IGNOU MST-004 Previous Year Paper Solution for PGDAST

IGNOU MST-004 Previous Year Paper Solution | PGDAST

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MST-004 Previous Year Paper Solution

 
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\(\sec ^2 \theta=1+\tan ^2 \theta\)

MST-004 Previous Year Paper Solution - Sample

MST-004 Dec 2023

Question:-01

1.State whether the following statements are True or False. Give reasons in support of your answers :
(a) If sample size of a survey has increased 3 times, then the standard error will be increased 3 times.

Answer:

The statement "If the sample size of a survey has increased 3 times, then the standard error will be increased 3 times" is false.
To justify why this statement is false, we can look at the relationship between the standard error of the mean and the sample size. The standard error of the mean (SEM) is given by the formula:
SEM = σ n SEM = σ n “SEM”=(sigma)/(sqrtn)\text{SEM} = \frac{\sigma}{\sqrt{n}}SEM=σn
where σ σ sigma\sigmaσ is the population standard deviation and n n nnn is the sample size. This formula shows that the standard error of the mean is inversely proportional to the square root of the sample size.
When the sample size n n nnn is increased by a factor of 3, the new sample size becomes 3 n 3 n 3n3n3n. Plugging this into the formula for the standard error gives:
SEM new = σ 3 n SEM new = σ 3 n “SEM”_(“new”)=(sigma)/(sqrt(3n))\text{SEM}_{\text{new}} = \frac{\sigma}{\sqrt{3n}}SEMnew=σ3n
To see how the standard error changes, compare the new SEM with the original SEM:
SEM new = σ 3 n = σ 3 n = 1 3 σ n = 1 3 SEM old SEM new = σ 3 n = σ 3 n = 1 3 σ n = 1 3 SEM old “SEM”_(“new”)=(sigma)/(sqrt(3n))=(sigma)/(sqrt3sqrtn)=(1)/(sqrt3)(sigma)/(sqrtn)=(1)/(sqrt3)”SEM”_(“old”)\text{SEM}_{\text{new}} = \frac{\sigma}{\sqrt{3n}} = \frac{\sigma}{\sqrt{3} \sqrt{n}} = \frac{1}{\sqrt{3}} \frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt{3}} \text{SEM}_{\text{old}}SEMnew=σ3n=σ3n=13σn=13SEMold
Since 1 3 ≈ 0.577 1 3 ≈ 0.577 (1)/(sqrt3)~~0.577\frac{1}{\sqrt{3}} \approx 0.57713≈0.577, the new standard error is approximately 0.577 0.577 0.5770.5770.577 times the original standard error. This means that the standard error does not increase three times; rather, it decreases by a factor of about 3 3 sqrt3\sqrt{3}3 or decreases to about 57.7% of its original value.
Therefore, the statement is false because increasing the sample size by three times results in the standard error decreasing, not increasing, and specifically decreasing by a factor of about 3 3 sqrt3\sqrt{3}3.
(b) If probability density function of a random variable X X X\mathrm{X}X follows F F F\mathrm{F}F-distribution
f ( x ) = 1 ( 1 + x ) 2 ; 0 < x < ∞ , f ( x ) = 1 ( 1 + x ) 2 ; 0 < x < ∞ , f(x)=(1)/((1+x)^(2));0 < x < oo,f(x)=\frac{1}{(1+x)^2} ; 0<x<\infty,f(x)=1(1+x)2;0<x<∞,
then the degree of freedom of the distribution will be ( 2 , 2 ) ( 2 , 2 ) (2,2)(2,2)(2,2).

Answer:

The statement "If the probability density function of a random variable X X XXX follows F F FFF-distribution f ( x ) = 1 ( 1 + x ) 2 ; 0 < x < ∞ f ( x ) = 1 ( 1 + x ) 2 ; 0 < x < ∞ f(x)=(1)/((1+x)^(2));0 < x < oof(x) = \frac{1}{(1+x)^2}; 0 < x < \inftyf(x)=1(1+x)2;0<x<∞, then the degree of freedom of the distribution will be ( 2 , 2 ) ( 2 , 2 ) (2,2)(2,2)(2,2)" is true.
To justify this, let’s analyze the given information:
The probability density function (PDF) provided is:
f ( x ) = 1 ( 1 + x ) 2 , 0 < x < ∞ f ( x ) = 1 ( 1 + x ) 2 , 0 < x < ∞ f(x)=(1)/((1+x)^(2)),quad0 < x < oof(x) = \frac{1}{(1+x)^2}, \quad 0 < x < \inftyf(x)=1(1+x)2,0<x<∞
The PDF of the F F FFF-distribution with degrees of freedom v 1 v 1 v_(1)v_1v1 and v 2 v 2 v_(2)v_2v2 is given by:
f ( x ) = ( v 1 / v 2 B ( v 1 2 , v 2 2 ) ) ( v 1 v 2 ) v 1 / 2 x ( v 1 / 2 ) − 1 ( 1 + v 1 v 2 x ) − ( v 1 + v 2 ) / 2 f ( x ) = v 1 / v 2 B v 1 2 , v 2 2 v 1 v 2 v 1 / 2 x ( v 1 / 2 ) − 1 1 + v 1 v 2 x − ( v 1 + v 2 ) / 2 f(x)=((v_(1)//v_(2))/(B((v_(1))/(2),(v_(2))/(2))))((v_(1))/(v_(2)))^(v_(1)//2)x^((v_(1)//2)-1)(1+(v_(1))/(v_(2))x)^(-(v_(1)+v_(2))//2)f(x) = \left( \frac{v_1 / v_2}{B\left( \frac{v_1}{2}, \frac{v_2}{2} \right)} \right) \left( \frac{v_1}{v_2} \right)^{v_1 / 2} x^{(v_1 / 2) – 1} \left( 1 + \frac{v_1}{v_2} x \right)^{-(v_1 + v_2) / 2}f(x)=(v1/v2B(v12,v22))(v1v2)v1/2x(v1/2)−1(1+v1v2x)−(v1+v2)/2
Given:
f ( x ) = 1 ( 1 + x ) 2 , 0 < x < ∞ f ( x ) = 1 ( 1 + x ) 2 , 0 < x < ∞ f(x)=(1)/((1+x)^(2)),quad0 < x < oof(x) = \frac{1}{(1+x)^2}, \quad 0 < x < \inftyf(x)=1(1+x)2,0<x<∞
We can rewrite the given PDF as:
f ( x ) = ( 2 / 2 ) 2 / 2 x ( 2 / 2 ) − 1 B ( 2 / 2 , 2 / 2 ) ( 1 + 2 2 x ) ( 2 + 2 ) / 2 f ( x ) = ( 2 / 2 ) 2 / 2 x ( 2 / 2 ) − 1 B ( 2 / 2 , 2 / 2 ) 1 + 2 2 x ( 2 + 2 ) / 2 f(x)=((2//2)^(2//2)x^((2//2)-1))/(B(2//2,2//2)(1+(2)/(2)x)^((2+2)//2))f(x) = \frac{(2/2)^{2/2} x^{(2/2) – 1}}{B(2/2, 2/2) \left(1 + \frac{2}{2} x \right)^{(2 + 2)/2}}f(x)=(2/2)2/2x(2/2)−1B(2/2,2/2)(1+22x)(2+2)/2
Comparing this with the standard form of the F F FFF-distribution PDF, we observe that the given PDF matches if v 1 = 2 v 1 = 2 v_(1)=2v_1 = 2v1=2 and v 2 = 2 v 2 = 2 v_(2)=2v_2 = 2v2=2.
Thus, the degrees of freedom for the given distribution are ( 2 , 2 ) ( 2 , 2 ) (2,2)(2,2)(2,2).
Therefore, the statement is true.
(c) For testing H 0 : θ = 2 H 0 : θ = 2 H_(0):theta=2\mathrm{H}_0: \theta=2H0:θ=2 against H 1 : θ = 3 H 1 : θ = 3 H_(1):theta=3\mathrm{H}_1: \theta=3H1:θ=3, the p d f p d f pdf\mathrm{pdf}pdf of the variable is given by
f ( x , θ ) = 1 θ ; 0 ≤ x ≤ θ f ( x , θ ) = 1 θ ; 0 ≤ x ≤ θ f(x,theta)=(1)/(theta);0 <= x <= thetaf(x, \theta)=\frac{1}{\theta} ; 0 \leq x \leq \thetaf(x,θ)=1θ;0≤x≤θ
If the critical region is x ≥ 0.6 x ≥ 0.6 x >= 0.6x \geq 0.6x≥0.6, the size of the test will be 0.6 .

Answer:

The statement "For testing H 0 : θ = 2 H 0 : θ = 2 H_(0):theta=2\mathrm{H}_0: \theta=2H0:θ=2 against H 1 : θ = 3 H 1 : θ = 3 H_(1):theta=3\mathrm{H}_1: \theta=3H1:θ=3, the p d f p d f pdf\mathrm{pdf}pdf of the variable is given by f ( x , θ ) = 1 θ ; 0 ≤ x ≤ θ f ( x , θ ) = 1 θ ; 0 ≤ x ≤ θ f(x,theta)=(1)/(theta);0 <= x <= thetaf(x, \theta)=\frac{1}{\theta} ; 0 \leq x \leq \thetaf(x,θ)=1θ;0≤x≤θ. If the critical region is x ≥ 0.6 x ≥ 0.6 x >= 0.6x \geq 0.6x≥0.6, the size of the test will be 0.6" is false.
To justify this, let’s understand the concepts of hypothesis testing and the size of a test.
The size of a test, also known as the significance level (α), is the probability of rejecting the null hypothesis H 0 H 0 H_(0)\mathrm{H}_0H0 when it is actually true. This is also known as the Type I error rate.
Given:
  • H 0 : θ = 2 H 0 : θ = 2 H_(0):theta=2\mathrm{H}_0: \theta=2H0:θ=2
  • H 1 : θ = 3 H 1 : θ = 3 H_(1):theta=3\mathrm{H}_1: \theta=3H1:θ=3
  • The probability density function (pdf) of the variable: f ( x , θ ) = 1 θ f ( x , θ ) = 1 θ f(x,theta)=(1)/(theta)f(x, \theta)=\frac{1}{\theta}f(x,θ)=1θ, for 0 ≤ x ≤ θ 0 ≤ x ≤ θ 0 <= x <= theta0 \leq x \leq \theta0≤x≤θ
The critical region is x ≥ 0.6 x ≥ 0.6 x >= 0.6x \geq 0.6x≥0.6.
First, let’s find the size of the test under the null hypothesis H 0 : θ = 2 H 0 : θ = 2 H_(0):theta=2\mathrm{H}_0: \theta=2H0:θ=2.
The pdf under H 0 H 0 H_(0)\mathrm{H}_0H0 is:
f ( x , θ = 2 ) = 1 2 , 0 ≤ x ≤ 2 f ( x , θ = 2 ) = 1 2 , 0 ≤ x ≤ 2 f(x,theta=2)=(1)/(2),quad0 <= x <= 2f(x, \theta=2) = \frac{1}{2}, \quad 0 \leq x \leq 2f(x,θ=2)=12,0≤x≤2
To find the size of the test, we need to calculate the probability that x ≥ 0.6 x ≥ 0.6 x >= 0.6x \geq 0.6x≥0.6 when θ = 2 θ = 2 theta=2\theta = 2θ=2:
Size of the test = P ( x ≥ 0.6 ∣ θ = 2 ) = ∫ 0.6 2 1 2 d x Size of the test = P ( x ≥ 0.6 ∣ θ = 2 ) = ∫ 0.6 2   1 2 d x “Size of the test”=P(x >= 0.6∣theta=2)=int_(0.6)^(2)(1)/(2)dx\text{Size of the test} = P(x \geq 0.6 \mid \theta=2) = \int_{0.6}^{2} \frac{1}{2} \, dxSize of the test=P(x≥0.6∣θ=2)=∫0.6212dx
Calculating the integral:
∫ 0.6 2 1 2 d x = 1 2 [ x ] 0.6 2 = 1 2 ( 2 − 0.6 ) = 1 2 × 1.4 = 0.7 ∫ 0.6 2   1 2 d x = 1 2 x 0.6 2 = 1 2 ( 2 − 0.6 ) = 1 2 × 1.4 = 0.7 int_(0.6)^(2)(1)/(2)dx=(1)/(2)[x]_(0.6)^(2)=(1)/(2)(2-0.6)=(1)/(2)xx1.4=0.7\int_{0.6}^{2} \frac{1}{2} \, dx = \frac{1}{2} \left[ x \right]_{0.6}^{2} = \frac{1}{2} (2 – 0.6) = \frac{1}{2} \times 1.4 = 0.7∫0.6212dx=12[x]0.62=12(2−0.6)=12×1.4=0.7
Therefore, the size of the test is 0.7, not 0.6.
Hence, the statement is false because the size of the test under the null hypothesis θ = 2 θ = 2 theta=2\theta = 2θ=2 with the critical region x ≥ 0.6 x ≥ 0.6 x >= 0.6x \geq 0.6x≥0.6 is 0.7, not 0.6.
(d) Kruskal-Wallis test is a non-parametric version of two-way analysis of variance.

Answer:

The statement "Kruskal-Wallis test is a non-parametric version of two-way analysis of variance" is false.
To justify this, let’s review the purposes and uses of the Kruskal-Wallis test and two-way analysis of variance (ANOVA).

Kruskal-Wallis Test:

  • The Kruskal-Wallis test is a non-parametric method used to test whether there are statistically significant differences between the medians of three or more independent groups.
  • It is the non-parametric alternative to one-way ANOVA and is used when the assumptions of one-way ANOVA (such as normality) are not met.
  • The Kruskal-Wallis test ranks all the data from all groups together and then compares the sum of ranks between the groups.

Two-Way ANOVA:

  • Two-way ANOVA is a parametric method used to determine if there are any significant differences between the means of three or more groups, considering two independent variables (factors).
  • It can test the interaction between the two factors, as well as the individual effects of each factor.

Key Differences:

  • Number of Factors: The Kruskal-Wallis test deals with one factor with multiple levels (one-way analysis), while two-way ANOVA involves two factors.
  • Type of Data: Kruskal-Wallis is non-parametric and uses ranks, making it suitable for ordinal data or data that do not meet the assumptions of normality. Two-way ANOVA is parametric and assumes normal distribution of the residuals.

Correct Non-Parametric Alternative:

  • The non-parametric alternative to two-way ANOVA is the Friedman test, which is used for analyzing differences in treatments across multiple test attempts (blocks) when the data is not normally distributed.
Therefore, the Kruskal-Wallis test is not a non-parametric version of two-way ANOVA; it is a non-parametric version of one-way ANOVA. The statement is false.
(e) A sample of size 4 is drawn randomly ( X 1 , X 2 , X 3 X 1 , X 2 , X 3 (X_(1),X_(2),X_(3):}\left(\mathrm{X}_1, \mathrm{X}_2, \mathrm{X}_3\right.(X1,X2,X3 and X 4 X 4 X_(4)\mathrm{X}_4X4 ) from a normal population with unknown mean μ μ mu\muμ, then X 1 + 2 X 2 + 3 X 3 + X 4 7 X 1 + 2 X 2 + 3 X 3 + X 4 7 (X_(1)+2X_(2)+3X_(3)+X_(4))/(7)\frac{\mathrm{X}_1+2 \mathrm{X}_2+3 \mathrm{X}_3+\mathrm{X}_4}{7}X1+2X2+3X3+X47 is an unbiased estimator of μ μ mu\muμ.

Answer:

To determine whether X 1 + 2 X 2 + 3 X 3 + X 4 7 X 1 + 2 X 2 + 3 X 3 + X 4 7 (X_(1)+2X_(2)+3X_(3)+X_(4))/(7)\frac{\mathrm{X}_1 + 2\mathrm{X}_2 + 3\mathrm{X}_3 + \mathrm{X}_4}{7}X1+2X2+3X3+X47 is an unbiased estimator of the population mean μ μ mu\muμ, we need to check if the expected value of this estimator equals μ μ mu\muμ.
Given that X 1 , X 2 , X 3 X 1 , X 2 , X 3 X_(1),X_(2),X_(3)\mathrm{X}_1, \mathrm{X}_2, \mathrm{X}_3X1,X2,X3, and X 4 X 4 X_(4)\mathrm{X}_4X4 are drawn from a normal population with unknown mean μ μ mu\muμ, the expected value of each X i X i X_(i)\mathrm{X}_iXi is μ μ mu\muμ.
The estimator in question is:
μ ^ = X 1 + 2 X 2 + 3 X 3 + X 4 7 μ ^ = X 1 + 2 X 2 + 3 X 3 + X 4 7 hat(mu)=(X_(1)+2X_(2)+3X_(3)+X_(4))/(7)\hat{\mu} = \frac{\mathrm{X}_1 + 2\mathrm{X}_2 + 3\mathrm{X}_3 + \mathrm{X}_4}{7}μ^=X1+2X2+3X3+X47
To check if this is an unbiased estimator of μ μ mu\muμ, we calculate the expected value of μ ^ μ ^ hat(mu)\hat{\mu}μ^:
E ( μ ^ ) = E ( X 1 + 2 X 2 + 3 X 3 + X 4 7 ) E μ ^ = E X 1 + 2 X 2 + 3 X 3 + X 4 7 E(( hat(mu)))=E((X_(1)+2X_(2)+3X_(3)+X_(4))/(7))E\left(\hat{\mu}\right) = E\left(\frac{\mathrm{X}_1 + 2\mathrm{X}_2 + 3\mathrm{X}_3 + \mathrm{X}_4}{7}\right)E(μ^)=E(X1+2X2+3X3+X47)
Using the linearity of expectation, we have:
E ( μ ^ ) = 1 7 E ( X 1 + 2 X 2 + 3 X 3 + X 4 ) E μ ^ = 1 7 E ( X 1 + 2 X 2 + 3 X 3 + X 4 ) E(( hat(mu)))=(1)/(7)E(X_(1)+2X_(2)+3X_(3)+X_(4))E\left(\hat{\mu}\right) = \frac{1}{7} E(\mathrm{X}_1 + 2\mathrm{X}_2 + 3\mathrm{X}_3 + \mathrm{X}_4)E(μ^)=17E(X1+2X2+3X3+X4)
E ( μ ^ ) = 1 7 ( E ( X 1 ) + 2 E ( X 2 ) + 3 E ( X 3 ) + E ( X 4 ) ) E μ ^ = 1 7 E ( X 1 ) + 2 E ( X 2 ) + 3 E ( X 3 ) + E ( X 4 ) E(( hat(mu)))=(1)/(7)(E(X_(1))+2E(X_(2))+3E(X_(3))+E(X_(4)))E\left(\hat{\mu}\right) = \frac{1}{7} \left( E(\mathrm{X}_1) + 2E(\mathrm{X}_2) + 3E(\mathrm{X}_3) + E(\mathrm{X}_4) \right)E(μ^)=17(E(X1)+2E(X2)+3E(X3)+E(X4))
Since E ( X i ) = μ E ( X i ) = μ E(X_(i))=muE(\mathrm{X}_i) = \muE(Xi)=μ for all i i iii, we get:
E ( μ ^ ) = 1 7 ( μ + 2 μ + 3 μ + μ ) E μ ^ = 1 7 μ + 2 μ + 3 μ + μ E(( hat(mu)))=(1)/(7)(mu+2mu+3mu+mu)E\left(\hat{\mu}\right) = \frac{1}{7} \left( \mu + 2\mu + 3\mu + \mu \right)E(μ^)=17(μ+2μ+3μ+μ)
E ( μ ^ ) = 1 7 ( 7 μ ) E μ ^ = 1 7 7 μ E(( hat(mu)))=(1)/(7)(7mu)E\left(\hat{\mu}\right) = \frac{1}{7} \left( 7\mu \right)E(μ^)=17(7μ)
E ( μ ^ ) = μ E μ ^ = μ E(( hat(mu)))=muE\left(\hat{\mu}\right) = \muE(μ^)=μ
Thus, the expected value of the estimator X 1 + 2 X 2 + 3 X 3 + X 4 7 X 1 + 2 X 2 + 3 X 3 + X 4 7 (X_(1)+2X_(2)+3X_(3)+X_(4))/(7)\frac{\mathrm{X}_1 + 2\mathrm{X}_2 + 3\mathrm{X}_3 + \mathrm{X}_4}{7}X1+2X2+3X3+X47 is equal to μ μ mu\muμ, which means it is an unbiased estimator of μ μ mu\muμ.
Therefore, the statement "A sample of size 4 is drawn randomly ( X 1 , X 2 , X 3 X 1 , X 2 , X 3 X_(1),X_(2),X_(3)\mathrm{X}_1, \mathrm{X}_2, \mathrm{X}_3X1,X2,X3, and X 4 X 4 X_(4)\mathrm{X}_4X4) from a normal population with unknown mean μ μ mu\muμ, then X 1 + 2 X 2 + 3 X 3 + X 4 7 X 1 + 2 X 2 + 3 X 3 + X 4 7 (X_(1)+2X_(2)+3X_(3)+X_(4))/(7)\frac{\mathrm{X}_1 + 2\mathrm{X}_2 + 3\mathrm{X}_3 + \mathrm{X}_4}{7}X1+2X2+3X3+X47 is an unbiased estimator of μ μ mu\muμ" is true.

Question:-02

2.The systolic blood pressure (SBP) of five women are given as follows :
120 , 110 , 130 , 140 , 100 120 , 110 , 130 , 140 , 100 120,110,130,140,100120,110,130,140,100120,110,130,140,100
(a) How many samples of size 2 can be drawn without replacement? Write them.
(b) Compute the mean of all samples of size 2 and set up the sampling distribution of the sample mean.
(c) Compute the expected value of the sample mean.
(d) How many samples of the same size 2 are possible with replacement ? Calculate expected value of the sample mean and compare it with the expected value calculated in the case of without replacement.

Answer:

Let’s address each part of the problem step by step.

Given Data:

Systolic Blood Pressure (SBP) of five women:
120 , 110 , 130 , 140 , 100 120 , 110 , 130 , 140 , 100 120,110,130,140,100120, 110, 130, 140, 100120,110,130,140,100

(a) How many samples of size 2 can be drawn without replacement? Write them.

The number of samples of size 2 that can be drawn without replacement from 5 items is given by the combination formula:
( 5 2 ) = 5 ! 2 ! ( 5 − 2 ) ! = 10 ( 5 2 ) = 5 ! 2 ! ( 5 − 2 ) ! = 10 ((5)/(2))=(5!)/(2!(5-2)!)=10\binom{5}{2} = \frac{5!}{2!(5-2)!} = 10(52)=5!2!(5−2)!=10
The possible samples without replacement are:
  1. (120, 110)
  2. (120, 130)
  3. (120, 140)
  4. (120, 100)
  5. (110, 130)
  6. (110, 140)
  7. (110, 100)
  8. (130, 140)
  9. (130, 100)
  10. (140, 100)

(b) Compute the mean of all samples of size 2 and set up the sampling distribution of the sample mean.

First, compute the means of each sample:
  1. (120, 110): 120 + 110 2 = 115 120 + 110 2 = 115 (120+110)/(2)=115\frac{120 + 110}{2} = 115120+1102=115
  2. (120, 130): 120 + 130 2 = 125 120 + 130 2 = 125 (120+130)/(2)=125\frac{120 + 130}{2} = 125120+1302=125
  3. (120, 140): 120 + 140 2 = 130 120 + 140 2 = 130 (120+140)/(2)=130\frac{120 + 140}{2} = 130120+1402=130
  4. (120, 100): 120 + 100 2 = 110 120 + 100 2 = 110 (120+100)/(2)=110\frac{120 + 100}{2} = 110120+1002=110
  5. (110, 130): 110 + 130 2 = 120 110 + 130 2 = 120 (110+130)/(2)=120\frac{110 + 130}{2} = 120110+1302=120
  6. (110, 140): 110 + 140 2 = 125 110 + 140 2 = 125 (110+140)/(2)=125\frac{110 + 140}{2} = 125110+1402=125
  7. (110, 100): 110 + 100 2 = 105 110 + 100 2 = 105 (110+100)/(2)=105\frac{110 + 100}{2} = 105110+1002=105
  8. (130, 140): 130 + 140 2 = 135 130 + 140 2 = 135 (130+140)/(2)=135\frac{130 + 140}{2} = 135130+1402=135
  9. (130, 100): 130 + 100 2 = 115 130 + 100 2 = 115 (130+100)/(2)=115\frac{130 + 100}{2} = 115130+1002=115
  10. (140, 100): 140 + 100 2 = 120 140 + 100 2 = 120 (140+100)/(2)=120\frac{140 + 100}{2} = 120140+1002=120
The sampling distribution of the sample mean is:
105 , 110 , 115 , 115 , 120 , 120 , 120 , 125 , 125 , 135 105 , 110 , 115 , 115 , 120 , 120 , 120 , 125 , 125 , 135 105,110,115,115,120,120,120,125,125,135105, 110, 115, 115, 120, 120, 120, 125, 125, 135105,110,115,115,120,120,120,125,125,135

(c) Compute the expected value of the sample mean.

The expected value of the sample mean X ¯ X ¯ bar(X)\bar{X}X¯ is the average of the sample means:
E ( X ¯ ) = 1 10 ∑ X ¯ i = 105 + 110 + 115 + 115 + 120 + 120 + 120 + 125 + 125 + 135 10 = 1190 10 = 119 E ( X ¯ ) = 1 10 ∑ X ¯ i = 105 + 110 + 115 + 115 + 120 + 120 + 120 + 125 + 125 + 135 10 = 1190 10 = 119 E( bar(X))=(1)/(10)sum bar(X)_(i)=(105+110+115+115+120+120+120+125+125+135)/(10)=(1190)/(10)=119E(\bar{X}) = \frac{1}{10} \sum \bar{X}_i = \frac{105 + 110 + 115 + 115 + 120 + 120 + 120 + 125 + 125 + 135}{10} = \frac{1190}{10} = 119E(X¯)=110∑X¯i=105+110+115+115+120+120+120+125+125+13510=119010=119

(d) How many samples of the same size 2 are possible with replacement? Calculate the expected value of the sample mean and compare it with the expected value calculated in the case of without replacement.

The number of samples of size 2 that can be drawn with replacement from 5 items is given by the formula for permutations with replacement:
5 2 = 25 5 2 = 25 5^(2)=255^2 = 2552=25
The possible samples with replacement are:
( 120 , 120 ) , ( 120 , 110 ) , ( 120 , 130 ) , ( 120 , 140 ) , ( 120 , 100 ) , ( 120 , 120 ) , ( 120 , 110 ) , ( 120 , 130 ) , ( 120 , 140 ) , ( 120 , 100 ) , (120,120),(120,110),(120,130),(120,140),(120,100),(120, 120), (120, 110), (120, 130), (120, 140), (120, 100),(120,120),(120,110),(120,130),(120,140),(120,100),
( 110 , 120 ) , ( 110 , 110 ) , ( 110 , 130 ) , ( 110 , 140 ) , ( 110 , 100 ) , ( 110 , 120 ) , ( 110 , 110 ) , ( 110 , 130 ) , ( 110 , 140 ) , ( 110 , 100 ) , (110,120),(110,110),(110,130),(110,140),(110,100),(110, 120), (110, 110), (110, 130), (110, 140), (110, 100),(110,120),(110,110),(110,130),(110,140),(110,100),
( 130 , 120 ) , ( 130 , 110 ) , ( 130 , 130 ) , ( 130 , 140 ) , ( 130 , 100 ) , ( 130 , 120 ) , ( 130 , 110 ) , ( 130 , 130 ) , ( 130 , 140 ) , ( 130 , 100 ) , (130,120),(130,110),(130,130),(130,140),(130,100),(130, 120), (130, 110), (130, 130), (130, 140), (130, 100),(130,120),(130,110),(130,130),(130,140),(130,100),
( 140 , 120 ) , ( 140 , 110 ) , ( 140 , 130 ) , ( 140 , 140 ) , ( 140 , 100 ) , ( 140 , 120 ) , ( 140 , 110 ) , ( 140 , 130 ) , ( 140 , 140 ) , ( 140 , 100 ) , (140,120),(140,110),(140,130),(140,140),(140,100),(140, 120), (140, 110), (140, 130), (140, 140), (140, 100),(140,120),(140,110),(140,130),(140,140),(140,100),
( 100 , 120 ) , ( 100 , 110 ) , ( 100 , 130 ) , ( 100 , 140 ) , ( 100 , 100 ) ( 100 , 120 ) , ( 100 , 110 ) , ( 100 , 130 ) , ( 100 , 140 ) , ( 100 , 100 ) (100,120),(100,110),(100,130),(100,140),(100,100)(100, 120), (100, 110), (100, 130), (100, 140), (100, 100)(100,120),(100,110),(100,130),(100,140),(100,100)
Compute the means of each sample:
X ¯ X ¯ bar(X)\bar{X}X¯
120 , 115 , 125 , 130 , 110 , 120 , 115 , 125 , 130 , 110 , 120,115,125,130,110,120, 115, 125, 130, 110,120,115,125,130,110,
115 , 110 , 120 , 125 , 105 , 115 , 110 , 120 , 125 , 105 , 115,110,120,125,105,115, 110, 120, 125, 105,115,110,120,125,105,
125 , 120 , 130 , 135 , 115 , 125 , 120 , 130 , 135 , 115 , 125,120,130,135,115,125, 120, 130, 135, 115,125,120,130,135,115,
130 , 125 , 135 , 140 , 120 , 130 , 125 , 135 , 140 , 120 , 130,125,135,140,120,130, 125, 135, 140, 120,130,125,135,140,120,
110 , 105 , 115 , 120 , 100 110 , 105 , 115 , 120 , 100 110,105,115,120,100110, 105, 115, 120, 100110,105,115,120,100
The expected value of the sample mean with replacement is:
E ( X ¯ ) = 1 25 ∑ X ¯ i = 120 + 115 + 125 + 130 + 110 + 115 + 110 + 120 + 125 + 105 + 125 + 120 + 130 + 135 + 115 + 130 + 125 + 135 + 140 + 120 + 110 + 105 + 115 + 120 + 100 25 E ( X ¯ ) = 1 25 ∑ X ¯ i = 120 + 115 + 125 + 130 + 110 + 115 + 110 + 120 + 125 + 105 + 125 + 120 + 130 + 135 + 115 + 130 + 125 + 135 + 140 + 120 + 110 + 105 + 115 + 120 + 100 25 E( bar(X))=(1)/(25)sum bar(X)_(i)=(120+115+125+130+110+115+110+120+125+105+125+120+130+135+115+130+125+135+140+120+110+105+115+120+100)/(25)E(\bar{X}) = \frac{1}{25} \sum \bar{X}_i = \frac{120 + 115 + 125 + 130 + 110 + 115 + 110 + 120 + 125 + 105 + 125 + 120 + 130 + 135 + 115 + 130 + 125 + 135 + 140 + 120 + 110 + 105 + 115 + 120 + 100}{25}E(X¯)=125∑X¯i=120+115+125+130+110+115+110+120+125+105+125+120+130+135+115+130+125+135+140+120+110+105+115+120+10025
Summing these values:
∑ X ¯ i = 2925 ∑ X ¯ i = 2925 sum bar(X)_(i)=2925\sum \bar{X}_i = 2925∑X¯i=2925
So,
E ( X ¯ ) = 2925 25 = 117 E ( X ¯ ) = 2925 25 = 117 E( bar(X))=(2925)/(25)=117E(\bar{X}) = \frac{2925}{25} = 117E(X¯)=292525=117

Comparison:

  • The expected value of the sample mean without replacement is 119.
  • The expected value of the sample mean with replacement is 117.
Thus, the expected value of the sample mean without replacement is slightly higher than that with replacement in this case.

Question:-03

3.(a) The following table gives the classification of 150 products according to types of tools and materials used to produce these products :
Tool Material
A B C
T 1 T 1 T_(1)\mathrm{T}_1T1 15 5 20
T 2 T 2 T_(2)\mathrm{~T}_2 T2 20 10 30
T 3 T 3 T_(3)\mathrm{~T}_3 T3 25 15 10
Tool Material A B C T_(1) 15 5 20 T_(2) 20 10 30 T_(3) 25 15 10| Tool | Material | | | | :—: | :—: | :—: | :—: | | | A | B | C | | $\mathrm{T}_1$ | 15 | 5 | 20 | | $\mathrm{~T}_2$ | 20 | 10 | 30 | | $\mathrm{~T}_3$ | 25 | 15 | 10 |
Test whether the tools and materials used are independent at 5 % 5 % 5%5 \%5% level of significance.

Answer:

To test whether the tools and materials used are independent at the 5 % 5 % 5%5\%5% level of significance, we can use the Chi-Square Test of Independence. Here are the steps:
  1. State the Hypotheses:
    • H 0 H 0 H_(0)H_0H0: The tools and materials used are independent.
    • H 1 H 1 H_(1)H_1H1: The tools and materials used are not independent.
  2. Observed Frequencies (O):
Tool A B C T 1 15 5 20 T 2 20 10 30 T 3 25 15 10 Tool A B C T 1 15 5 20 T 2 20 10 30 T 3 25 15 10 {:[“Tool”,”A”,”B”,”C”],[T_(1),15,5,20],[T_(2),20,10,30],[T_(3),25,15,10]:}\begin{array}{|c|c|c|c|} \hline \text{Tool} & \text{A} & \text{B} & \text{C} \\ \hline T_1 & 15 & 5 & 20 \\ T_2 & 20 & 10 & 30 \\ T_3 & 25 & 15 & 10 \\ \hline \end{array}ToolABCT115520T2201030T3251510
  1. Calculate the row and column totals and the grand total:
Tool A B C Row Total T 1 15 5 20 40 T 2 20 10 30 60 T 3 25 15 10 50 Column Total 60 30 60 150 Tool A B C Row Total T 1 15 5 20 40 T 2 20 10 30 60 T 3 25 15 10 50 Column Total 60 30 60 150 {:[“Tool”,”A”,”B”,”C”,”Row Total”],[T_(1),15,5,20,40],[T_(2),20,10,30,60],[T_(3),25,15,10,50],[“Column Total”,60,30,60,150]:}\begin{array}{|c|c|c|c|c|} \hline \text{Tool} & \text{A} & \text{B} & \text{C} & \text{Row Total} \\ \hline T_1 & 15 & 5 & 20 & 40 \\ T_2 & 20 & 10 & 30 & 60 \\ T_3 & 25 & 15 & 10 & 50 \\ \hline \text{Column Total} & 60 & 30 & 60 & 150 \\ \hline \end{array}ToolABCRow TotalT11552040T220103060T325151050Column Total603060150
  1. Calculate the Expected Frequencies (E):
E i j = ( Row Total of i ) × ( Column Total of j ) Grand Total E i j = ( Row Total of i ) × ( Column Total of j ) Grand Total E_(ij)=((“Row Total of “i)xx(“Column Total of “j))/(“Grand Total”)E_{ij} = \frac{(\text{Row Total of } i) \times (\text{Column Total of } j)}{\text{Grand Total}}Eij=(Row Total of i)×(Column Total of j)Grand Total
Tool A B C T 1 40 × 60 150 = 16 40 × 30 150 = 8 40 × 60 150 = 16 T 2 60 × 60 150 = 24 60 × 30 150 = 12 60 × 60 150 = 24 T 3 50 × 60 150 = 20 50 × 30 150 = 10 50 × 60 150 = 20 Tool A B C T 1 40 × 60 150 = 16 40 × 30 150 = 8 40 × 60 150 = 16 T 2 60 × 60 150 = 24 60 × 30 150 = 12 60 × 60 150 = 24 T 3 50 × 60 150 = 20 50 × 30 150 = 10 50 × 60 150 = 20 {:[“Tool”,”A”,”B”,”C”],[T_(1),(40 xx60)/(150)=16,(40 xx30)/(150)=8,(40 xx60)/(150)=16],[T_(2),(60 xx60)/(150)=24,(60 xx30)/(150)=12,(60 xx60)/(150)=24],[T_(3),(50 xx60)/(150)=20,(50 xx30)/(150)=10,(50 xx60)/(150)=20]:}\begin{array}{|c|c|c|c|} \hline \text{Tool} & \text{A} & \text{B} & \text{C} \\ \hline T_1 & \frac{40 \times 60}{150} = 16 & \frac{40 \times 30}{150} = 8 & \frac{40 \times 60}{150} = 16 \\ T_2 & \frac{60 \times 60}{150} = 24 & \frac{60 \times 30}{150} = 12 & \frac{60 \times 60}{150} = 24 \\ T_3 & \frac{50 \times 60}{150} = 20 & \frac{50 \times 30}{150} = 10 & \frac{50 \times 60}{150} = 20 \\ \hline \end{array}ToolABCT140×60150=1640×30150=840×60150=16T260×60150=2460×30150=1260×60150=24T350×60150=2050×30150=1050×60150=20
  1. Compute the Chi-Square Test Statistic:
χ 2 = ∑ ( O i j − E i j ) 2 E i j χ 2 = ∑ ( O i j − E i j ) 2 E i j chi^(2)=sum((O_(ij)-E_(ij))^(2))/(E_(ij))\chi^2 = \sum \frac{(O_{ij} – E_{ij})^2}{E_{ij}}χ2=∑(Oij−Eij)2Eij
Tool A B C T 1 ( 15 − 16 ) 2 16 = 1 16 ( 5 − 8 ) 2 8 = 9 8 ( 20 − 16 ) 2 16 = 16 16 T 2 ( 20 − 24 ) 2 24 = 16 24 ( 10 − 12 ) 2 12 = 4 12 ( 30 − 24 ) 2 24 = 36 24 T 3 ( 25 − 20 ) 2 20 = 25 20 ( 15 − 10 ) 2 10 = 25 10 ( 10 − 20 ) 2 20 = 100 20 Tool A B C T 1 ( 15 − 16 ) 2 16 = 1 16 ( 5 − 8 ) 2 8 = 9 8 ( 20 − 16 ) 2 16 = 16 16 T 2 ( 20 − 24 ) 2 24 = 16 24 ( 10 − 12 ) 2 12 = 4 12 ( 30 − 24 ) 2 24 = 36 24 T 3 ( 25 − 20 ) 2 20 = 25 20 ( 15 − 10 ) 2 10 = 25 10 ( 10 − 20 ) 2 20 = 100 20 {:[“Tool”,”A”,”B”,”C”],[T_(1),((15-16)^(2))/(16)=(1)/(16),((5-8)^(2))/(8)=(9)/(8),((20-16)^(2))/(16)=(16)/(16)],[T_(2),((20-24)^(2))/(24)=(16)/(24),((10-12)^(2))/(12)=(4)/(12),((30-24)^(2))/(24)=(36)/(24)],[T_(3),((25-20)^(2))/(20)=(25)/(20),((15-10)^(2))/(10)=(25)/(10),((10-20)^(2))/(20)=(100)/(20)]:}\begin{array}{|c|c|c|c|} \hline \text{Tool} & \text{A} & \text{B} & \text{C} \\ \hline T_1 & \frac{(15-16)^2}{16} = \frac{1}{16} & \frac{(5-8)^2}{8} = \frac{9}{8} & \frac{(20-16)^2}{16} = \frac{16}{16} \\ T_2 & \frac{(20-24)^2}{24} = \frac{16}{24} & \frac{(10-12)^2}{12} = \frac{4}{12} & \frac{(30-24)^2}{24} = \frac{36}{24} \\ T_3 & \frac{(25-20)^2}{20} = \frac{25}{20} & \frac{(15-10)^2}{10} = \frac{25}{10} & \frac{(10-20)^2}{20} = \frac{100}{20} \\ \hline \end{array}ToolABCT1(15−16)216=116(5−8)28=98(20−16)216=1616T2(20−24)224=1624(10−12)212=412(30−24)224=3624T3(25−20)220=2520(15−10)210=2510(10−20)220=10020
Calculate each term and sum them up:
χ 2 = 1 16 + 9 8 + 16 16 + 16 24 + 4 12 + 36 24 + 25 20 + 25 10 + 100 20 χ 2 = 1 16 + 9 8 + 16 16 + 16 24 + 4 12 + 36 24 + 25 20 + 25 10 + 100 20 chi^(2)=(1)/(16)+(9)/(8)+(16)/(16)+(16)/(24)+(4)/(12)+(36)/(24)+(25)/(20)+(25)/(10)+(100)/(20)\chi^2 = \frac{1}{16} + \frac{9}{8} + \frac{16}{16} + \frac{16}{24} + \frac{4}{12} + \frac{36}{24} + \frac{25}{20} + \frac{25}{10} + \frac{100}{20}χ2=116+98+1616+1624+412+3624+2520+2510+10020
χ 2 = 0.0625 + 1.125 + 1 + 0.6667 + 0.3333 + 1.5 + 1.25 + 2.5 + 5 χ 2 = 0.0625 + 1.125 + 1 + 0.6667 + 0.3333 + 1.5 + 1.25 + 2.5 + 5 chi^(2)=0.0625+1.125+1+0.6667+0.3333+1.5+1.25+2.5+5\chi^2 = 0.0625 + 1.125 + 1 + 0.6667 + 0.3333 + 1.5 + 1.25 + 2.5 + 5χ2=0.0625+1.125+1+0.6667+0.3333+1.5+1.25+2.5+5
χ 2 = 13.4375 χ 2 = 13.4375 chi^(2)=13.4375\chi^2 = 13.4375χ2=13.4375
  1. Degrees of Freedom:
df = ( r − 1 ) ( c − 1 ) = ( 3 − 1 ) ( 3 − 1 ) = 2 × 2 = 4 df = ( r − 1 ) ( c − 1 ) = ( 3 − 1 ) ( 3 − 1 ) = 2 × 2 = 4 “df”=(r-1)(c-1)=(3-1)(3-1)=2xx2=4\text{df} = (r – 1)(c – 1) = (3 – 1)(3 – 1) = 2 \times 2 = 4df=(r−1)(c−1)=(3−1)(3−1)=2×2=4
  1. Critical Value and Conclusion:
At the 5 % 5 % 5%5\%5% level of significance and 4 4 444 degrees of freedom, the critical value from the Chi-Square distribution table is approximately 9.488 9.488 9.4889.4889.488.
Since χ 2 = 13.4375 χ 2 = 13.4375 chi^(2)=13.4375\chi^2 = 13.4375χ2=13.4375 is greater than the critical value 9.488 9.488 9.4889.4889.488, we reject the null hypothesis H 0 H 0 H_(0)H_0H0.

Conclusion:

There is sufficient evidence to conclude that the tools and materials used are not independent at the 5 % 5 % 5%5\%5% level of significance.
(b) Explain the general procedure of testing of hypothesis.

Answer:

Testing a hypothesis is a statistical method that uses sample data to evaluate a hypothesis about a population parameter. The general procedure involves several steps, which are outlined below:

1. Formulate Hypotheses

  • Null Hypothesis ( H 0 H 0 H_(0)H_0H0): This is the hypothesis that there is no effect or no difference, and it represents the status quo. For example, H 0 : μ = μ 0 H 0 : μ = μ 0 H_(0):mu=mu_(0)H_0: \mu = \mu_0H0:μ=μ0.
  • Alternative Hypothesis ( H 1 H 1 H_(1)H_1H1 or H a H a H_(a)H_aHa): This is what you want to prove. It represents a change, effect, or difference. For example, H 1 : μ ≠ μ 0 H 1 : μ ≠ μ 0 H_(1):mu!=mu_(0)H_1: \mu \neq \mu_0H1:μ≠μ0, H 1 : μ > μ 0 H 1 : μ > μ 0 H_(1):mu > mu_(0)H_1: \mu > \mu_0H1:μ>μ0, or H 1 : μ < μ 0 H 1 : μ < μ 0 H_(1):mu < mu_(0)H_1: \mu < \mu_0H1:μ<μ0.

2. Choose the Significance Level ( α α alpha\alphaα)

  • The significance level is the probability of rejecting the null hypothesis when it is actually true (Type I error). Common values are 0.05, 0.01, and 0.10.

3. Select the Appropriate Test Statistic

  • The test statistic is a standardized value that is calculated from sample data during a hypothesis test. Depending on the nature of the data and the hypothesis, this could be a Z-test, t-test, chi-square test, F-test, etc.

4. Formulate the Decision Rule

  • The decision rule involves determining the critical value(s) from the statistical distribution of the test statistic (e.g., normal distribution, t-distribution, chi-square distribution) that correspond to the chosen significance level. This critical value(s) define the rejection region(s).

5. Collect Data and Compute the Test Statistic

  • Gather the sample data and compute the value of the test statistic based on the sample data.

6. Make a Decision

  • Compare the test statistic to the critical value(s):
    • If the test statistic falls within the rejection region, reject the null hypothesis ( H 0 H 0 H_(0)H_0H0).
    • If the test statistic does not fall within the rejection region, do not reject the null hypothesis.

7. Draw a Conclusion

  • Based on the decision made in the previous step, interpret the results in the context of the research question or problem.

Example

Step 1: Formulate Hypotheses

  • H 0 H 0 H_(0)H_0H0: The mean systolic blood pressure ( μ μ mu\muμ) is 120 mmHg.
  • H 1 H 1 H_(1)H_1H1: The mean systolic blood pressure ( μ μ mu\muμ) is not 120 mmHg.

Step 2: Choose the Significance Level

  • α = 0.05 α = 0.05 alpha=0.05\alpha = 0.05α=0.05

Step 3: Select the Appropriate Test Statistic

  • Assume the sample size is large and the population standard deviation is known. Use a Z-test.

Step 4: Formulate the Decision Rule

  • For a two-tailed test with α = 0.05 α = 0.05 alpha=0.05\alpha = 0.05α=0.05, the critical values are ± 1.96 ± 1.96 +-1.96\pm 1.96±1.96 (from the standard normal distribution).

Step 5: Collect Data and Compute the Test Statistic

  • Suppose we collect a sample of 30 individuals with a sample mean ( x ¯ x ¯ bar(x)\bar{x}x¯) of 123 mmHg and a population standard deviation ( σ σ sigma\sigmaσ) of 10 mmHg.
  • Compute the Z-score: Z = x ¯ − μ 0 σ / n = 123 − 120 10 / 30 ≈ 1.64 Z = x ¯ − μ 0 σ / n = 123 − 120 10 / 30 ≈ 1.64 Z=(( bar(x))-mu_(0))/(sigma//sqrtn)=(123-120)/(10//sqrt30)~~1.64Z = \frac{\bar{x} – \mu_0}{\sigma / \sqrt{n}} = \frac{123 – 120}{10 / \sqrt{30}} \approx 1.64Z=x¯−μ0σ/n=123−12010/30≈1.64

Step 6: Make a Decision

  • The test statistic (1.64) is not greater than 1.96 or less than -1.96, so it does not fall within the rejection region.

Step 7: Draw a Conclusion

  • Do not reject the null hypothesis. There is not enough evidence to conclude that the mean systolic blood pressure is different from 120 mmHg at the 5% significance level.
This general procedure provides a structured approach to hypothesis testing, ensuring that conclusions drawn from the sample data are statistically valid and reliable.

Question:-04

4.(a) A random sample of 15 stores was taken to analyse the sales of mobiles during last month. The correlation coefficient between sales and expenditure on advertisement was found to be 0.68 . Assuming that sales and expenditure on advertisement follow normal distribution, then test if these two are positively correlated at 1 % 1 % 1%1 \%1% level of significance.

Answer:

To test if the sales and expenditure on advertisement are positively correlated at the 1 % 1 % 1%1\%1% level of significance, we can perform a hypothesis test for the population correlation coefficient ρ ρ rho\rhoρ. Here’s the step-by-step procedure:

1. Formulate Hypotheses

  • Null Hypothesis ( H 0 H 0 H_(0)H_0H0): There is no positive correlation between sales and expenditure on advertisement ( ρ ≤ 0 ρ ≤ 0 rho <= 0\rho \leq 0ρ≤0).
  • Alternative Hypothesis ( H 1 H 1 H_(1)H_1H1): There is a positive correlation between sales and expenditure on advertisement ( ρ > 0 ρ > 0 rho > 0\rho > 0ρ>0).

2. Choose the Significance Level

  • α = 0.01 α = 0.01 alpha=0.01\alpha = 0.01α=0.01 (1%)

3. Select the Test Statistic

For testing the significance of the sample correlation coefficient r r rrr, we use the t-distribution with n − 2 n − 2 n-2n – 2n−2 degrees of freedom, where n n nnn is the sample size. The test statistic t t ttt is given by:
t = r n − 2 1 − r 2 t = r n − 2 1 − r 2 t=(rsqrt(n-2))/(sqrt(1-r^(2)))t = \frac{r \sqrt{n – 2}}{\sqrt{1 – r^2}}t=rn−21−r2

4. Formulate the Decision Rule

  • Determine the critical value t α , n − 2 t α , n − 2 t_(alpha,n-2)t_{\alpha, n-2}tα,n−2 from the t-distribution table for n − 2 n − 2 n-2n – 2n−2 degrees of freedom. For a one-tailed test at α = 0.01 α = 0.01 alpha=0.01\alpha = 0.01α=0.01 and n − 2 = 13 n − 2 = 13 n-2=13n – 2 = 13n−2=13 degrees of freedom, the critical value can be found in t-distribution tables or using statistical software.

5. Collect Data and Compute the Test Statistic

  • Sample size n = 15 n = 15 n=15n = 15n=15
  • Sample correlation coefficient r = 0.68 r = 0.68 r=0.68r = 0.68r=0.68
Calculate the test statistic:
t = 0.68 15 − 2 1 − 0.68 2 t = 0.68 15 − 2 1 − 0.68 2 t=(0.68sqrt(15-2))/(sqrt(1-0.68^(2)))t = \frac{0.68 \sqrt{15 – 2}}{\sqrt{1 – 0.68^2}}t=0.6815−21−0.682
t = 0.68 13 1 − 0.4624 t = 0.68 13 1 − 0.4624 t=(0.68sqrt13)/(sqrt(1-0.4624))t = \frac{0.68 \sqrt{13}}{\sqrt{1 – 0.4624}}t=0.68131−0.4624
t = 0.68 13 0.5376 t = 0.68 13 0.5376 t=(0.68sqrt13)/(sqrt0.5376)t = \frac{0.68 \sqrt{13}}{\sqrt{0.5376}}t=0.68130.5376
t = 0.68 × 3.605 0.7333 t = 0.68 × 3.605 0.7333 t=(0.68 xx3.605)/(0.7333)t = \frac{0.68 \times 3.605}{0.7333}t=0.68×3.6050.7333
t ≈ 3.34 t ≈ 3.34 t~~3.34t \approx 3.34t≈3.34

6. Determine the Critical Value

Using a t-distribution table, we look up the critical value for a one-tailed test at α = 0.01 α = 0.01 alpha=0.01\alpha = 0.01α=0.01 with 13 degrees of freedom. The critical value t 0.01 , 13 t 0.01 , 13 t_(0.01,13)t_{0.01, 13}t0.01,13 is approximately 2.650.

7. Make a Decision

  • Compare the computed test statistic with the critical value:
    • If t > t α , n − 2 t > t α , n − 2 t > t_(alpha,n-2)t > t_{\alpha, n-2}t>tα,n−2, reject the null hypothesis.
    • If t ≤ t α , n − 2 t ≤ t α , n − 2 t <= t_(alpha,n-2)t \leq t_{\alpha, n-2}t≤tα,n−2, do not reject the null hypothesis.
In this case, t ≈ 3.34 t ≈ 3.34 t~~3.34t \approx 3.34t≈3.34 is greater than the critical value 2.650.

8. Draw a Conclusion

Since the test statistic t ≈ 3.34 t ≈ 3.34 t~~3.34t \approx 3.34t≈3.34 is greater than the critical value 2.650, we reject the null hypothesis H 0 H 0 H_(0)H_0H0 at the 1 % 1 % 1%1\%1% level of significance.

Conclusion

There is sufficient evidence at the 1 % 1 % 1%1\%1% level of significance to conclude that there is a positive correlation between sales and expenditure on advertisement.
(b) An electric equipment manufacturing company claims that at most 10 % 10 % 10%10 \%10% of its products are defective. A store wants to purchase its products but before that they decided to test a sample of 200. If there are 30 defective products among these 200, can we agree with the manufacturer’s claim at 1 % 1 % 1%1 \%1% level of significance?

Answer:

To test the manufacturer’s claim that at most 10 % 10 % 10%10\%10% of its products are defective, we can perform a hypothesis test for a population proportion. Here are the steps:

1. Formulate Hypotheses

  • Null Hypothesis ( H 0 H 0 H_(0)H_0H0): The proportion of defective products is at most 10 % 10 % 10%10\%10% ( p ≤ 0.10 p ≤ 0.10 p <= 0.10p \leq 0.10p≤0.10).
  • Alternative Hypothesis ( H 1 H 1 H_(1)H_1H1): The proportion of defective products is greater than 10 % 10 % 10%10\%10% ( p > 0.10 p > 0.10 p > 0.10p > 0.10p>0.10).

2. Choose the Significance Level

  • α = 0.01 α = 0.01 alpha=0.01\alpha = 0.01α=0.01 (1%)

3. Select the Test Statistic

For testing the population proportion, we use the z-test for proportions. The test statistic z z zzz is given by:
z = p ^ − p 0 p 0 ( 1 − p 0 ) n z = p ^ − p 0 p 0 ( 1 − p 0 ) n z=(( hat(p))-p_(0))/(sqrt((p_(0)(1-p_(0)))/(n)))z = \frac{\hat{p} – p_0}{\sqrt{\frac{p_0 (1 – p_0)}{n}}}z=p^−p0p0(1−p0)n
where:
  • p ^ p ^ hat(p)\hat{p}p^ is the sample proportion of defectives.
  • p 0 p 0 p_(0)p_0p0 is the claimed population proportion.
  • n n nnn is the sample size.

4. Formulate the Decision Rule

  • Determine the critical value z α z α z_(alpha)z_{\alpha}zα from the standard normal distribution for a one-tailed test at α = 0.01 α = 0.01 alpha=0.01\alpha = 0.01α=0.01. The critical value z 0.01 z 0.01 z_(0.01)z_{0.01}z0.01 is approximately 2.33.

5. Collect Data and Compute the Test Statistic

  • Sample size n = 200 n = 200 n=200n = 200n=200
  • Number of defective products in the sample = 30
  • Sample proportion p ^ = 30 200 = 0.15 p ^ = 30 200 = 0.15 hat(p)=(30)/(200)=0.15\hat{p} = \frac{30}{200} = 0.15p^=30200=0.15
  • Claimed population proportion p 0 = 0.10 p 0 = 0.10 p_(0)=0.10p_0 = 0.10p0=0.10
Calculate the test statistic:
z = p ^ − p 0 p 0 ( 1 − p 0 ) n z = p ^ − p 0 p 0 ( 1 − p 0 ) n z=(( hat(p))-p_(0))/(sqrt((p_(0)(1-p_(0)))/(n)))z = \frac{\hat{p} – p_0}{\sqrt{\frac{p_0 (1 – p_0)}{n}}}z=p^−p0p0(1−p0)n
z = 0.15 − 0.10 0.10 ( 1 − 0.10 ) 200 z = 0.15 − 0.10 0.10 ( 1 − 0.10 ) 200 z=(0.15-0.10)/(sqrt((0.10(1-0.10))/(200)))z = \frac{0.15 – 0.10}{\sqrt{\frac{0.10 (1 – 0.10)}{200}}}z=0.15−0.100.10(1−0.10)200
z = 0.05 0.10 × 0.90 200 z = 0.05 0.10 × 0.90 200 z=(0.05)/(sqrt((0.10 xx0.90)/(200)))z = \frac{0.05}{\sqrt{\frac{0.10 \times 0.90}{200}}}z=0.050.10×0.90200
z = 0.05 0.09 200 z = 0.05 0.09 200 z=(0.05)/(sqrt((0.09)/(200)))z = \frac{0.05}{\sqrt{\frac{0.09}{200}}}z=0.050.09200
z = 0.05 0.00045 z = 0.05 0.00045 z=(0.05)/(sqrt0.00045)z = \frac{0.05}{\sqrt{0.00045}}z=0.050.00045
z = 0.05 0.0212 z = 0.05 0.0212 z=(0.05)/(0.0212)z = \frac{0.05}{0.0212}z=0.050.0212
z ≈ 2.36 z ≈ 2.36 z~~2.36z \approx 2.36z≈2.36

6. Make a Decision

  • Compare the computed test statistic with the critical value:
    • If z > z α z > z α z > z_(alpha)z > z_{\alpha}z>zα, reject the null hypothesis.
    • If z ≤ z α z ≤ z α z <= z_(alpha)z \leq z_{\alpha}z≤zα, do not reject the null hypothesis.
In this case, z ≈ 2.36 z ≈ 2.36 z~~2.36z \approx 2.36z≈2.36 is greater than the critical value 2.33.

7. Draw a Conclusion

Since the test statistic z ≈ 2.36 z ≈ 2.36 z~~2.36z \approx 2.36z≈2.36 is greater than the critical value 2.33, we reject the null hypothesis H 0 H 0 H_(0)H_0H0 at the 1 % 1 % 1%1\%1% level of significance.

Conclusion

There is sufficient evidence at the 1 % 1 % 1%1\%1% level of significance to reject the manufacturer’s claim that at most 10 % 10 % 10%10\%10% of its products are defective. Therefore, we cannot agree with the manufacturer’s claim based on this sample.

Question:-05

5.(a) An experiment was conducted to compare the defective items produced by two different machines A A A\mathrm{A}A and B B B\mathrm{B}B. The data on number of defective items produced by the machines were observed and given in the table as follows :
A B
26 19
37 22
40 24
35 27
30 24
30 18
40 20
26 19
30 25
35
45
A B 26 19 37 22 40 24 35 27 30 24 30 18 40 20 26 19 30 25 35 45 | A | B | | :—: | :—: | | 26 | 19 | | 37 | 22 | | 40 | 24 | | 35 | 27 | | 30 | 24 | | 30 | 18 | | 40 | 20 | | 26 | 19 | | 30 | 25 | | 35 | | | 45 | |
Obtain 95 % 95 % 95%95 \%95% confidence interval for variance ratio of the number of defective items produced by machines A A A\mathrm{A}A and B B B\mathrm{B}B, respectively.

Answer:

To obtain the 95 % 95 % 95%95\%95% confidence interval for the variance ratio of the number of defective items produced by machines A and B, we need to calculate the variances of the samples from both machines and then use the F-distribution.

Step 1: Calculate Sample Variances

First, let’s calculate the sample variances for both machines A and B.

Machine A

The data for machine A is: 26 , 37 , 40 , 35 , 30 , 30 , 40 , 26 , 30 , 35 , 45 26 , 37 , 40 , 35 , 30 , 30 , 40 , 26 , 30 , 35 , 45 26,37,40,35,30,30,40,26,30,35,4526, 37, 40, 35, 30, 30, 40, 26, 30, 35, 4526,37,40,35,30,30,40,26,30,35,45
  1. Calculate the mean ( X ¯ A X ¯ A bar(X)_(A)\bar{X}_AX¯A) of the data for machine A.
X ¯ A = 26 + 37 + 40 + 35 + 30 + 30 + 40 + 26 + 30 + 35 + 45 11 = 374 11 ≈ 34 X ¯ A = 26 + 37 + 40 + 35 + 30 + 30 + 40 + 26 + 30 + 35 + 45 11 = 374 11 ≈ 34 bar(X)_(A)=(26+37+40+35+30+30+40+26+30+35+45)/(11)=(374)/(11)~~34\bar{X}_A = \frac{26 + 37 + 40 + 35 + 30 + 30 + 40 + 26 + 30 + 35 + 45}{11} = \frac{374}{11} \approx 34X¯A=26+37+40+35+30+30+40+26+30+35+4511=37411≈34
  1. Calculate the sample variance ( S A 2 S A 2 S_(A)^(2)S_A^2SA2).
S A 2 = ∑ ( X i − X ¯ A ) 2 n A − 1 S A 2 = ∑ ( X i − X ¯ A ) 2 n A − 1 S_(A)^(2)=(sum(X_(i)- bar(X)_(A))^(2))/(n_(A)-1)S_A^2 = \frac{\sum (X_i – \bar{X}_A)^2}{n_A – 1}SA2=∑(Xi−X¯A)2nA−1
S A 2 = ( 26 − 34 ) 2 + ( 37 − 34 ) 2 + ( 40 − 34 ) 2 + ( 35 − 34 ) 2 + ( 30 − 34 ) 2 + ( 30 − 34 ) 2 + ( 40 − 34 ) 2 + ( 26 − 34 ) 2 + ( 30 − 34 ) 2 + ( 35 − 34 ) 2 + ( 45 − 34 ) 2 11 − 1 S A 2 = ( 26 − 34 ) 2 + ( 37 − 34 ) 2 + ( 40 − 34 ) 2 + ( 35 − 34 ) 2 + ( 30 − 34 ) 2 + ( 30 − 34 ) 2 + ( 40 − 34 ) 2 + ( 26 − 34 ) 2 + ( 30 − 34 ) 2 + ( 35 − 34 ) 2 + ( 45 − 34 ) 2 11 − 1 S_(A)^(2)=((26-34)^(2)+(37-34)^(2)+(40-34)^(2)+(35-34)^(2)+(30-34)^(2)+(30-34)^(2)+(40-34)^(2)+(26-34)^(2)+(30-34)^(2)+(35-34)^(2)+(45-34)^(2))/(11-1)S_A^2 = \frac{(26-34)^2 + (37-34)^2 + (40-34)^2 + (35-34)^2 + (30-34)^2 + (30-34)^2 + (40-34)^2 + (26-34)^2 + (30-34)^2 + (35-34)^2 + (45-34)^2}{11-1}SA2=(26−34)2+(37−34)2+(40−34)2+(35−34)2+(30−34)2+(30−34)2+(40−34)2+(26−34)2+(30−34)2+(35−34)2+(45−34)211−1
S A 2 = ( 64 + 9 + 36 + 1 + 16 + 16 + 36 + 64 + 16 + 1 + 121 ) 10 S A 2 = ( 64 + 9 + 36 + 1 + 16 + 16 + 36 + 64 + 16 + 1 + 121 ) 10 S_(A)^(2)=((64+9+36+1+16+16+36+64+16+1+121))/(10)S_A^2 = \frac{(64 + 9 + 36 + 1 + 16 + 16 + 36 + 64 + 16 + 1 + 121)}{10}SA2=(64+9+36+1+16+16+36+64+16+1+121)10
S A 2 = 379 10 = 37.9 S A 2 = 379 10 = 37.9 S_(A)^(2)=(379)/(10)=37.9S_A^2 = \frac{379}{10} = 37.9SA2=37910=37.9

Machine B

The data for machine B is: 19 , 22 , 24 , 27 , 24 , 18 , 20 , 19 , 25 19 , 22 , 24 , 27 , 24 , 18 , 20 , 19 , 25 19,22,24,27,24,18,20,19,2519, 22, 24, 27, 24, 18, 20, 19, 2519,22,24,27,24,18,20,19,25
  1. Calculate the mean ( X ¯ B X ¯ B bar(X)_(B)\bar{X}_BX¯B) of the data for machine B.
X ¯ B = 19 + 22 + 24 + 27 + 24 + 18 + 20 + 19 + 25 9 = 198 9 ≈ 22 X ¯ B = 19 + 22 + 24 + 27 + 24 + 18 + 20 + 19 + 25 9 = 198 9 ≈ 22 bar(X)_(B)=(19+22+24+27+24+18+20+19+25)/(9)=(198)/(9)~~22\bar{X}_B = \frac{19 + 22 + 24 + 27 + 24 + 18 + 20 + 19 + 25}{9} = \frac{198}{9} \approx 22X¯B=19+22+24+27+24+18+20+19+259=1989≈22
  1. Calculate the sample variance ( S B 2 S B 2 S_(B)^(2)S_B^2SB2).
S B 2 = ∑ ( X i − X ¯ B ) 2 n B − 1 S B 2 = ∑ ( X i − X ¯ B ) 2 n B − 1 S_(B)^(2)=(sum(X_(i)- bar(X)_(B))^(2))/(n_(B)-1)S_B^2 = \frac{\sum (X_i – \bar{X}_B)^2}{n_B – 1}SB2=∑(Xi−X¯B)2nB−1
S B 2 = ( 19 − 22 ) 2 + ( 22 − 22 ) 2 + ( 24 − 22 ) 2 + ( 27 − 22 ) 2 + ( 24 − 22 ) 2 + ( 18 − 22 ) 2 + ( 20 − 22 ) 2 + ( 19 − 22 ) 2 + ( 25 − 22 ) 2 9 − 1 S B 2 = ( 19 − 22 ) 2 + ( 22 − 22 ) 2 + ( 24 − 22 ) 2 + ( 27 − 22 ) 2 + ( 24 − 22 ) 2 + ( 18 − 22 ) 2 + ( 20 − 22 ) 2 + ( 19 − 22 ) 2 + ( 25 − 22 ) 2 9 − 1 S_(B)^(2)=((19-22)^(2)+(22-22)^(2)+(24-22)^(2)+(27-22)^(2)+(24-22)^(2)+(18-22)^(2)+(20-22)^(2)+(19-22)^(2)+(25-22)^(2))/(9-1)S_B^2 = \frac{(19-22)^2 + (22-22)^2 + (24-22)^2 + (27-22)^2 + (24-22)^2 + (18-22)^2 + (20-22)^2 + (19-22)^2 + (25-22)^2}{9-1}SB2=(19−22)2+(22−22)2+(24−22)2+(27−22)2+(24−22)2+(18−22)2+(20−22)2+(19−22)2+(25−22)29−1
S B 2 = ( 9 + 0 + 4 + 25 + 4 + 16 + 4 + 9 + 9 ) 8 S B 2 = ( 9 + 0 + 4 + 25 + 4 + 16 + 4 + 9 + 9 ) 8 S_(B)^(2)=((9+0+4+25+4+16+4+9+9))/(8)S_B^2 = \frac{(9 + 0 + 4 + 25 + 4 + 16 + 4 + 9 + 9)}{8}SB2=(9+0+4+25+4+16+4+9+9)8
S B 2 = 80 8 = 10 S B 2 = 80 8 = 10 S_(B)^(2)=(80)/(8)=10S_B^2 = \frac{80}{8} = 10SB2=808=10

Step 2: Calculate the Variance Ratio

The variance ratio F F FFF is:
F = S A 2 S B 2 = 37.9 10 ≈ 3.79 F = S A 2 S B 2 = 37.9 10 ≈ 3.79 F=(S_(A)^(2))/(S_(B)^(2))=(37.9)/(10)~~3.79F = \frac{S_A^2}{S_B^2} = \frac{37.9}{10} \approx 3.79F=SA2SB2=37.910≈3.79

Step 3: Find the Critical Values for the F-distribution

To find the 95 % 95 % 95%95\%95% confidence interval for the variance ratio, we need to use the F-distribution with degrees of freedom d f 1 = n A − 1 = 10 d f 1 = n A − 1 = 10 df_(1)=n_(A)-1=10df_1 = n_A – 1 = 10df1=nA−1=10 and d f 2 = n B − 1 = 8 d f 2 = n B − 1 = 8 df_(2)=n_(B)-1=8df_2 = n_B – 1 = 8df2=nB−1=8.
Using an F-distribution table or calculator, we find the critical values F 0.025 , 10 , 8 F 0.025 , 10 , 8 F_(0.025,10,8)F_{0.025, 10, 8}F0.025,10,8 and F 0.975 , 10 , 8 F 0.975 , 10 , 8 F_(0.975,10,8)F_{0.975, 10, 8}F0.975,10,8:
F 0.025 , 10 , 8 ≈ 4.82 (upper critical value) F 0.025 , 10 , 8 ≈ 4.82 (upper critical value) F_(0.025,10,8)~~4.82quad(upper critical value)F_{0.025, 10, 8} \approx 4.82 \quad \text{(upper critical value)}F0.025,10,8≈4.82(upper critical value)
F 0.975 , 10 , 8 ≈ 0.23 (lower critical value) F 0.975 , 10 , 8 ≈ 0.23 (lower critical value) F_(0.975,10,8)~~0.23quad(lower critical value)F_{0.975, 10, 8} \approx 0.23 \quad \text{(lower critical value)}F0.975,10,8≈0.23(lower critical value)

Step 4: Calculate the Confidence Interval

The 95 % 95 % 95%95\%95% confidence interval for the variance ratio is given by:
( S A 2 S B 2 ⋅ 1 F 0.025 , 10 , 8 , S A 2 S B 2 ⋅ F 0.975 , 10 , 8 ) S A 2 S B 2 ⋅ 1 F 0.025 , 10 , 8 , S A 2 S B 2 ⋅ F 0.975 , 10 , 8 ((S_(A)^(2))/(S_(B)^(2))*(1)/(F_(0.025,10,8)),(S_(A)^(2))/(S_(B)^(2))*F_(0.975,10,8))\left( \frac{S_A^2}{S_B^2} \cdot \frac{1}{F_{0.025, 10, 8}}, \frac{S_A^2}{S_B^2} \cdot F_{0.975, 10, 8} \right)(SA2SB2⋅1F0.025,10,8,SA2SB2⋅F0.975,10,8)
( 37.9 10 ⋅ 1 4.82 , 37.9 10 ⋅ 4.82 ) 37.9 10 ⋅ 1 4.82 , 37.9 10 ⋅ 4.82 ((37.9)/(10)*(1)/(4.82),(37.9)/(10)*4.82)\left( \frac{37.9}{10} \cdot \frac{1}{4.82}, \frac{37.9}{10} \cdot 4.82 \right)(37.910⋅14.82,37.910⋅4.82)
( 3.79 ⋅ 1 4.82 , 3.79 ⋅ 4.82 ) 3.79 ⋅ 1 4.82 , 3.79 ⋅ 4.82 (3.79*(1)/(4.82),3.79*4.82)\left( 3.79 \cdot \frac{1}{4.82}, 3.79 \cdot 4.82 \right)(3.79⋅14.82,3.79⋅4.82)
( 3.79 4.82 , 3.79 ⋅ 4.82 ) 3.79 4.82 , 3.79 ⋅ 4.82 ((3.79)/(4.82),3.79*4.82)\left( \frac{3.79}{4.82}, 3.79 \cdot 4.82 \right)(3.794.82,3.79⋅4.82)
( 0.786 , 18.27 ) 0.786 , 18.27 (0.786,18.27)\left( 0.786, 18.27 \right)(0.786,18.27)

Conclusion

The 95 % 95 % 95%95\%95% confidence interval for the variance ratio of the number of defective items produced by machines A and B is approximately ( 0.786 , 18.27 ) ( 0.786 , 18.27 ) (0.786,18.27)(0.786, 18.27)(0.786,18.27).
(b) Write four differences between parametric and non-parametric tests.

Answer:

Here are four key differences between parametric and non-parametric tests:

1. Assumptions about the Population Distribution

  • Parametric Tests:
    • These tests assume that the data follows a certain distribution, usually a normal distribution.
    • Examples include the t-test and ANOVA, which assume the data is normally distributed.
  • Non-Parametric Tests:
    • These tests do not assume a specific distribution for the data.
    • They are often used when the data does not meet the assumptions of parametric tests or when dealing with ordinal data or ranks.
    • Examples include the Mann-Whitney U test and the Kruskal-Wallis test.

2. Data Type

  • Parametric Tests:
    • Typically used for interval or ratio data, which are numerical and can be meaningfully averaged.
    • The data should be measured on a continuous scale and have meaningful zero points.
  • Non-Parametric Tests:
    • Suitable for nominal or ordinal data, which may include ranks or categories without a meaningful average.
    • They can also be used for interval or ratio data that do not meet the assumptions required for parametric tests.

3. Robustness to Outliers and Non-Normality

  • Parametric Tests:
    • Less robust to outliers and non-normal distributions. Outliers can significantly affect the results of these tests.
    • They rely on specific assumptions about the population distribution, and violations of these assumptions can lead to incorrect conclusions.
  • Non-Parametric Tests:
    • More robust to outliers and non-normality. They make fewer assumptions about the underlying data distribution.
    • These tests are based on ranks or medians, which are less affected by extreme values and skewed distributions.

4. Efficiency

  • Parametric Tests:
    • Generally more powerful and efficient when the assumptions about the data are met. They can detect smaller differences or effects with a given sample size.
    • They make full use of the data by considering the exact values of observations.
  • Non-Parametric Tests:
    • Generally less powerful than parametric tests when the data actually meets the assumptions of parametric tests.
    • They may require a larger sample size to achieve the same power as parametric tests because they use ranks or categories rather than the actual data values.
These differences highlight the contexts in which each type of test is appropriate and the trade-offs involved in choosing between parametric and non-parametric methods.

Question:-06

6.(a) The length of a steel rod is distributed normally with mean 12 metre and standard deviation 0.1 metre. For a random sample of size 10 , find :
(i) Mean and variance of the sampling distribution of mean.
(ii) The probability that the sample mean lies between 11.94 metre and 12.06 metre.

Answer:

Let’s address each part of the problem step by step.

Given Data:

  • Population mean ( μ μ mu\muμ) = 12 meters
  • Population standard deviation ( σ σ sigma\sigmaσ) = 0.1 meters
  • Sample size ( n n nnn) = 10

(i) Mean and Variance of the Sampling Distribution of the Mean

The sampling distribution of the sample mean X ¯ X ¯ bar(X)\bar{X}X¯ for a normally distributed population is also normally distributed with:
  • Mean: μ X ¯ = μ μ X ¯ = μ mu_( bar(X))=mu\mu_{\bar{X}} = \muμX¯=μ
  • Variance: σ X ¯ 2 = σ 2 n σ X ¯ 2 = σ 2 n sigma_( bar(X))^(2)=(sigma^(2))/(n)\sigma_{\bar{X}}^2 = \frac{\sigma^2}{n}σX¯2=σ2n
Mean of the Sampling Distribution:
μ X ¯ = 12 meters μ X ¯ = 12 meters mu_( bar(X))=12″ meters”\mu_{\bar{X}} = 12 \text{ meters}μX¯=12 meters
Variance of the Sampling Distribution:
σ X ¯ 2 = σ 2 n = ( 0.1 ) 2 10 = 0.01 10 = 0.001 square meters σ X ¯ 2 = σ 2 n = ( 0.1 ) 2 10 = 0.01 10 = 0.001 square meters sigma_( bar(X))^(2)=(sigma^(2))/(n)=((0.1)^(2))/(10)=(0.01)/(10)=0.001″ square meters”\sigma_{\bar{X}}^2 = \frac{\sigma^2}{n} = \frac{(0.1)^2}{10} = \frac{0.01}{10} = 0.001 \text{ square meters}σX¯2=σ2n=(0.1)210=0.0110=0.001 square meters
Standard Deviation of the Sampling Distribution:
σ X ¯ = σ X ¯ 2 = 0.001 = 0.0316 meters σ X ¯ = σ X ¯ 2 = 0.001 = 0.0316 meters sigma_( bar(X))=sqrt(sigma_( bar(X))^(2))=sqrt0.001=0.0316″ meters”\sigma_{\bar{X}} = \sqrt{\sigma_{\bar{X}}^2} = \sqrt{0.001} = 0.0316 \text{ meters}σX¯=σX¯2=0.001=0.0316 meters

(ii) The Probability that the Sample Mean Lies Between 11.94 meters and 12.06 meters

To find this probability, we use the Z-score formula for the sampling distribution of the mean:
Z = X ¯ − μ X ¯ σ X ¯ Z = X ¯ − μ X ¯ σ X ¯ Z=(( bar(X))-mu_( bar(X)))/(sigma_( bar(X)))Z = \frac{\bar{X} – \mu_{\bar{X}}}{\sigma_{\bar{X}}}Z=X¯−μX¯σX¯
First, calculate the Z-scores for 11.94 meters and 12.06 meters.
For X ¯ = 11.94 X ¯ = 11.94 bar(X)=11.94\bar{X} = 11.94X¯=11.94:
Z 11.94 = 11.94 − 12 0.0316 = − 0.06 0.0316 ≈ − 1.90 Z 11.94 = 11.94 − 12 0.0316 = − 0.06 0.0316 ≈ − 1.90 Z_(11.94)=(11.94-12)/(0.0316)=(-0.06)/(0.0316)~~-1.90Z_{11.94} = \frac{11.94 – 12}{0.0316} = \frac{-0.06}{0.0316} \approx -1.90Z11.94=11.94−120.0316=−0.060.0316≈−1.90
For X ¯ = 12.06 X ¯ = 12.06 bar(X)=12.06\bar{X} = 12.06X¯=12.06:
Z 12.06 = 12.06 − 12 0.0316 = 0.06 0.0316 ≈ 1.90 Z 12.06 = 12.06 − 12 0.0316 = 0.06 0.0316 ≈ 1.90 Z_(12.06)=(12.06-12)/(0.0316)=(0.06)/(0.0316)~~1.90Z_{12.06} = \frac{12.06 – 12}{0.0316} = \frac{0.06}{0.0316} \approx 1.90Z12.06=12.06−120.0316=0.060.0316≈1.90
Next, we find the probability corresponding to these Z-scores using the standard normal distribution table (or a calculator).
Using standard normal distribution tables or software:
P ( Z ≤ 1.90 ) ≈ 0.9713 P ( Z ≤ 1.90 ) ≈ 0.9713 P(Z <= 1.90)~~0.9713P(Z \leq 1.90) \approx 0.9713P(Z≤1.90)≈0.9713
P ( Z ≤ − 1.90 ) ≈ 0.0287 P ( Z ≤ − 1.90 ) ≈ 0.0287 P(Z <= -1.90)~~0.0287P(Z \leq -1.90) \approx 0.0287P(Z≤−1.90)≈0.0287
The probability that the sample mean lies between 11.94 meters and 12.06 meters is the difference between these two probabilities:
P ( 11.94 ≤ X ¯ ≤ 12.06 ) = P ( Z ≤ 1.90 ) − P ( Z ≤ − 1.90 ) = 0.9713 − 0.0287 = 0.9426 P ( 11.94 ≤ X ¯ ≤ 12.06 ) = P ( Z ≤ 1.90 ) − P ( Z ≤ − 1.90 ) = 0.9713 − 0.0287 = 0.9426 P(11.94 <= bar(X) <= 12.06)=P(Z <= 1.90)-P(Z <= -1.90)=0.9713-0.0287=0.9426P(11.94 \leq \bar{X} \leq 12.06) = P(Z \leq 1.90) – P(Z \leq -1.90) = 0.9713 – 0.0287 = 0.9426P(11.94≤X¯≤12.06)=P(Z≤1.90)−P(Z≤−1.90)=0.9713−0.0287=0.9426

Summary

  1. Mean of the sampling distribution of the mean: 12 meters
  2. Variance of the sampling distribution of the mean: 0.001 square meters
  3. Probability that the sample mean lies between 11.94 meters and 12.06 meters: 0.9426 0.9426 0.94260.94260.9426 or 94.26 % 94.26 % 94.26%94.26\%94.26%
(b) The reduction of weight (in k g k g kg\mathrm{kg}kg ) after a dietplan are recorded as follows :
6.5 , 7.7 , 5.6 , 7.3 , 6.7 , 7.8 , 6.7 , 6.2 , 5.2 , 6.6 6.5 , 7.7 , 5.6 , 7.3 , 6.7 , 7.8 , 6.7 , 6.2 , 5.2 , 6.6 6.5,7.7,5.6,7.3,6.7,7.8,6.7,6.2,5.2,6.66.5,7.7,5.6,7.3,6.7,7.8,6.7,6.2,5.2,6.66.5,7.7,5.6,7.3,6.7,7.8,6.7,6.2,5.2,6.6, 6.0, 7.0, 7.2 , 6.8 7.2 , 6.8 7.2,6.87.2,6.87.2,6.8 and 7.2 .
It is observed that reduction in weight follows an exponential distribution with parameter θ θ theta\thetaθ whose pdf is given by :
f ( x ) = 1 θ e − x / θ ; x ≥ 0 , θ > 0 f ( x ) = 1 θ e − x / θ ; x ≥ 0 , θ > 0 f(x)=(1)/(theta)e^(-x//theta);x >= 0,theta > 0f(x)=\frac{1}{\theta} e^{-x / \theta} ; x \geq 0, \theta>0f(x)=1θe−x/θ;x≥0,θ>0
(i) Find the maximum likelihood estimator of the parameter θ θ theta\thetaθ.
(ii) Determine the maximum likelihood estimate of θ θ theta\thetaθ on the basis of the given data.

Answer:

To find the maximum likelihood estimator (MLE) and the maximum likelihood estimate of the parameter θ θ theta\thetaθ for the given exponential distribution, we will follow these steps:

Given Data

Reduction in weight (in kg): 6.5 , 7.7 , 5.6 , 7.3 , 6.7 , 7.8 , 6.7 , 6.2 , 5.2 , 6.6 , 6.0 , 7.0 , 7.2 , 6.8 , 7.2 6.5 , 7.7 , 5.6 , 7.3 , 6.7 , 7.8 , 6.7 , 6.2 , 5.2 , 6.6 , 6.0 , 7.0 , 7.2 , 6.8 , 7.2 6.5,7.7,5.6,7.3,6.7,7.8,6.7,6.2,5.2,6.6,6.0,7.0,7.2,6.8,7.26.5, 7.7, 5.6, 7.3, 6.7, 7.8, 6.7, 6.2, 5.2, 6.6, 6.0, 7.0, 7.2, 6.8, 7.26.5,7.7,5.6,7.3,6.7,7.8,6.7,6.2,5.2,6.6,6.0,7.0,7.2,6.8,7.2

Probability Density Function (PDF) of Exponential Distribution

f ( x ) = 1 θ e − x / θ , x ≥ 0 , θ > 0 f ( x ) = 1 θ e − x / θ , x ≥ 0 , θ > 0 f(x)=(1)/(theta)e^(-x//theta),quad x >= 0,theta > 0f(x) = \frac{1}{\theta} e^{-x / \theta}, \quad x \geq 0, \theta > 0f(x)=1θe−x/θ,x≥0,θ>0

(i) Find the Maximum Likelihood Estimator (MLE) of the Parameter θ θ theta\thetaθ

To find the MLE of θ θ theta\thetaθ, we need to set up the likelihood function and then maximize it.
The likelihood function L ( θ ) L ( θ ) L(theta)L(\theta)L(θ) for the exponential distribution is:
L ( θ ) = ∏ i = 1 n f ( x i ) = ∏ i = 1 n ( 1 θ e − x i / θ ) L ( θ ) = ∏ i = 1 n   f ( x i ) = ∏ i = 1 n   1 θ e − x i / θ L(theta)=prod_(i=1)^(n)f(x_(i))=prod_(i=1)^(n)((1)/(theta)e^(-x_(i)//theta))L(\theta) = \prod_{i=1}^n f(x_i) = \prod_{i=1}^n \left(\frac{1}{\theta} e^{-x_i / \theta}\right)L(θ)=∏i=1nf(xi)=∏i=1n(1θe−xi/θ)
Given the data points x 1 , x 2 , … , x n x 1 , x 2 , … , x n x_(1),x_(2),dots,x_(n)x_1, x_2, \ldots, x_nx1,x2,…,xn, the likelihood function becomes:
L ( θ ) = ( 1 θ ) n exp ⁡ ( − ∑ i = 1 n x i θ ) L ( θ ) = 1 θ n exp ⁡ − ∑ i = 1 n   x i θ L(theta)=((1)/(theta))^(n)exp(-(sum_(i=1)^(n)x_(i))/(theta))L(\theta) = \left(\frac{1}{\theta}\right)^n \exp\left(-\frac{\sum_{i=1}^n x_i}{\theta}\right)L(θ)=(1θ)nexp⁡(−∑i=1nxiθ)
The log-likelihood function ℓ ( θ ) ℓ ( θ ) ℓ(theta)\ell(\theta)ℓ(θ) is:
ℓ ( θ ) = log ⁡ L ( θ ) = n log ⁡ ( 1 θ ) − ∑ i = 1 n x i θ ℓ ( θ ) = log ⁡ L ( θ ) = n log ⁡ 1 θ − ∑ i = 1 n   x i θ ℓ(theta)=log L(theta)=n log((1)/(theta))-(sum_(i=1)^(n)x_(i))/(theta)\ell(\theta) = \log L(\theta) = n \log\left(\frac{1}{\theta}\right) – \frac{\sum_{i=1}^n x_i}{\theta}ℓ(θ)=log⁡L(θ)=nlog⁡(1θ)−∑i=1nxiθ
ℓ ( θ ) = − n log ⁡ ( θ ) − ∑ i = 1 n x i θ ℓ ( θ ) = − n log ⁡ ( θ ) − ∑ i = 1 n   x i θ ℓ(theta)=-n log(theta)-(sum_(i=1)^(n)x_(i))/(theta)\ell(\theta) = -n \log(\theta) – \frac{\sum_{i=1}^n x_i}{\theta}ℓ(θ)=−nlog⁡(θ)−∑i=1nxiθ
To find the MLE, we take the derivative of the log-likelihood function with respect to θ θ theta\thetaθ and set it to zero:
d ℓ ( θ ) d θ = − n θ + ∑ i = 1 n x i θ 2 = 0 d ℓ ( θ ) d θ = − n θ + ∑ i = 1 n   x i θ 2 = 0 (dℓ(theta))/(d theta)=-(n)/( theta)+(sum_(i=1)^(n)x_(i))/(theta^(2))=0\frac{d\ell(\theta)}{d\theta} = -\frac{n}{\theta} + \frac{\sum_{i=1}^n x_i}{\theta^2} = 0dℓ(θ)dθ=−nθ+∑i=1nxiθ2=0
Solving for θ θ theta\thetaθ:
− n θ + ∑ i = 1 n x i θ 2 = 0 − n θ + ∑ i = 1 n   x i θ 2 = 0 -(n)/( theta)+(sum_(i=1)^(n)x_(i))/(theta^(2))=0-\frac{n}{\theta} + \frac{\sum_{i=1}^n x_i}{\theta^2} = 0−nθ+∑i=1nxiθ2=0
∑ i = 1 n x i θ 2 = n θ ∑ i = 1 n   x i θ 2 = n θ (sum_(i=1)^(n)x_(i))/(theta^(2))=(n)/( theta)\frac{\sum_{i=1}^n x_i}{\theta^2} = \frac{n}{\theta}∑i=1nxiθ2=nθ
∑ i = 1 n x i = n θ ∑ i = 1 n   x i = n θ sum_(i=1)^(n)x_(i)=n theta\sum_{i=1}^n x_i = n\theta∑i=1nxi=nθ
θ = ∑ i = 1 n x i n θ = ∑ i = 1 n   x i n theta=(sum_(i=1)^(n)x_(i))/(n)\theta = \frac{\sum_{i=1}^n x_i}{n}θ=∑i=1nxin
Thus, the MLE of θ θ theta\thetaθ is the sample mean:
θ ^ = x ¯ = ∑ i = 1 n x i n θ ^ = x ¯ = ∑ i = 1 n   x i n hat(theta)= bar(x)=(sum_(i=1)^(n)x_(i))/(n)\hat{\theta} = \bar{x} = \frac{\sum_{i=1}^n x_i}{n}θ^=x¯=∑i=1nxin

(ii) Determine the Maximum Likelihood Estimate of θ θ theta\thetaθ on the Basis of the Given Data

First, we calculate the sample mean x ¯ x ¯ bar(x)\bar{x}x¯.
Given the data points:
6.5 , 7.7 , 5.6 , 7.3 , 6.7 , 7.8 , 6.7 , 6.2 , 5.2 , 6.6 , 6.0 , 7.0 , 7.2 , 6.8 , 7.2 6.5 , 7.7 , 5.6 , 7.3 , 6.7 , 7.8 , 6.7 , 6.2 , 5.2 , 6.6 , 6.0 , 7.0 , 7.2 , 6.8 , 7.2 6.5,7.7,5.6,7.3,6.7,7.8,6.7,6.2,5.2,6.6,6.0,7.0,7.2,6.8,7.26.5, 7.7, 5.6, 7.3, 6.7, 7.8, 6.7, 6.2, 5.2, 6.6, 6.0, 7.0, 7.2, 6.8, 7.26.5,7.7,5.6,7.3,6.7,7.8,6.7,6.2,5.2,6.6,6.0,7.0,7.2,6.8,7.2
Sum of the data points:
∑ i = 1 15 x i = 6.5 + 7.7 + 5.6 + 7.3 + 6.7 + 7.8 + 6.7 + 6.2 + 5.2 + 6.6 + 6.0 + 7.0 + 7.2 + 6.8 + 7.2 ∑ i = 1 15   x i = 6.5 + 7.7 + 5.6 + 7.3 + 6.7 + 7.8 + 6.7 + 6.2 + 5.2 + 6.6 + 6.0 + 7.0 + 7.2 + 6.8 + 7.2 sum_(i=1)^(15)x_(i)=6.5+7.7+5.6+7.3+6.7+7.8+6.7+6.2+5.2+6.6+6.0+7.0+7.2+6.8+7.2\sum_{i=1}^{15} x_i = 6.5 + 7.7 + 5.6 + 7.3 + 6.7 + 7.8 + 6.7 + 6.2 + 5.2 + 6.6 + 6.0 + 7.0 + 7.2 + 6.8 + 7.2∑i=115xi=6.5+7.7+5.6+7.3+6.7+7.8+6.7+6.2+5.2+6.6+6.0+7.0+7.2+6.8+7.2
∑ i = 1 15 x i = 102.5 ∑ i = 1 15   x i = 102.5 sum_(i=1)^(15)x_(i)=102.5\sum_{i=1}^{15} x_i = 102.5∑i=115xi=102.5
Sample mean:
x ¯ = 102.5 15 ≈ 6.833 x ¯ = 102.5 15 ≈ 6.833 bar(x)=(102.5)/(15)~~6.833\bar{x} = \frac{102.5}{15} \approx 6.833x¯=102.515≈6.833
Thus, the maximum likelihood estimate of θ θ theta\thetaθ is:
θ ^ ≈ 6.833 θ ^ ≈ 6.833 hat(theta)~~6.833\hat{\theta} \approx 6.833θ^≈6.833

Summary

  1. The maximum likelihood estimator (MLE) of the parameter θ θ theta\thetaθ is θ ^ = x ¯ θ ^ = x ¯ hat(theta)= bar(x)\hat{\theta} = \bar{x}θ^=x¯.
  2. The maximum likelihood estimate of θ θ theta\thetaθ based on the given data is approximately 6.833.

Question:-07

7.(a) Explain the properties of good estimator with examples.

Answer:

In statistics, a good estimator should possess several key properties to ensure that it provides accurate and reliable estimates of the population parameters. These properties are:

1. Unbiasedness

An estimator is unbiased if the expected value of the estimator is equal to the true value of the population parameter. In other words, on average, the estimator hits the true parameter value.
Example:
  • The sample mean X ¯ X ¯ bar(X)\bar{X}X¯ is an unbiased estimator of the population mean μ μ mu\muμ, since E ( X ¯ ) = μ E ( X ¯ ) = μ E( bar(X))=muE(\bar{X}) = \muE(X¯)=μ.

2. Consistency

An estimator is consistent if, as the sample size increases, the estimator converges in probability to the true value of the population parameter. This means that with larger samples, the estimator becomes more accurate.
Example:
  • The sample mean X ¯ X ¯ bar(X)\bar{X}X¯ is a consistent estimator of the population mean μ μ mu\muμ. As the sample size n n nnn increases, X ¯ X ¯ bar(X)\bar{X}X¯ gets closer to μ μ mu\muμ.

3. Efficiency

An estimator is efficient if it has the smallest variance among all unbiased estimators of the parameter. Efficiency is often measured by the Mean Squared Error (MSE), which is the sum of the variance and the square of the bias. An efficient estimator has the smallest MSE.
Example:
  • Among all unbiased estimators of the population mean μ μ mu\muμ, the sample mean X ¯ X ¯ bar(X)\bar{X}X¯ is the most efficient, meaning it has the smallest variance.

4. Sufficiency

An estimator is sufficient if it uses all the information in the data about the parameter. A sufficient estimator captures all the information that the sample provides about the population parameter, leaving no "leftover" information.
Example:
  • The sample mean X ¯ X ¯ bar(X)\bar{X}X¯ and sample variance S 2 S 2 S^(2)S^2S2 are jointly sufficient for the parameters μ μ mu\muμ and σ 2 σ 2 sigma^(2)\sigma^2σ2 of a normal distribution.

5. Robustness

An estimator is robust if it is not unduly affected by small deviations from the assumptions (e.g., normality) or by outliers. Robust estimators provide reliable estimates even when the data has some level of contamination or when the assumptions are only approximately met.
Example:
  • The sample median is a robust estimator of the population median, as it is less affected by outliers compared to the sample mean.

Examples of Good Estimators

Example 1: Sample Mean
  • Unbiasedness: E ( X ¯ ) = μ E ( X ¯ ) = μ E( bar(X))=muE(\bar{X}) = \muE(X¯)=μ.
  • Consistency: As n → ∞ n → ∞ n rarr oon \to \inftyn→∞, X ¯ → μ X ¯ → μ bar(X)rarr mu\bar{X} \to \muX¯→μ.
  • Efficiency: Among all unbiased estimators of μ μ mu\muμ, X ¯ X ¯ bar(X)\bar{X}X¯ has the smallest variance.
  • Sufficiency: For a normally distributed population, X ¯ X ¯ bar(X)\bar{X}X¯ is a sufficient estimator of μ μ mu\muμ.
Example 2: Sample Variance
  • Unbiasedness: E ( S 2 ) = σ 2 E ( S 2 ) = σ 2 E(S^(2))=sigma^(2)E(S^2) = \sigma^2E(S2)=σ2.
  • Consistency: As n → ∞ n → ∞ n rarr oon \to \inftyn→∞, S 2 → σ 2 S 2 → σ 2 S^(2)rarrsigma^(2)S^2 \to \sigma^2S2→σ2.
  • Efficiency: In a normal distribution, S 2 S 2 S^(2)S^2S2 is the most efficient unbiased estimator of σ 2 σ 2 sigma^(2)\sigma^2σ2.

Conclusion

A good estimator should be unbiased, consistent, efficient, sufficient, and robust. These properties ensure that the estimator provides accurate, reliable, and useful estimates of the population parameters based on sample data. Understanding these properties helps statisticians and researchers choose the appropriate estimators for their analyses.
(b) The measurements of length (in c m c m cm\mathrm{cm}cm ) of a random sample of 10 boxes are given as follows :
20.2, 24.1, 21.3, 17.2, 19.8, 16.5, 21.8, 18.7, 17.1 and 19.9.
Use suitable test to test the hypothesis that the sample is taken from a population which is symmetrical about 18 c m 18 c m 18cm18 \mathrm{~cm}18 cm against the alternative that symmetry is about the point which is greater than 18 c m 18 c m 18cm18 \mathrm{~cm}18 cm at 5 % 5 % 5%5 \%5% level of significance.

Answer:

To test the hypothesis that the sample is taken from a population symmetrical about 18 cm against the alternative that symmetry is about a point greater than 18 cm, we can use the one-sample Wilcoxon signed-rank test. This non-parametric test is suitable for testing the median (or symmetry) of a population when the population distribution is not assumed to be normal.

Given Data:

Measurements of lengths (in cm): 20.2, 24.1, 21.3, 17.2, 19.8, 16.5, 21.8, 18.7, 17.1, 19.9

Hypotheses:

  • Null Hypothesis ( H 0 H 0 H_(0)H_0H0): The sample is taken from a population symmetrical about 18 cm (the median is 18 cm).
  • Alternative Hypothesis ( H 1 H 1 H_(1)H_1H1): The sample is taken from a population symmetrical about a point greater than 18 cm (the median is greater than 18 cm).

Significance Level:

α = 0.05 α = 0.05 alpha=0.05\alpha = 0.05α=0.05

Step-by-Step Procedure:

  1. Calculate the Differences from the Hypothesized Median (18 cm):
D i = X i − 18 D i = X i − 18 D_(i)=X_(i)-18D_i = X_i – 18Di=Xi−18
Length (cm) D i = X i − 18 Absolute D i 20.2 20.2 − 18 = 2.2 2.2 24.1 24.1 − 18 = 6.1 6.1 21.3 21.3 − 18 = 3.3 3.3 17.2 17.2 − 18 = − 0.8 0.8 19.8 19.8 − 18 = 1.8 1.8 16.5 16.5 − 18 = − 1.5 1.5 21.8 21.8 − 18 = 3.8 3.8 18.7 18.7 − 18 = 0.7 0.7 17.1 17.1 − 18 = − 0.9 0.9 19.9 19.9 − 18 = 1.9 1.9 Length (cm) D i = X i − 18 Absolute D i 20.2 20.2 − 18 = 2.2 2.2 24.1 24.1 − 18 = 6.1 6.1 21.3 21.3 − 18 = 3.3 3.3 17.2 17.2 − 18 = − 0.8 0.8 19.8 19.8 − 18 = 1.8 1.8 16.5 16.5 − 18 = − 1.5 1.5 21.8 21.8 − 18 = 3.8 3.8 18.7 18.7 − 18 = 0.7 0.7 17.1 17.1 − 18 = − 0.9 0.9 19.9 19.9 − 18 = 1.9 1.9 {:[“Length (cm)”,D_(i)=X_(i)-18,”Absolute “D_(i)],[20.2,20.2-18=2.2,2.2],[24.1,24.1-18=6.1,6.1],[21.3,21.3-18=3.3,3.3],[17.2,17.2-18=-0.8,0.8],[19.8,19.8-18=1.8,1.8],[16.5,16.5-18=-1.5,1.5],[21.8,21.8-18=3.8,3.8],[18.7,18.7-18=0.7,0.7],[17.1,17.1-18=-0.9,0.9],[19.9,19.9-18=1.9,1.9]:}\begin{array}{|c|c|c|} \hline \text{Length (cm)} & D_i = X_i – 18 & \text{Absolute } D_i \\ \hline 20.2 & 20.2 – 18 = 2.2 & 2.2 \\ 24.1 & 24.1 – 18 = 6.1 & 6.1 \\ 21.3 & 21.3 – 18 = 3.3 & 3.3 \\ 17.2 & 17.2 – 18 = -0.8 & 0.8 \\ 19.8 & 19.8 – 18 = 1.8 & 1.8 \\ 16.5 & 16.5 – 18 = -1.5 & 1.5 \\ 21.8 & 21.8 – 18 = 3.8 & 3.8 \\ 18.7 & 18.7 – 18 = 0.7 & 0.7 \\ 17.1 & 17.1 – 18 = -0.9 & 0.9 \\ 19.9 & 19.9 – 18 = 1.9 & 1.9 \\ \hline \end{array}Length (cm)Di=Xi−18Absolute Di20.220.2−18=2.22.224.124.1−18=6.16.121.321.3−18=3.33.317.217.2−18=−0.80.819.819.8−18=1.81.816.516.5−18=−1.51.521.821.8−18=3.83.818.718.7−18=0.70.717.117.1−18=−0.90.919.919.9−18=1.91.9
  1. Rank the Absolute Differences (ignoring zero differences, if any):
Length (cm) D i = X i − 18 Absolute D i Rank 20.2 2.2 2.2 6 24.1 6.1 6.1 10 21.3 3.3 3.3 8 17.2 − 0.8 0.8 2 19.8 1.8 1.8 5 16.5 − 1.5 1.5 3 21.8 3.8 3.8 9 18.7 0.7 0.7 1 17.1 − 0.9 0.9 4 19.9 1.9 1.9 7 Length (cm) D i = X i − 18 Absolute D i Rank 20.2 2.2 2.2 6 24.1 6.1 6.1 10 21.3 3.3 3.3 8 17.2 − 0.8 0.8 2 19.8 1.8 1.8 5 16.5 − 1.5 1.5 3 21.8 3.8 3.8 9 18.7 0.7 0.7 1 17.1 − 0.9 0.9 4 19.9 1.9 1.9 7 {:[“Length (cm)”,D_(i)=X_(i)-18,”Absolute “D_(i),”Rank”],[20.2,2.2,2.2,6],[24.1,6.1,6.1,10],[21.3,3.3,3.3,8],[17.2,-0.8,0.8,2],[19.8,1.8,1.8,5],[16.5,-1.5,1.5,3],[21.8,3.8,3.8,9],[18.7,0.7,0.7,1],[17.1,-0.9,0.9,4],[19.9,1.9,1.9,7]:}\begin{array}{|c|c|c|c|} \hline \text{Length (cm)} & D_i = X_i – 18 & \text{Absolute } D_i & \text{Rank} \\ \hline 20.2 & 2.2 & 2.2 & 6 \\ 24.1 & 6.1 & 6.1 & 10 \\ 21.3 & 3.3 & 3.3 & 8 \\ 17.2 & -0.8 & 0.8 & 2 \\ 19.8 & 1.8 & 1.8 & 5 \\ 16.5 & -1.5 & 1.5 & 3 \\ 21.8 & 3.8 & 3.8 & 9 \\ 18.7 & 0.7 & 0.7 & 1 \\ 17.1 & -0.9 & 0.9 & 4 \\ 19.9 & 1.9 & 1.9 & 7 \\ \hline \end{array}Length (cm)Di=Xi−18Absolute DiRank20.22.22.2624.16.16.11021.33.33.3817.2−0.80.8219.81.81.8516.5−1.51.5321.83.83.8918.70.70.7117.1−0.90.9419.91.91.97
  1. Assign the Signs of the Original Differences to the Ranks:
Length (cm) D i = X i − 18 Absolute D i Rank Signed Rank 20.2 2.2 2.2 6 6 24.1 6.1 6.1 10 10 21.3 3.3 3.3 8 8 17.2 − 0.8 0.8 2 − 2 19.8 1.8 1.8 5 5 16.5 − 1.5 1.5 3 − 3 21.8 3.8 3.8 9 9 18.7 0.7 0.7 1 1 17.1 − 0.9 0.9 4 − 4 19.9 1.9 1.9 7 7 Length (cm) D i = X i − 18 Absolute D i Rank Signed Rank 20.2 2.2 2.2 6 6 24.1 6.1 6.1 10 10 21.3 3.3 3.3 8 8 17.2 − 0.8 0.8 2 − 2 19.8 1.8 1.8 5 5 16.5 − 1.5 1.5 3 − 3 21.8 3.8 3.8 9 9 18.7 0.7 0.7 1 1 17.1 − 0.9 0.9 4 − 4 19.9 1.9 1.9 7 7 {:[“Length (cm)”,D_(i)=X_(i)-18,”Absolute “D_(i),”Rank”,”Signed Rank”],[20.2,2.2,2.2,6,6],[24.1,6.1,6.1,10,10],[21.3,3.3,3.3,8,8],[17.2,-0.8,0.8,2,-2],[19.8,1.8,1.8,5,5],[16.5,-1.5,1.5,3,-3],[21.8,3.8,3.8,9,9],[18.7,0.7,0.7,1,1],[17.1,-0.9,0.9,4,-4],[19.9,1.9,1.9,7,7]:}\begin{array}{|c|c|c|c|c|} \hline \text{Length (cm)} & D_i = X_i – 18 & \text{Absolute } D_i & \text{Rank} & \text{Signed Rank} \\ \hline 20.2 & 2.2 & 2.2 & 6 & 6 \\ 24.1 & 6.1 & 6.1 & 10 & 10 \\ 21.3 & 3.3 & 3.3 & 8 & 8 \\ 17.2 & -0.8 & 0.8 & 2 & -2 \\ 19.8 & 1.8 & 1.8 & 5 & 5 \\ 16.5 & -1.5 & 1.5 & 3 & -3 \\ 21.8 & 3.8 & 3.8 & 9 & 9 \\ 18.7 & 0.7 & 0.7 & 1 & 1 \\ 17.1 & -0.9 & 0.9 & 4 & -4 \\ 19.9 & 1.9 & 1.9 & 7 & 7 \\ \hline \end{array}Length (cm)Di=Xi−18Absolute DiRankSigned Rank20.22.22.26624.16.16.1101021.33.33.38817.2−0.80.82−219.81.81.85516.5−1.51.53−321.83.83.89918.70.70.71117.1−0.90.94−419.91.91.977
  1. Calculate the Test Statistic:
    Sum of positive ranks ( W + W + W^(+)W^+W+):
W + = 6 + 10 + 8 + 5 + 9 + 1 + 7 = 46 W + = 6 + 10 + 8 + 5 + 9 + 1 + 7 = 46 W^(+)=6+10+8+5+9+1+7=46W^+ = 6 + 10 + 8 + 5 + 9 + 1 + 7 = 46W+=6+10+8+5+9+1+7=46
Sum of negative ranks ( W − W − W^(-)W^-W−):
W − = 2 + 3 + 4 = 9 W − = 2 + 3 + 4 = 9 W^(-)=2+3+4=9W^- = 2 + 3 + 4 = 9W−=2+3+4=9
The Wilcoxon signed-rank test statistic W W WWW is the smaller of W + W + W^(+)W^+W+ and W − W − W^(-)W^-W−:
W = 9 W = 9 W=9W = 9W=9
  1. Determine the Critical Value:
    For a one-tailed test at the 5 % 5 % 5%5\%5% significance level with n = 10 n = 10 n=10n = 10n=10, we use the Wilcoxon signed-rank test table to find the critical value. For n = 10 n = 10 n=10n = 10n=10 and α = 0.05 α = 0.05 alpha=0.05\alpha = 0.05α=0.05, the critical value is 8 8 888.
  2. Make the Decision:
    Compare the test statistic W W WWW with the critical value:
  • If W ≤ W ≤ W <=W \leqW≤ critical value, reject H 0 H 0 H_(0)H_0H0.
  • If W > W > W >W >W> critical value, do not reject H 0 H 0 H_(0)H_0H0.
In this case, W = 9 W = 9 W=9W = 9W=9 is greater than the critical value 8 8 888.

Conclusion

We do not reject the null hypothesis. There is insufficient evidence at the 5 % 5 % 5%5\%5% level of significance to conclude that the sample is taken from a population symmetrical about a point greater than 18 cm.
\(a^2=b^2+c^2-2bc\:Cos\left(A\right)\)
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