 20% of students are below this mark. So we need the 20th percentile.

$P(Z<z)=0.20$

From the standard normal table,

$z=-0.8416$

Hence,

$X=\mu+z\sigma=55+(-0.8416)(10)$

$X=55-8.416=46.584$

$\boxed{X \approx 46.58 \text{ marks}}$

So, the highest marks of the lowest 20% students is about 46.58 marks.

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(b) Test whether average height of sailor is greater than soldier

Let:

• Soldier: $n_1=10$, $\bar{x}_1=167$, $\sum (x_1-\bar{x}_1)^2=153$
• Sailor: $n_2=6$, $\bar{x}_2=169$, $\sum (x_2-\bar{x}_2)^2=60$

We use an independent sample $t$-test.

Hypotheses

$H_0:\mu_2 \le \mu_1 \quad \text{(average height of sailor is not greater than soldier)}$

$H_1:\mu_2 > \mu_1 \quad \text{(average height of sailor is greater than soldier)}$

where $\mu_1$ = mean height of soldier and $\mu_2$ = mean height of sailor.

Level of significance:

$\alpha=0.05$

Sample variances

$s_1^2=\frac{\sum (x_1-\bar{x}_1)^2}{n_1-1}=\frac{153}{9}=17$

$s_2^2=\frac{\sum (x_2-\bar{x}_2)^2}{n_2-1}=\frac{60}{5}=12$

Pooled variance

$s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}$

$s_p^2=\frac{153+60}{10+6-2}=\frac{213}{14}=15.2143$

Standard error of difference

$SE=\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}$

$SE=\sqrt{15.2143\left(\frac{1}{10}+\frac{1}{6}\right)}$

$SE=\sqrt{15.2143(0.2667)}=\sqrt{4.0571}=2.0142$

Test statistic

$t=\frac{\bar{x}_2-\bar{x}_1}{SE}=\frac{169-167}{2.0142}$

$t=\frac{2}{2.0142}=0.993$

Critical value

Degrees of freedom:

$df=n_1+n_2-2=10+6-2=14$

For a one-tailed test at $\alpha=0.05$ and $df=14$,

$t_{0.05,14}=1.761$

Decision

Since

$t_{\text{cal}}=0.993 < t_{\text{crit}}=1.761$

we fail to reject $H_0$.

Conclusion

At the 5% level of significance, there is not sufficient evidence to suggest that the average height of sailors is greater than that of soldiers. Although the sample mean height of sailors is greater than soldiers, the difference is not statistically significant.

Total students $=20$, girls $=14$, boys $=6$.

Two students are selected at random, so total ways: $n(S)={}^{20}C_2=\frac{20\times19}{2}=190$

(a) Probability that both are girls

Favourable ways: $n(A)={}^{14}C_2=\frac{14\times13}{2}=91$

$P(\text{both girls})=\frac{91}{190}\approx 0.479$

(b) Probability that one is a girl and one is a boy

Favourable ways: $n(B)={}^{14}C_1\times{}^{6}C_1=14\times6=84$

$P(\text{one girl and one boy})=\frac{84}{190}=\frac{42}{95}\approx 0.442$

Answers: $(a)\ \frac{91}{190}\qquad (b)\ \frac{42}{95}$

Given simple correlation coefficients:

$r_{12}=0.51,\quad r_{13}=0.45,\quad r_{23}=0.75$

The formula for partial correlation between (X_1) and (X_2) keeping (X_3) constant is:

$r_{12.3}=\frac{r_{12}-r_{13}r_{23}}{\sqrt{(1-r_{13}^2)(1-r_{23}^2)}}$

Substituting the values:

$r_{12.3}=\frac{0.51-(0.45)(0.75)}{\sqrt{(1-0.45^2)(1-0.75^2)}}$

$=\frac{0.51-0.3375}{\sqrt{(0.7975)(0.4375)}}$

$=\frac{0.1725}{0.59068}$

$\boxed{r_{12.3}=0.2920}$

So, the partial correlation between (X_1) and (X_2) keeping (X_3) constant is approximately 0.292.

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The formula for partial correlation between (X_1) and (X_3) keeping (X_2) constant is:

$r_{13.2}=\frac{r_{13}-r_{12}r_{23}}{\sqrt{(1-r_{12}^2)(1-r_{23}^2)}}$

Substituting the values:

$r_{13.2}=\frac{0.45-(0.51)(0.75)}{\sqrt{(1-0.51^2)(1-0.75^2)}}$

$=\frac{0.45-0.3825}{\sqrt{(0.7399)(0.4375)}}$

$=\frac{0.0675}{0.56895}$

$\boxed{r_{13.2}=0.1186}$

So, the partial correlation between (X_1) and (X_3) keeping (X_2) constant is approximately 0.119.

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Final Answers:

(i) (r_{12.3}=0.292)

(ii) (r_{13.2}=0.119)

Given:

• Sample size, $n = 128$
• Population standard deviation, $\sigma = 40$
• Sample mean, $\bar{x} = 160$

Since the population standard deviation is known, use the $z$-test for confidence limits:

$\text{Confidence limits} = \bar{x} \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

Standard error:

$SE = \frac{\sigma}{\sqrt{n}} = \frac{40}{\sqrt{128}} = \frac{40}{11.3137} = 3.5355$

95% Confidence Limits

For 95%, $z_{0.025} = 1.96$

$160 \pm 1.96(3.5355) = 160 \pm 6.93$

$\text{Limits} = (153.07,\ 166.93)$

99% Confidence Limits

For 99%, $z_{0.005} = 2.576$

$160 \pm 2.576(3.5355) = 160 \pm 9.11$

$\text{Limits} = (150.89,\ 169.11)$

Answer:

• 95% confidence limits for the population mean: 153.07 to 166.93
• 99% confidence limits for the population mean: 150.89 to 169.11

(If $z = 2.58$ is used for 99%, the limits are approximately 150.88 to 169.12.)

Testing a population proportion (one-sample z-test)

Given:

• Sample size, $n = 250$
• Number of graduates in the sample, $x = 60$
• Sample proportion, $\hat{p} = \dfrac{x}{n} = \dfrac{60}{250} = 0.24$
• Hypothesised population proportion, $p_0 = 0.25$
• Level of significance, $\alpha = 0.05$

Hypotheses: $H_0: p = 0.25 \quad \text{(the population proportion of graduates is 25\%)}$ $H_1: p \neq 0.25 \quad \text{(two-tailed test)}$

Test statistic: $z = \frac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}$

Substitution: $z = \frac{0.24 - 0.25}{\sqrt{\dfrac{0.25 \times 0.75}{250}}} = \frac{-0.01}{\sqrt{0.00075}} = \frac{-0.01}{0.027386} = -0.365$

Critical value: For a two-tailed test at $\alpha = 0.05$, the critical values are $z_{0.025} = \pm 1.96$

Decision: Since $|z| = 0.365 < 1.96$, the calculated value does not fall in the critical (rejection) region. We fail to reject $H_0$.

Conclusion: At the 5% level of significance, there is not enough evidence to conclude that the population proportion of graduates in the company differs from 25%. The sample result is consistent with the claim that 25% of the employees are graduates.

Given: For a Poisson variate $X$, mean $m = \lambda = 2$.

The Poisson probability mass function is $P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}=\frac{e^{-2}2^x}{x!}$

(a) $P(X<3)$

$P(X<3)=P(X=0)+P(X=1)+P(X=2)$

$=e^{-2}\left(\frac{2^0}{0!}+\frac{2^1}{1!}+\frac{2^2}{2!}\right)$

$=e^{-2}(1+2+2)=5e^{-2}$

$=5(0.135335)=0.6767$

$\boxed{P(X<3)=0.6767}$

(b) $P(X>2)$

$P(X>2)=1-P(X\le 2)$

$=1-P(X<3)$

$=1-0.6767=0.3233$

$\boxed{P(X>2)=0.3233}$

Standard deviation

For a Poisson distribution, $\sigma=\sqrt{\lambda}=\sqrt{2}=1.4142$

$\boxed{\sigma=1.4142}$

Final Answers:

• $P(X<3)=0.6767$
• $P(X>2)=0.3233$
• Standard deviation $=1.4142$

(a) Sampling Errors

Meaning. A sampling error is the difference between the value of a statistic computed from a sample and the true value of the corresponding population parameter. It arises because observation is confined to a part of the population rather than the whole of it. If $\bar{x}$ is the sample mean and $\mu$ the population mean, the sampling error is $\bar{x}-\mu$.

Nature and causes.

• It is inherent in the sampling method itself and cannot be avoided completely in a sample survey; it can only be reduced.
• It is not the result of mistakes in collection, recording or tabulation — those are non-sampling errors.
• It decreases as the sample size increases, and the standard error varies inversely with the square root of the sample size. For simple random sampling without replacement, $SE(\bar{x})=\frac{\sigma}{\sqrt{n}}\sqrt{\frac{N-n}{N-1}}.$ For simple random sampling with replacement, $SE(\bar{x})=\sigma/\sqrt{n}$.
• It also depends on the variability of the population ($\sigma$) and on the design of the sample.

Control measures. Increase the sample size; use a more efficient sampling design where appropriate (e.g., stratified or systematic sampling in place of simple random sampling); and select the sample strictly at random so that bias does not enter. Since the sampling distribution of $\bar{x}$ is approximately normal for large $n$, the probable limits within which the population mean lies can be stated with a known confidence coefficient, e.g. $\bar{x}\pm1.96\,SE(\bar{x})$ at 95% confidence.

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(b) Stratified Random Sampling

Meaning. Stratified random sampling is a probability sampling design in which the heterogeneous population is first divided into homogeneous, mutually exclusive and collectively exhaustive groups called strata, and then a simple random sample is drawn independently from each stratum. The strata are formed on the basis of some characteristic closely related to the variable under study — income, age, sex, region, etc.

Allocation of sample size. If $N$ is the population size, $N_i$ the size of the $i$-th stratum and $n$ the total sample size:

• Proportional allocation: $n_i=n\cdot\dfrac{N_i}{N}$ — the sample is distributed in proportion to stratum size.
• Neyman allocation: $n_i\propto N_i\sigma_i$ — larger samples are taken from strata with greater variability, giving the smallest variance for a given $n$ when sampling costs are equal. For unequal costs, optimum allocation is $n_i\propto N_i\sigma_i/\sqrt{c_i}$, where $c_i$ is the cost per unit in stratum $i$.

Estimate and precision. The stratified estimate of the population mean is

$$\bar{x}_{st}=\frac{1}{N}\sum_{i=1}^{k}N_i\bar{x}_i,\qquad V(\bar{x}_{st})=\sum_{i=1}^{k}\left(\frac{N_i}{N}\right)^{2}\frac{\sigma_i^{2}}{n_i}\left(\frac{N_i-n_i}{N_i-1}\right).$$

The corresponding estimate of the population total is $N\bar{x}_{st}=\sum_{i=1}^{k}N_i\bar{x}_i$.

Advantages.

1. More precise (smaller standard error) than simple random sampling when strata are internally homogeneous.
2. Ensures representation of every important section of the population, so no group is left out.
3. Separate estimates for each stratum, and hence for sub-groups, can be obtained.
4. Convenient and administratively easier, since field work can be organised stratum-wise.

Limitations. It requires prior knowledge of the population to form suitable strata; stratification is less useful if the strata are not homogeneous; and it involves more time, cost and computation than simple random sampling.