Statistics II Solved Paper 2013 FBISE

Statistics II Solved Paper 2013 FBISE
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In this article, titled “Statistics II Solved Paper 2013 FBISE,” we will walk through a comprehensive analysis of the solved paper, addressing various aspects of statistics as encountered in the 2013 FBISE exam. The post will cover multiple-choice questions (MCQs), short-answer questions, numerical problems, and extensive questions on fundamental topics such as probability distribution, binomial and hypergeometric distributions, normal distribution, sampling distribution, inferential statistics (hypothesis testing and estimation), and association using the Chi-square test.

Each topic will be explained in detail, with solutions to guide students through the process of answering such questions efficiently. Whether you’re preparing for an exam or just brushing up on your statistics knowledge, this post serves as a helpful resource for mastering key statistical concepts and methodologies. This topic is equally important for the students of statistics across all the major Boards and Universities such as FBISE, BISERWP, BISELHR, MU, DU, PU, NCERT, CBSE & others & across all the statistics, business & finance disciplines.

Table of Contents

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Statistics II Solved Paper 2013 FBISE

MCQS

Fill the relevant bubble against each question according to curriculum:
1Neither tail nor head comes up on the upper side of the coin, is an example of________set.
A)FiniteB)Infinite
C)NullD)None of these
2If three six-faced dice are rolled. the possible outcomes are:
A)2B)216
C)36D)18
3If P(A) = 2/3 , P(B) = 1/3. B is a complementary event of event A.
A)YesB)No
C)IrrelevantD)None of these
4If E(X)=10 a=2 and b=5 then E(ax-b)=
A)20B)15
C)25D)None of these
5Median of the Binomial distribution (½+ ½) will be:
A)1/2B)1/4
C)5D)10
6For a symmetrical binomial distribution p and q are
A)EqualB)Unequal
C)IrrelevantD)None of these
7In normal distribution the 4th mean moment about mean is equal to
A) 

    \[ 13\sigma^{4}\ \]

B) 

    \[ 3\sigma^{1}\ \]

C) 

    \[ 3\sigma^{2}\ \]

D)None of these
8If a normal distribution σ = 10, then M.D will approximately be:
A)8B)10
C)12D)14
9SE of X̅ for without replacement sampling is:
A) 

    \[ \frac{\mathbf{\sigma}}{\mathbf{n}}\ \]

B) 

    \[ \frac{\sigma}{\sqrt{n}}\ \]

C) 

    \[ \frac{\sigma}{\sqrt{n}}.\sqrt{\frac{N - n}{N - 1}}\ \]

D)None of these
10Population proportion is a:
A)VariableB)Statistic
C)ParameterD)None of these
11[ x̅±σ/√n ] is the confidence interval for μ ;when the level of confidence is:
A)80%B)90%
C)95%D)99%
12Rejecting Ho , when Ho is actually false is a
A)Type-I errorB)Type-II error
C)Wrong decisionD)Correct decision
13For a = 0.05, the critical value of Z for two tailed test is
A)±2.33B)±1.96
C)±2.58D)None of these
14Which test-statistic should be preferred to test the population mean when the population variance is known?
A)t-statisticB)z-statistic
C)χ²-statisticD)None of these
15The calculated value of the Chi-square could NOTbe:
A)PositiveB)Negative
C)ZeroD)None of these
16If a contingency table consists of four rows and three columns, the d. f will be
A)6B)7
C)9D)12
17Chi-square distribution is a, _______ distribution.
A)SymmetricalB)Negatively skewed
C)Positively skewedD)None of these

Short Questions

(i) Show that in a single throw with two dice, the probability of throwing more than 7 is equal to that of throwing less than 7.

Solution

Sample Space η(S) = 6² = 36

Event: Sum less than seven η(A) = 15

Event: Sum more than seven η(B) = 15

Sum Table

1, 1 = 22, 1 = 33, 1 = 44, 1 = 55, 1 = 66, 1 = 7
1, 2 = 32, 2 = 43, 2 = 54, 2 = 65, 2 = 76, 2 = 8
1, 3 = 42, 3 = 53, 3 = 64, 3 = 75, 3 = 86, 3 = 9
1, 4 = 52, 4 = 63, 4 = 74, 4 = 85, 4 = 96, 4 = 10
1, 5 = 62, 5 = 73, 5 = 84, 5 = 95, 5 = 106, 5 = 11
1, 6 = 72, 6 = 83, 6 = 94, 6 = 105, 6 = 116, 6 = 12

    \[ \mathbf{P}\left( \mathbf{A} \right)\mathbf{= \ }\frac{\mathbf{\eta(A)}}{\mathbf{\eta(S)}}\mathbf{= \ }\frac{\mathbf{15}}{\mathbf{36}}\ \]

    \[ \mathbf{P}\left( \mathbf{B} \right)\mathbf{= \ }\frac{\mathbf{\eta(B)}}{\mathbf{\eta(S)}}\mathbf{= \ }\frac{\mathbf{15}}{\mathbf{36}}\ \]

So P(A) = P(B)

(ii) Find the probability that on a single draw from a pack of playing cards, we draw a diamond card or picture card or both.

Solution

    \[ \mathbf{Total\ Cards\ \eta(S)\ = \ 52}\ \]

    \[ \mathbf{Picture\ Cards\ \eta(F)\ = \ 12}\ \]

    \[ \mathbf{Diamond\ Cards\ \eta(D)\ = \ 13}\ \]

    \[ \mathbf{\eta}\left( \mathbf{F \cap D} \right)\mathbf{= 3}\ \]

    \[ \mathbf{P}\left( \mathbf{F} \right)\mathbf{=}\frac{\mathbf{\eta(F)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{12}}{\mathbf{52}}\ \]

    \[ \mathbf{P}\left( \mathbf{F} \right)\mathbf{=}\frac{\mathbf{\eta(D)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{13}}{\mathbf{52}}\ \]

    \[ \mathbf{P}\left( \mathbf{F \cap D} \right)\mathbf{=}\frac{\mathbf{\eta}\left( \mathbf{F \cap D} \right)}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{52}}\ \]

Three diamond cards are already included above in picture cards.

    \[ \mathbf{P}\left( \mathbf{P \cup D} \right)\mathbf{= P}\left( \mathbf{F} \right)\mathbf{+ P}\left( \mathbf{D} \right)\mathbf{- P(F \cap D)}\ \]

    \[ \mathbf{P}\left( \mathbf{P \cup D} \right)\mathbf{=}\frac{\mathbf{12}}{\mathbf{52}}\mathbf{+}\frac{\mathbf{13}}{\mathbf{52}}\mathbf{-}\frac{\mathbf{3}}{\mathbf{52}}\ \]

    \[ \mathbf{P}\left( \mathbf{P \cup D} \right)\mathbf{=}\frac{\mathbf{22}}{\mathbf{52}}\mathbf{=}\frac{\mathbf{11}}{\mathbf{26}}\ \]

(iii) Write down the properties of Mathematical Expectation.

Answer:

E(Constant) = Constant

E(aX) + aE(X), if “a” is a constant but a ≠ 0

E(X + a) = E(X) + a, if “a” is a constant

E(aX + b) = aE(X) + b if “a” & “b” are constants

E(X±Y) = E(X) ± E(Y)

E(XY) = E(X).E(Y), if X & Y are independent random variables

E[X – E(X)] = 0

(iv) If for f(x) = (6-⎮7- x⎮)/36 for x 2, 3, 4, 5, 6, ……..12, find the variance of the random variable “X”.

Solution

xf(x)xf(x)x²f(x)
21/362/364/36
32/366/3618/36
43/3612/3648/36
54/3620/36100/36
65/3630/36180/36
76/3642/36294/36
85/3640/36320/36
94/3636/36324/36
103/3630/36300/36
112/3622/36242/36
121/3612/36144/36
  ∑xf(x)∑x²f(x)
  252/361974/36

µ =∑xf(x) = 252/36 = 7

Var(X) = E(X²) – [E(X)]²

Var(X) = ∑x²f(x)– [∑xf(x)]²

Var(X) = 54.84 – (7)²

Var(X) = 54.84 – 49 = 5.84

(v) What is Binomial Distribution and what are its properties?

Answer: Binomial distribution is a function which is used to calculate the binomial experiment.

Function is given below:

    \[ \mathbf{P(X\ = \ x)\ = \ }\)\textbf{b(x, n, p)} \(\begin{pmatrix} \mathbf{n} \\ \mathbf{x} \\ \end{pmatrix}\mathbf{\ }\mathbf{p}^{\mathbf{x}}\mathbf{\ }\mathbf{q}^{\mathbf{n - x}}\ \]

for x = 0, 1, 2, 3 …..n.

Where n = sample size, x = number of successes, n – x = number of failure

P =success and q = failure = 1 – p

Properties

  1. The mean of binomial distribution µ = np
  2. Variance 𝜎² = npq
  3. Standard Deviation σ= √npq
  4. ∑px = 1
  5. If p = q = ½ then probability distribution is symmetrical
  6. For negative skewness P>q
  7. For positive skewness P<q

(vi) A random variable x is binomial distributed with mean 3 & variance 2. Compute P(x = 6).

Solution

Mean = np = 3 & Var = npq = 2

    \[ \mathbf{npq\ = \ npq}\ \]

    \[ \mathbf{3}\mathbf{q\ = \ 2\ \ }\ \]

    \[ \mathbf{q\ =}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{\ = \ 0.67}\ \]

    \[ \mathbf{If\ q\ = \ 0.67\ then\ p\ = \ 1\ -\ q\ = \ 1\ -\ 0.67\ = \ 0.33}\ \]

    \[ \mathbf{npq\ = \ npq}\ \]

    \[ \mathbf{n(0.33)(0.67)\ = \ 2}\ \]

    \[ \mathbf{n\ = \ 2/0.2211\ = \ 9\ Approx}\ \]

So now n = 9, p = 0.33, q = 0.67 and x = 6

    \[ \mathbf{P(X\ = \ x)\ = \ b(x,\ n,\ p)\ }\begin{pmatrix}\mathbf{n} \\\mathbf{x} \\\end{pmatrix}\mathbf{\ }\mathbf{p}^{\mathbf{x}}\mathbf{\ }\mathbf{q}^{\mathbf{n - x}}\mathbf{\ \ \ for\ x\ = \ 0}\ \]

    \[ \mathbf{P(X\ = \ 6)\ = \ b(6,\ 9,\ 0.33)\ }\begin{pmatrix}\mathbf{9} \\\mathbf{6} \\\end{pmatrix}\mathbf{\ }\mathbf{(0.33)}^{\mathbf{6}}\mathbf{\ }\mathbf{(0.67)}^{\mathbf{9 - 6}}\ \]

    \[ \mathbf{P(X\ = \ 6)\ = \ 84\ \times \ 0.001291\ \times \ 0.300763\ = \ 0.0326}\ \]

(vii) If X is binomial random-variable with n = 5 & P = 0.6 then find E(2X – 3) and Var(2X – 3).

Solution

N = 5, p = 0.6 & q = 0.4 so

E(X) = np = 5×0.6 = 3

Var(X) = npq = 5 x 0.6 x 0.4 = 1.2

So E(2X – 3) = 2E(X) – 3 = 2(3) – 3 = 6 – 3 = 3

Variance (X) = E(X)² – [E(X)]²

1.2 = E(X)² – (3)²

E(X)² = 1.2 + 9

E(X)² = 10.2

Var(2x – 3) = E(2X – 3)² – [E(2X – 3)]²

Where E(2X – 3) = 2E(X) – 3 = 2(3) – 3 = 6 – 3 = 3 &

E(2X – 3)² = E(4X² – 12X + 9) = 4(10.2) – 12(3) + 9 = 40.8 – 36 + 9 = 13.8

Var(2x – 5) = 13.8 – (3)² = 13.8 – 9 = 4.8

(viii) Write down the properties of the Normal distribution.

Answer:

  1. The mean, median and mode are equal.
  2. Mean deviation is 0.7979 of its standard deviation.
  3. Quartile Deviation is 0.6745 of its standard deviation.
  4. The curve is symmetrical about ordinate at X̅ = µ.
  5. The two quartiles are at equal distance from the mean and are at distance of 0.6745 𝜎.

(ix) The two quartiles of the normal distribution are 9 & 18 respectively. Find the Mean and Standard Deviation of the distribution.

Solution

Here Q1 = 9 & Q2 = 18

    \[ \mathbf{\mu = \ }\frac{\mathbf{Q}\mathbf{1 + Q}\mathbf{3}}{\mathbf{2}}\mathbf{= \ \mu = \ }\frac{\mathbf{9 + 18}}{\mathbf{2}}\mathbf{= \ \mu = \ }\frac{\mathbf{27}}{\mathbf{2}}\mathbf{= 13.5}\ \]

Q1 =µ – 0.6745 𝜎

9 = 13.5 – 0.6745 𝜎 or

13.5 – 0.6745 𝜎 = 9

– 0.6745 𝜎 = 9 – 13.5

– 0.6745 𝜎 = – 4.5

    \[ \mathbf{\sigma = \ }\frac{\mathbf{4.5}}{\mathbf{0.6745}}\mathbf{= 6.671}\ \]

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(x) Define Standardized Normal Variate. Also write equation of the normal curve for standardized normal variate.

Answer:

The standardized Normal Variate or SNV which converts or transforms a normal random variable in to standard normal form. It is denoted by Z and it follows the standard normal distribution having mean zero (0) and standard deviation (1). Its formula is given below:

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

Where:

Z = Standardized normal variate

X = Observed value

μ = Mean of the distribution

σ = Standard deviation of the distribution

Equation of the Normal Curve for SNV:

The equation for the standard normal curve (probability density function) is:

    \[ \mathbf{f}\left( \mathbf{Z} \right)\mathbf{= \ }\frac{\mathbf{1}}{\sqrt{\mathbf{2}\mathbf{\pi}}}\mathbf{\ }\mathbf{e}^{\mathbf{-}\frac{\mathbf{z}^{\mathbf{2}}}{\mathbf{2}}}\ \]

Where:

Z = Standardized normal variate

e = Euler’s number (approximately 2.71828)

π = Mathematical constant (approximately 3.14159)

(xi) Differentiate between Probability & Non-probability Sampling.

Answer:

Probability Sampling

Probability sampling which is also called random sampling is a sampling technique in which sample selection is purely chance based and it has no human control while choosing. In probability sampling every unit of the population has equal chance of being selected or non-zero probability. Common examples of probability sampling are:

  • Simple Random Sampling
  • Stratified Random Sampling
  • Systematic Random Sampling

Non Probability Sampling

In contrast non-probability sampling or non-random sampling is a technique in which sample of the population is not purely selected by chance but determined by some judgement or person’s predesigned intention. In some cases, this technique is considered as helpful in which judgement is evident but in most of cases non-probability sampling is not considered suitable because we cannot determine the error out of this.

Statistics II Solved Paper 2013 FBISE

(xii) What is the value of the finite population correction factor (f.p.c) when n = 18 & N = 125?

Solution

    \[ \mathbf{fpc = \ }\sqrt{\left\lbrack \frac{\mathbf{N - n}}{\mathbf{N - 1}} \right\rbrack}\ \]

    \[ \mathbf{fpc = \ }\sqrt{\left\lbrack \frac{\mathbf{125 - 18}}{\mathbf{125 - 1}} \right\rbrack}\ \]

    \[ \mathbf{fpc = \ }\sqrt{\left\lbrack \frac{\mathbf{107}}{\mathbf{124}} \right\rbrack}\mathbf{= 0.9289}\ \]

(xiii) Differentiate between Point-estimate and interval estimate.

Answer:

Point Estimate

A point estimate of a population parameter is a single value used to estimate the population parameter. For example, the sample mean x̅ is a point estimate of the population mean μ.

Interval Estimate

An interval estimate is defined by two numbers, between which a population parameter is said to lie. For example, a < μ < b is an interval estimate for the population mean μ. It indicates that the population mean is greater than a, but less than b.

(xiv) Given n = 500, p̂= 0.08, Z = 0.005 = 2.58. Find the 99% Confidence Interval for the population proportion p.

Solution

A 100(1 – α) % confidence interval for p is

Data

    \[ \mathbf{Here\ n\ = \ 500,\ }\widehat{\mathbf{p}}\mathbf{\ = \ 0.08,\ }\widehat{\mathbf{q}}\mathbf{\ = \ 0.92\ \&\ }\mathbf{Z}_{\frac{\mathbf{\alpha}}{\mathbf{2}}}\mathbf{\ = \ }\mathbf{Z}_{\mathbf{0.005\ }}\mathbf{= \ 2.58\ }\ \]

Test Statistic

    \[ \widehat{\mathbf{p}}\mathbf{\pm}\mathbf{Z}_{\mathbf{\alpha/2}}\sqrt{\frac{\widehat{\mathbf{p}}\widehat{\mathbf{q}}}{\mathbf{n}}}\mathbf{\ \ }\ \]

Calculation

    \[ \widehat{\mathbf{p}}\mathbf{-}\mathbf{Z}_{\frac{\mathbf{\alpha}}{\mathbf{2}}}\sqrt{\frac{\widehat{\mathbf{p}}\widehat{\mathbf{q}}}{\mathbf{n}}}\mathbf{\ \ \ \ < p\ < (\ }\widehat{\mathbf{p}}\mathbf{) +}\mathbf{Z}_{\mathbf{\alpha/2}}\sqrt{\frac{\widehat{\mathbf{p}}\widehat{\mathbf{q}}}{\mathbf{n}}}\mathbf{\ \ }\ \]

    \[ \mathbf{0.08 - 2.58}\sqrt{\frac{\left( \mathbf{0.08} \right)\left( \mathbf{0.92} \right)}{\mathbf{500}}}\mathbf{\ \ \ \ < p\ < 0.08 + 2.58}\sqrt{\frac{\left( \mathbf{0.08} \right)\left( \mathbf{0.92} \right)}{\mathbf{500}}}\mathbf{\ \ \ \ }\ \]

Conclusion

    \[ \mathbf{0.0487 < p < 0.1113}\ \]

(xv) Describe the difference between One-sided & Two sided tests.

Answer:

One-Tailed or One Sided Test

It is also called directional test. This type of test is used to test the significance and determine if there is a relationship between the variables in one direction.

Two-Tailed or Two Sided Test

It is also called non-directional test. This type of test is used to test the significance and determine if there is a relationship between the variables in either direction. In two tailed test α is divided by two and taking half on each side of the bell curve.

(xvi) Given Ho: µ1-µ2 = 0 vs H1: µ1-µ2 ≠ 0 when n1 = 11, n2 = 14, X̅1 = 75, X̅2 = 60, (n1 – 1) S1² = 372.1, (n2 – 1) S2² = 365.34

Solution

Step 1 Hypothesis

Ho: µ1-µ2 = 0

H1: µ1-µ2 ≠ 0

Step 2 Level of Significance

Level of Significance α = 0.05

Step 3 Test Statistic

    \[ \mathbf{t =}\frac{\left( \overline{\mathbf{X}}\mathbf{1 -}\overline{\mathbf{X}}\mathbf{2} \right)\mathbf{- \ (\mu 1 - \ \mu 2)}}{\mathbf{sp}\sqrt{\left\lbrack \frac{\mathbf{1}}{\mathbf{n}\mathbf{1}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}\mathbf{2}} \right\rbrack}}\ \]

Step 4 Critical Region

    \[ \left| \mathbf{t} \right|_{\mathbf{\alpha}\mathbf{/2(df)}}\mathbf{= \ \ }\left| \mathbf{t} \right|_{\mathbf{0.025\ (23)}}\mathbf{\ = 2.069}\ \]

Where d.f = v = n1+n2-2=11 + 14 – 2 = 23

Step 5 Calculation

    \[ \mathbf{S}\mathbf{p}^{\mathbf{2}}\mathbf{=}\frac{\left( \mathbf{n}\mathbf{1 - 1} \right)\mathbf{s²1 +}\left( \mathbf{n}\mathbf{2 - 1} \right)\mathbf{s²2}}{\mathbf{n}\mathbf{1 + n}\mathbf{2 - 2}}\ \]

    \[ \mathbf{S}\mathbf{p}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{372.1 + 365.34}}{\mathbf{11 + 14 - 2}}\ \]

    \[ \mathbf{S}\mathbf{p}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{737.44}}{\mathbf{23}}\mathbf{= 32.062}\ \]

    \[ \mathbf{sp = \ }\sqrt{\mathbf{s}\mathbf{p}^{\mathbf{2}}}\ \]

    \[ \mathbf{sp = \ }\sqrt{\mathbf{32.062}}\ \]

    \[ \mathbf{sp = 5.662}\ \]

    \[ \mathbf{t =}\frac{\left( \overline{\mathbf{X}}\mathbf{1 -}\overline{\mathbf{X}}\mathbf{2} \right)\mathbf{- \ (\mu 1 - \ \mu 2)}}{\mathbf{sp}\sqrt{\left\lbrack \frac{\mathbf{1}}{\mathbf{n}\mathbf{1}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}\mathbf{2}} \right\rbrack}}\ \]

    \[ \mathbf{t =}\frac{\left( \mathbf{75 - 60} \right)\mathbf{- \ (0)}}{\mathbf{5.662}\sqrt{\left\lbrack \frac{\mathbf{1}}{\mathbf{11}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{14}} \right\rbrack}}\ \]

    \[ \mathbf{t =}\frac{\mathbf{15}}{\mathbf{2.281}}\mathbf{= 6.576}\ \]

Step 6 Conclusion

Statistical Decision: Calculated value falls in critical region or exceeds than tabulated value so reject Null Hypothesis Ho & accepts H1 that µ1-µ2 ≠ 0.

(xvii) What is the difference between Simple and Composite Hypothesis?

Answer:

Simple Hypothesis

In Simple hypothesis, all the parameters of the distribution are known or specified. For Example, if per gallon mileage coverage of trucks is normally distributed with σ² = 9, the hypothesis that its population mean μ is say, 50 km, that is Ho:μ=50, we have stated a simple hypothesis because the mean and variance together specify a normal distribution completely.

Composite Hypothesis

In contrast to simple hypothesis, a hypothesis which is not simple or we can say in which all the parameters of the distribution are not specified is called composite hypothesis. For example, if per gallon mileage coverage of trucks is normally distributed with σ² = 9, the hypothesis that its population mean μ is greater than, 50 km, that is Ho:μ >50, or Ho:μ =50, σ² < 9, the hypothesis becomes a composite hypothesis because we do not know the exact distribution of the population in either case.

(xviii) Given the following information (α) = 54, (αβ) = 16, (B) = 490 & n = 1000. Show whether attributes A & B are positively associated, negatively associated or independent.

Solution

(αβ) = 16

n = (B) + (β)

1000 = 490 + (β)

(β) = 1000 – 490 = 510

n = (A) + (α)

1000 = (A) + 54

(A) = 1000 – 54 = 946

(β) = (Aβ) + (αβ)

510 = (Aβ) + 16

(Aβ) = 510 – 16 = 494

(A) = (AB) + (Aβ)

946 = (AB) + 494

(AB) = 946 – 494 = 452

(α) = (αB) + (αβ)

54 = (αB) + 16

(αB) = 54 – 16 = 38

Now (αβ) = 16, (AB) =452, (Aβ) =494 & (αB) =38

    \[ \mathbf{Q =}\frac{\left( \mathbf{AB} \right)\left( \mathbf{\alpha\beta} \right)\mathbf{- \ (A\beta)(\alpha B)}}{\left( \mathbf{AB} \right)\left( \mathbf{\alpha\beta} \right)\mathbf{+ \ (A\beta)(\alpha B)}}\ \]

    \[ \mathbf{Q = \ }\frac{\left( \mathbf{452} \right)\left( \mathbf{16} \right)\mathbf{- \ (494)(38)}}{\left( \mathbf{452} \right)\left( \mathbf{16} \right)\mathbf{+ \ (494)(38)}}\ \]

    \[ \mathbf{Q = \ }\frac{\mathbf{7232 - \ 18772}}{\mathbf{7232 + \ 18772}}\mathbf{\ }\ \]

    \[ \mathbf{Q =}\frac{\mathbf{- \ 11540}}{\mathbf{26004}}\mathbf{= - 0.4438\ Negative\ Association}\ \]

(xix) What is meant by Association of attributes?

Answer:

Attributes are characteristics of qualitative variable to which we assign some numbers to characteristics so the association of attributes is simply their relation with each other which we can measure through association or Chi Square.

Extensive Questions

Q.3 a. A random variable X has the following probability distribution:

X-3-2-101
P(X=x)1/161/4K1/41/16

Find: (i) The value of K (ii) P(X<0) and P(X>1)

Solution:

Solution (a)

XP(X)Rearranged P(X)
-31/161/16
-244/16
-1KK (6/16)
01/44/16
11/161/16

    \[ \left( \mathbf{i} \right)\mathbf{\ As\ we\ know\ P(X) = \ 1\ so}\ \]

    \[ \frac{\mathbf{1}}{\mathbf{16}}\mathbf{+}\frac{\mathbf{4}}{\mathbf{16}}\mathbf{+ K +}\frac{\mathbf{4}}{\mathbf{16}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{16}}\mathbf{= 1}\ \]

    \[ \mathbf{K = 1 -}\frac{\mathbf{1}}{\mathbf{16}}\mathbf{+}\frac{\mathbf{4}}{\mathbf{16}}\mathbf{+}\frac{\mathbf{4}}{\mathbf{16}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{16}}\ \]

    \[ \mathbf{K =}\frac{\mathbf{6}}{\mathbf{16}}\ \]

    \[ \mathbf{P}\left( \mathbf{X < 0} \right)\mathbf{= P}\left( \mathbf{- 1} \right)\mathbf{+ P}\left( \mathbf{- 2} \right)\mathbf{+ P( - 3)}\ \]

    \[ \mathbf{P}\left( \mathbf{X < 0} \right)\mathbf{=}\frac{\mathbf{6}}{\mathbf{16}}\mathbf{+}\frac{\mathbf{4}}{\mathbf{16}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{16}}\ \]

    \[ \mathbf{P}\left( \mathbf{X < 0} \right)\mathbf{=}\frac{\mathbf{11}}{\mathbf{16}}\ \]

    \[ \mathbf{P}\left( \mathbf{X > 1} \right)\mathbf{= Not\ Possible}\ \]

b. A finite population consists of numbers 2, 2, 4, 6 and 5, written on 5 tags of different colours. Draw all possible random samples of size 2 without replacement from this population and find their means. Construct the sampling distribution of the sample mean and verify that:

    \[ \left( \mathbf{i} \right)\mathbf{\ \mu}\overline{\mathbf{x}}\mathbf{= \ }\mathbf{\mu}\ \]

    \[ \left( \mathbf{ii} \right)\mathbf{\ }\mathbf{\sigma}^{\mathbf{2}}\overline{\mathbf{x}}\mathbf{= \ }\frac{\mathbf{\sigma}^{\mathbf{2}}}{\mathbf{n}}\left\lbrack \frac{\mathbf{N}\mathbf{-}\mathbf{n}}{\mathbf{N}\mathbf{-}\mathbf{1}} \right\rbrack\ \]

Solution:

Population: 2, 2, 4, 6, 5

N = 5, n = 2, Without Replacement

    \[ \mathbf{Sample\ Space}\mathbf{\ }\mathbf{\eta}\left( \mathbf{s}\right)\mathbf{= \ }\begin{bmatrix}\mathbf{N} \\\mathbf{C} \\\mathbf{n} \\\end{bmatrix}\mathbf{= \ }\begin{bmatrix}\mathbf{5} \\\mathbf{C} \\\mathbf{2} \\\end{bmatrix}\mathbf{= 10}\ \]

All Possible Samples

2, 22, 42, 62, 52, 42, 62, 54, 64, 56, 5

Means of Sampling Distribution

SamplesSum of Sample MeansX̅
2, 242
2, 463
2, 684
2, 573.5
2, 463
2, 684
2, 573.5
4, 6105
4, 594.5
6, 5115.5
Sampling Distribution of X̅
X̅ffX̅fX̅²
2124
32618
3.52724.5
42832
4.514.520.25
51525
5.515.530.25
Sum1038154
 ∑f=∑fX̅ =∑fX̅² =

Mean, Variance & S.D of Sampling Distribution

    \[ \mathbf{µ}\overline{\mathbf{x}}\mathbf{=}\frac{\mathbf{\sum fx\overline{}}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{38}}{\mathbf{10}}\mathbf{= 3.8}\ \]

    \[ \mathbf{\sigma ²}_{\mathbf{x\overline{}}}\mathbf{=}\frac{\mathbf{\sum fx\overline{}²}}{\mathbf{\sum f}}\mathbf{-}\left( \frac{\mathbf{\sum fx\overline{}}}{\mathbf{\sum f}} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{\sigma ²}_{\mathbf{x\overline{}}}\mathbf{=}\frac{\mathbf{154}}{\mathbf{10}}\mathbf{-}\left( \frac{\mathbf{38}}{\mathbf{10}} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{\sigma ²}_{\mathbf{x\overline{}}}\mathbf{= 15.4 - 14.44}\ \]

    \[ \mathbf{\sigma ²}_{\mathbf{x\overline{}}}\mathbf{= 0.96}\ \]

Population Parameters

XX²
24
24
416
636
525
  
∑X = 19∑X̅² = 85

    \[ \mathbf{µ =}\frac{\mathbf{\sum X}}{\mathbf{N}}\mathbf{=}\frac{\mathbf{19}}{\mathbf{5}}\mathbf{= 3.8}\ \]

    \[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{\sum}\mathbf{X}^{\mathbf{2}}}{\mathbf{N}}\mathbf{-}\left( \frac{\mathbf{\sum X}}{\mathbf{N}} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{85}}{\mathbf{5}}\mathbf{-}\left( \frac{\mathbf{19}}{\mathbf{5}} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{= 17 - 14.44}\ \]

    \[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{= 2.56}\ \]

Verification

    \[ \mathbf{\mu}\overline{\mathbf{x}}\mathbf{= \mu}\ \]

    \[ \mathbf{3.8\ = \ 3.8}\ \]

    \[ \mathbf{\sigma ²}_{\mathbf{x\overline{}}}\mathbf{=}\frac{\mathbf{\sigma}^{\mathbf{2}}}{\mathbf{n}}\left\lbrack \frac{\mathbf{N - n}}{\mathbf{N - 1}} \right\rbrack\ \]

    \[ \mathbf{0.96 =}\frac{\mathbf{2.56}}{\mathbf{2}}\left\lbrack \frac{\mathbf{5 - 2}}{\mathbf{5 - 1}} \right\rbrack\ \]

    \[ \mathbf{0.96 = 0.96}\ \]

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Q.4 a. If X ̴ N (30, 31, 36), then find: (i) P(X>20), (i) P(X≤25)

Solution:

To solve the problem, we first interpret the information provided:

X∼N (μ=30, σ²=31, n=36)

μ=30: Mean of the distribution.

σ²=31: Variance of the population distribution.

n=36: Sample size.

Test Statistic

    \[ \mathbf{Z =}\frac{\mathbf{X - \mu}}{\frac{\mathbf{\sigma}}{\sqrt{\mathbf{n}}}}\ \]

Calculation (i) P(X>20)

    \[ \mathbf{Z =}\frac{\mathbf{X - \mu}}{\frac{\mathbf{\sigma}}{\sqrt{\mathbf{n}}}}\ \]

    \[ \mathbf{Z =}\frac{\mathbf{20 - 30}}{\frac{\mathbf{5.5677}}{\sqrt{\mathbf{36}}}}\ \]

    \[ \mathbf{Z =}\frac{\mathbf{- 10}}{\mathbf{0.92795}}\mathbf{= - 10.77}\ \]

Calculation (ii) P(X≤25)

    \[ \mathbf{Z =}\frac{\mathbf{X - \mu}}{\frac{\mathbf{\sigma}}{\sqrt{\mathbf{n}}}}\ \]

    \[ \mathbf{Z =}\frac{\mathbf{25 - 30}}{\frac{\mathbf{5.5677}}{\sqrt{\mathbf{36}}}}\ \]

    \[ \mathbf{Z =}\frac{\mathbf{- 5}}{\mathbf{0.92795}}\mathbf{= - 5.388}\ \]

b. In a test given to two groups of students, the marks obtained are given below: Test the hypothesis that µ1 = µ2 at 5% level of significance, assuming that σ1² = σ2²

G-I91113111591214
G-II101210149810 

Solution:

Supporting Calculations W1

X1X1²X2X2²
98110100
1112112144
1316910100
1112114196
15225981
981864
1214410100
14196  
∑X1=94∑X1²=1138∑X1=73∑X2²=785

    \[ {\overline{\mathbf{X}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{\sum X}_{\mathbf{1}}}{\mathbf{n}_{\mathbf{1}}}\mathbf{=}\frac{\mathbf{94}}{\mathbf{8}}\mathbf{= 11.75}\ \]

    \[ \mathbf{S}_{\mathbf{1}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{n}_{\mathbf{1}}\mathbf{- 1}}\mathbf{\ }\left\lbrack \mathbf{\sum}\mathbf{X}_{\mathbf{1}}^{\mathbf{2}}\mathbf{-}\frac{\left( \mathbf{\sum}\mathbf{X}_{\mathbf{1}} \right)^{\mathbf{2}}}{\mathbf{n}_{\mathbf{1}}} \right\rbrack\ \]

    \[ \mathbf{S}_{\mathbf{1}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{8 - 1}}\mathbf{\ }\left\lbrack \mathbf{1138 -}\frac{\left( \mathbf{94} \right)^{\mathbf{2}}}{\mathbf{8}} \right\rbrack\ \]

    \[ \mathbf{S}_{\mathbf{1}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{7}}\mathbf{(33.5)}\ \]

    \[ \mathbf{S}_{\mathbf{1}}^{\mathbf{2}}\mathbf{= 4.785}\ \]

    \[ {\overline{\mathbf{X}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{\sum X}_{\mathbf{2}}}{\mathbf{2}}\mathbf{=}\frac{\mathbf{73}}{\mathbf{7}}\mathbf{= 10.43}\ \]

    \[ \mathbf{S}_{\mathbf{2}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{n}_{\mathbf{2}}\mathbf{- 1}}\mathbf{\ }\left\lbrack \mathbf{\sum}\mathbf{X}_{\mathbf{2}}^{\mathbf{2}}\mathbf{-}\frac{\left( \mathbf{\sum}\mathbf{X}_{\mathbf{2}} \right)^{\mathbf{2}}}{\mathbf{n}_{\mathbf{2}}} \right\rbrack\ \]

    \[ \mathbf{S}_{\mathbf{2}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{7 - 1}}\mathbf{\ }\left\lbrack \mathbf{785 -}\frac{\left( \mathbf{73} \right)^{\mathbf{2}}}{\mathbf{7}} \right\rbrack\ \]

    \[ \mathbf{S}_{\mathbf{2}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{7}}\mathbf{(23.714)}\ \]

    \[ \mathbf{S}_{\mathbf{2}}^{\mathbf{2}}\mathbf{= 3.39}\ \]

    \[ \mathbf{S}\mathbf{p}^{\mathbf{2}}\mathbf{=}\frac{\left( \mathbf{n}_{\mathbf{1}}\mathbf{- 1} \right)\mathbf{S}_{\mathbf{1}}^{\mathbf{2}}\mathbf{+}\left( \mathbf{n}_{\mathbf{2}}\mathbf{- 1} \right)\mathbf{S}_{\mathbf{2}}^{\mathbf{2}}}{\mathbf{n}\mathbf{1 + n}\mathbf{2 - 2}}\ \]

    \[ \mathbf{S}\mathbf{p}^{\mathbf{2}}\mathbf{=}\frac{\left( \mathbf{8 - 1} \right)\mathbf{4.785 +}\left( \mathbf{7 - 1} \right)\mathbf{3.39}}{\mathbf{8 + 7 - 2}}\ \]

    \[ \mathbf{S}\mathbf{p}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{53.835}}{\mathbf{13}}\mathbf{= 4.14}\ \]

    \[ \mathbf{sp = \ }\sqrt{\mathbf{s}\mathbf{p}^{\mathbf{2}}}\ \]

    \[ \mathbf{sp = \ }\sqrt{\mathbf{4.14}}\ \]

    \[ \mathbf{sp = 2.034}\ \]

Step 1 Hypothesis

Ho: µ1=µ2

H1: µ1≠µ2

Step 2 Level of Significance

Level of Significance α = 0.05

Step 3 Test Statistic

    \[ \mathbf{t =}\frac{\left( {\overline{\mathbf{X}}}_{\mathbf{1}}\mathbf{-}{\overline{\mathbf{X}}}_{\mathbf{2}} \right)\mathbf{- \ (}\mathbf{\mu}_{\mathbf{1}}\mathbf{- \ }\mathbf{\mu}_{\mathbf{2}}\mathbf{)}}{\mathbf{sp}\sqrt{\left\lbrack \frac{\mathbf{1}}{\mathbf{n}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}_{\mathbf{2}}} \right\rbrack}}\ \]

Step 4 Critical Region

    \[ \left| \mathbf{t} \right|_{\mathbf{\alpha}\mathbf{/2(df)}}\mathbf{= \ \ }\left| \mathbf{t} \right|_{\mathbf{0.025\ (13)}}\mathbf{\ = 2.160}\ \]

Where d.f = v = n1+n2-2=8 + 7 – 2 = 13

Step 5 Calculation

    \[ \mathbf{t =}\frac{\left( {\overline{\mathbf{X}}}_{\mathbf{1}}\mathbf{-}{\overline{\mathbf{X}}}_{\mathbf{2}} \right)\mathbf{- \ (}\mathbf{\mu}_{\mathbf{1}}\mathbf{- \ }\mathbf{\mu}_{\mathbf{2}}\mathbf{)}}{\mathbf{sp}\sqrt{\left\lbrack \frac{\mathbf{1}}{\mathbf{n}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}_{\mathbf{2}}} \right\rbrack}}\ \]

    \[ \mathbf{t =}\frac{\left( \mathbf{11.75 - 10.43} \right)\mathbf{- \ (0)}}{\mathbf{2.034}\sqrt{\left\lbrack \frac{\mathbf{1}}{\mathbf{8}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{7}} \right\rbrack}}\ \]

    \[ \mathbf{t =}\frac{\mathbf{1.32}}{\mathbf{1.0526}}\mathbf{= 1.254}\ \]

Step 6 Conclusion

Statistical Decision: Calculated value 1.254 falls in acceptance region 2.160 so we accept Null Hypothesis Ho & reject H1 and we can conclude that µ1=µ2.

Q.5 a. A sample of 120 observations from a population known to be non-normal yielded the Sample

Values, X̅ = 576, S² = 2475. Find an approximate 90% Confidence Interval for mean of the population.

Solution:

Data

    \[ n = \ 120,\mathbf{\ }\overline{X}\ = \ 576,\ S^{2} = \ 2475,S = 49.74,\ C.I\ = \ 90\%\ \]

Test Statistic:

    \[ \overline{\mathbf{X}}\mathbf{\pm}\mathbf{Z}_{\mathbf{\propto}\mathbf{/}\mathbf{2}}\frac{\mathbf{s}}{\sqrt{\mathbf{n}}}\ \]

Calculation:

    \[ \mathbf{576}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{645}\frac{\mathbf{49.74}}{\sqrt{\mathbf{120}}}\mathbf{=}\mathbf{576}\mathbf{-}\mathbf{7.4693}\mathbf{=}\mathbf{568.53}\ \]

    \[ \mathbf{576}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{645}\frac{\mathbf{49.74}}{\sqrt{\mathbf{120}}}\mathbf{=}\mathbf{576}\mathbf{+}\mathbf{7.4693}\mathbf{=}\mathbf{583.46}\ \]

Conclusion

568.53<µ<583.46  

b. The following table shows the distribution of 200 school children according to their Physical defect and Speech defect. Use α= 0.01

Speech DefectPhysical Defect
P1P2P3
S1342224
S2251421
S3212415

Do the data suggest any association between Physical defect and Speech defect?

Solution

Speech DefectPhysical DefectTotal
P1P2P3
S134222480
S225142160
S321241560
Total806060200

Step 1 Testing the Hypothesis

Ho: There is no Association between Speech Defect & Physical Defect.

H1: There is Association between Speech Defect & Physical Defect.

Step 2 Level of Significance

Level of Significance = α=0.05

Step 3 Test Statistic

    \[ \mathbf{\chi ² = \ \sum}\frac{\mathbf{(fo - fe)}^{\mathbf{2}}}{\mathbf{fe}}\ \]

Step 4 Critical Region

Critical Region: Degree of Freedom d.f= (R-1)(C-1)

So d.f= (3-1)(3-1)=4

The Value of Tabulated χ²(0.01,4)=13.277

The Critical Region χ²cal>13.277

Step 5 Calculation of Expected Frequencies

    \[ \frac{80 \times 80}{200} = 32,\ \frac{80 \times 60}{200} = 24,\ \frac{80 \times 60}{200} = 24\ \]

    \[ \frac{60 \times 80}{200} = 24,\ \frac{60 \times 60}{200} = 18,\ \frac{60 \times 60}{200} = 18\ \]

    \[ \frac{60 \times 80}{200} = 24,\ \frac{60 \times 60}{200} = 18,\ \frac{60 \times 60}{200} = 18\ \ \]

Step 6 Calculation of χ²:

Table B. Computation of χ²
fofefo-fe(fo-fe)²

    \[ \mathbf{\sum}\left\lbrack \frac{\left( \mathbf{fo - fe} \right)^{\mathbf{2}}}{\mathbf{fe}} \right\rbrack\mathbf{=}\ \]

3432240.125
2224-240.167
2424000
2524110.04167
1418-4160.89
2118390.5
2124-390.375
24186362
1518-390.5
    4.59867
    

    \[ \mathbf{\sum}\left\lbrack \frac{\left( \mathbf{fo - fe} \right)^{\mathbf{2}}}{\mathbf{fe}} \right\rbrack\mathbf{=}\ \]


Step 7 Conclusion

The calculated value of χ² is 4.59867 is less than the tabulated value of χ² 13.277 or 4.59867 falls in the acceptance region. We accept the Null Hypothesis Ho that there is no association between Speech Defect & Physical Defect.

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