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IGNOU MST-013 Solved Assignment 2023 | MSCAST

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

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IGNOU MST-013 Assignment Question Paper 2023

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  1. State whether the following statements are true or false and also give the reason in support of your answer:
(a) V opt ( x ¯ st ) V opt  x ¯ st  V_(“opt “)( bar(x)_(“st “))V_{\text {opt }}\left(\bar{x}_{\text {st }}\right)Vopt (x¯st ) lies between V prop ( x ¯ st ) V prop  x ¯ st  V_(“prop “)( bar(x)_(“st “))V_{\text {prop }}\left(\bar{x}_{\text {st }}\right)Vprop (x¯st ) and V Random ( x ¯ st ) V Random  x ¯ st  V_(“Random “)( bar(x)_(“st “))V_{\text {Random }}\left(\bar{x}_{\text {st }}\right)VRandom (x¯st ).
(b) The total number of all possible samples of size 3 without replacement from a population of size 7 is 21 .
(c) While analysing the data of a 5 × 5 5 × 5 5xx55 \times 55×5 Latin Square design the d.f. for ESS is equal to 16 .
(d) In a Two-way Analysis of Variance test with 5 observations per cell having 4 blocks and 4 treatments the degree of freedom for the total variation is 64 .
(e) The probability of selection of a sample of n n n\mathrm{n}n from the population by SRSWOR is 1 / N 1 / N 1//N1 / \mathrm{N}1/N.
2(a) A sample of 100 employees is to be drawn from a population of collages A A AAA and B B BBB. The population means and population mean squares of their monthly wages are given below:
Village N i N i N_(i)\mathbf{N}_{\mathbf{i}}Ni X ¯ i X ¯ i bar(X)_(i)\overline{\mathbf{X}}_{\mathbf{i}}X¯i S i 2 S i 2 S_(i)^(2)\mathbf{S}_{\mathbf{i}}^{\mathbf{2}}Si2
Collage A 400 60 20
Collage B 200 120 80
Village N_(i) bar(X)_(i) S_(i)^(2) Collage A 400 60 20 Collage B 200 120 80| Village | $\mathbf{N}_{\mathbf{i}}$ | $\overline{\mathbf{X}}_{\mathbf{i}}$ | $\mathbf{S}_{\mathbf{i}}^{\mathbf{2}}$ | | :— | :—: | :—: | :—: | | Collage A | 400 | 60 | 20 | | Collage B | 200 | 120 | 80 |
Draw the samples using Proportional and Neyman allocation techniques and compare. Obtain the sample mean and variances for the Proportional Allocation and SRSWOR for the given information. Then Find the percentage gain in precision of variances of sample mean under the Proportional Allocation over the that of SRSWOR.
(b) A population consists of 10 villages with a total of 212 households. The second column of the accompanying table shows the number of households corresponding to each village. Select a PPS with replacement sample of 6 villages by using the Cumulative Total method:
Village 1 1 1\mathbf{1}1 2 2 2\mathbf{2}2 3 3 3\mathbf{3}3 4 4 4\mathbf{4}4 5 5 5\mathbf{5}5 6 6 6\mathbf{6}6 7 7 7\mathbf{7}7 8 8 8\mathbf{8}8 9 9 9\mathbf{9}9 1 0 1 0 10\mathbf{1 0}10
No. of Households  No. of   Households  {:[” No. of “],[” Households “]:}\begin{array}{l}\text { No. of } \\ \text { Households }\end{array} No. of  Households  35 28 20 25 30 19 10 12 18 15
Village 1 2 3 4 5 6 7 8 9 10 ” No. of Households ” 35 28 20 25 30 19 10 12 18 15| Village | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ | $\mathbf{5}$ | $\mathbf{6}$ | $\mathbf{7}$ | $\mathbf{8}$ | $\mathbf{9}$ | $\mathbf{1 0}$ | | :— | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | | $\begin{array}{l}\text { No. of } \\ \text { Households }\end{array}$ | 35 | 28 | 20 | 25 | 30 | 19 | 10 | 12 | 18 | 15 |
3(a) In a population of size N = 5 N = 5 N=5\mathrm{N}=5N=5, the values of the population characteristics are 1, 3, 5, 7, 9 , a sample of size 2 is drawn. Verify that y ¯ y ¯ bar(y)\bar{y}y¯ is an unbiased estimate of Y ¯ Y ¯ bar(Y)\bar{Y}Y¯ and V ( y ¯ ) V ( y ¯ ) V( bar(y))V(\bar{y})V(y¯) is equal to
V ( y ¯ ) = N − n N ⋅ n S 2 V ( y ¯ ) = N − n N ⋅ n S 2 V( bar(y))=(N-n)/(N*n)S^(2)\mathrm{V}(\overline{\mathrm{y}})=\frac{\mathrm{N}-\mathrm{n}}{\mathrm{N} \cdot \mathrm{n}} \mathrm{S}^2V(y¯)=N−nN⋅nS2
(b) In order to compare the mileage yields of 3 kinds of Gasoline, several tests were run, and the following results were obtained:
Gasoline A: 19 21 20 18 21 21
Gasoline B: 23 20 22 20 24 23
Gasoline C: 20 17 21 19 20 17
Gasoline A: 19 21 20 18 21 21 Gasoline B: 23 20 22 20 24 23 Gasoline C: 20 17 21 19 20 17| Gasoline A: | 19 | 21 | 20 | 18 | 21 | 21 | | :— | :— | :— | :— | :— | :— | :— | | Gasoline B: | 23 | 20 | 22 | 20 | 24 | 23 | | Gasoline C: | 20 | 17 | 21 | 19 | 20 | 17 |
Carry out the Analysis of Variance test and test whether there is significant differences between the average mileage of 3 kinds of gasoline at 5 % 5 % 5%5 \%5% level of significance.
4(a) A manurial trial with six levels of Farmyard Manure (FYM) was carried out in a randomised block design with 4 replications at the experimental station Junagarh with a new study the rate of decomposition of organic matters in soil and its synthetic capacity in soil on cotton crop. The yield per plot in k g k g kg\mathrm{kg}kg for different levels of FYM and replications is given below:
Cotton Yield Per Plot (in Kg)
Levels of FYM  Levels   of FYM  {:[” Levels “],[” of FYM “]:}\begin{array}{c}\text { Levels } \\ \text { of FYM }\end{array} Levels  of FYM  Replications
I II III IV
1 6.90 4.60 4.40 4.81
2 6.48 5.57 4.28 4.45
3 6.52 7.60 5.30 5.30
4 6.90 6.65 6.75 7.75
5 6.00 6.18 6.50 5.50
6 7.90 7.57 6.80 6.62
” Levels of FYM ” Replications I II III IV 1 6.90 4.60 4.40 4.81 2 6.48 5.57 4.28 4.45 3 6.52 7.60 5.30 5.30 4 6.90 6.65 6.75 7.75 5 6.00 6.18 6.50 5.50 6 7.90 7.57 6.80 6.62| $\begin{array}{c}\text { Levels } \\ \text { of FYM }\end{array}$ | Replications | | | | | :—: | :—: | :—: | :—: | :—: | | | I | II | III | IV | | 1 | 6.90 | 4.60 | 4.40 | 4.81 | | 2 | 6.48 | 5.57 | 4.28 | 4.45 | | 3 | 6.52 | 7.60 | 5.30 | 5.30 | | 4 | 6.90 | 6.65 | 6.75 | 7.75 | | 5 | 6.00 | 6.18 | 6.50 | 5.50 | | 6 | 7.90 | 7.57 | 6.80 | 6.62 |
Carry out the Analysis of Variance test and draw the conclusions.
(b) For the given data the yield of the treatment 2 in 3 rd 3 rd  3^(“rd “)3^{\text {rd }}3rd  block is missing. Estimate the missing value and analyse the data.
Treatments Blocks
I I I\mathbf{I}I II III IV
1 1 1\mathbf{1}1 105 114 108 109
2 2 2\mathbf{2}2 112 113 Y Y Y\mathrm{Y}Y 112
3 3 3\mathbf{3}3 106 114 105 109
Treatments Blocks I II III IV 1 105 114 108 109 2 112 113 Y 112 3 106 114 105 109| Treatments | Blocks | | | | | :—: | :—: | :—: | :—: | :—: | | | $\mathbf{I}$ | II | III | IV | | $\mathbf{1}$ | 105 | 114 | 108 | 109 | | $\mathbf{2}$ | 112 | 113 | $\mathrm{Y}$ | 112 | | $\mathbf{3}$ | 106 | 114 | 105 | 109 |
5(a) The following data is the data pertaining to a feeding trial on sheep. Treatments
A: Grazing only
B: Grazing + Maize Supplements
C: Grazing + Maize + Protein Supplement P 1 P 1 P_(1)P_1P1
D: Grazing + Maize + Protein Supplement P 2 P 2 P_(2)P_2P2
E E E\mathrm{E}E : Grazing + Maize + Protein Supplement P 3 P 3 P_(3)\mathrm{P}_3P3
Layout and Wool Yield (100 g m ) g m ) gm)\mathrm{gm})gm) is given as:
32 ( D ) 32 ( D ) 32(D)32(\mathrm{D})32(D) 33 ( E ) 33 ( E ) 33(E)33(\mathrm{E})33(E) 30 ( C ) 30 ( C ) 30(C)30(\mathrm{C})30(C) 28 ( B ) 28 ( B ) 28(B)28(\mathrm{~B})28( B) 24 ( A ) 24 ( A ) 24(A)24(\mathrm{~A})24( A)
51 ( C ) 51 ( C ) 51(C)51(\mathrm{C})51(C) 45 ( D ) 45 ( D ) 45(D)45(\mathrm{D})45(D) 41 ( A ) 41 ( A ) 41(A)41(\mathrm{~A})41( A) 45 ( E ) 45 ( E ) 45(E)45(\mathrm{E})45(E) 29 ( B ) 29 ( B ) 29(B)29(\mathrm{~B})29( B)
41 ( E ) 41 ( E ) 41(E)41(\mathrm{E})41(E) 29 ( A ) 29 ( A ) 29(A)29(\mathrm{~A})29( A) 24 ( B ) 24 ( B ) 24(B)24(\mathrm{~B})24( B) 36 ( D ) 36 ( D ) 36(D)36(\mathrm{D})36(D) 35 ( C ) 35 ( C ) 35(C)35(\mathrm{C})35(C)
38 ( B ) 38 ( B ) 38(B)38(\mathrm{~B})38( B) 39 ( C ) 39 ( C ) 39(C)39(\mathrm{C})39(C) 42 ( E ) 42 ( E ) 42(E)42(\mathrm{E})42(E) 23 ( A ) 23 ( A ) 23(A)23(\mathrm{~A})23( A) 37 ( D ) 37 ( D ) 37(D)37(\mathrm{D})37(D)
38 ( A ) 38 ( A ) 38(A)38(\mathrm{~A})38( A) 24 ( B ) 24 ( B ) 24(B)24(\mathrm{~B})24( B) 21 ( D ) 21 ( D ) 21(D)21(\mathrm{D})21(D) 29 ( C ) 29 ( C ) 29(C)29(\mathrm{C})29(C) 26 ( E ) 26 ( E ) 26(E)26(\mathrm{E})26(E)
32(D) 33(E) 30(C) 28(B) 24(A) 51(C) 45(D) 41(A) 45(E) 29(B) 41(E) 29(A) 24(B) 36(D) 35(C) 38(B) 39(C) 42(E) 23(A) 37(D) 38(A) 24(B) 21(D) 29(C) 26(E)| $32(\mathrm{D})$ | $33(\mathrm{E})$ | $30(\mathrm{C})$ | $28(\mathrm{~B})$ | $24(\mathrm{~A})$ | | :—: | :—: | :—: | :—: | :—: | | $51(\mathrm{C})$ | $45(\mathrm{D})$ | $41(\mathrm{~A})$ | $45(\mathrm{E})$ | $29(\mathrm{~B})$ | | $41(\mathrm{E})$ | $29(\mathrm{~A})$ | $24(\mathrm{~B})$ | $36(\mathrm{D})$ | $35(\mathrm{C})$ | | $38(\mathrm{~B})$ | $39(\mathrm{C})$ | $42(\mathrm{E})$ | $23(\mathrm{~A})$ | $37(\mathrm{D})$ | | $38(\mathrm{~A})$ | $24(\mathrm{~B})$ | $21(\mathrm{D})$ | $29(\mathrm{C})$ | $26(\mathrm{E})$ |
Analyse the design with appropriate method and calculate the Critical Difference for the treatment mean yield.
(b) In a class of Statistics, total number of students is 30 . Select the linear and circular systematic random samples of 12 students. The age of 30 students is given below:
Age: 22 25 22 21 22 25 24 23 22 21 20 21 22 23 25 23 24 22 24 24 21 20 23 21 22 20 20 21 22 25  Age:  22 25 22 21 22 25 24 23 22 21 20 21 22 23 25 23 24 22 24 24 21 20 23 21 22 20 20 21 22 25 ” Age: “{:[22,25,22,21,22,25,24,23,22,21,20],[21,22,23,25,23,24,22,24,24,21,20],[23,21,22,20,20,21,22,25,,,]:}\text { Age: } \begin{array}{rllllllllll} 22 & 25 & 22 & 21 & 22 & 25 & 24 & 23 & 22 & 21 & 20 \\ 21 & 22 & 23 & 25 & 23 & 24 & 22 & 24 & 24 & 21 & 20 \\ 23 & 21 & 22 & 20 & 20 & 21 & 22 & 25 & & & \end{array} Age: 222522212225242322212021222325232422242421202321222020212225
\(2\:sin\:\theta \:cos\:\phi =sin\:\left(\theta +\phi \right)+sin\:\left(\theta -\phi \right)\)

MST-013 Sample Solution 2023

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Question:-01

  1. State whether the following statements are true or false and also give the reason in support of your answer:
(a) V opt ( x ¯ st ) V opt  x ¯ st  V_(“opt “)( bar(x)_(“st “))V_{\text {opt }}\left(\bar{x}_{\text {st }}\right)Vopt (x¯st ) lies between V prop ( x ¯ st ) V prop  x ¯ st  V_(“prop “)( bar(x)_(“st “))V_{\text {prop }}\left(\bar{x}_{\text {st }}\right)Vprop (x¯st ) and V Random ( x ¯ st ) V Random  x ¯ st  V_(“Random “)( bar(x)_(“st “))V_{\text {Random }}\left(\bar{x}_{\text {st }}\right)VRandom (x¯st ).
Answer:

Introduction

We are given three different expressions representing variances or measures of dispersion, namely V opt V opt V_(“opt”)V_{\text{opt}}Vopt, V prop V prop V_(“prop”)V_{\text{prop}}Vprop, and V ran V ran V_(“ran”)V_{\text{ran}}Vran, and we are asked to verify the statement:
V opt ( x ¯ st ) V opt  x ¯ st  V_(“opt “)( bar(x)_(“st “))V_{\text {opt }}\left(\bar{x}_{\text {st }}\right)Vopt (x¯st ) lies between V prop ( x ¯ st ) V prop  x ¯ st  V_(“prop “)( bar(x)_(“st “))V_{\text {prop }}\left(\bar{x}_{\text {st }}\right)Vprop (x¯st ) and V Random ( x ¯ st ) V Random  x ¯ st  V_(“Random “)( bar(x)_(“st “))V_{\text {Random }}\left(\bar{x}_{\text {st }}\right)VRandom (x¯st ).

Justification

Given the provided equations and derivations, we have:
  1. For V ran − V prop V ran − V prop V_(“ran”)-V_(“prop”)V_{\text{ran}} – V_{\text{prop}}Vran−Vprop:
    V ran − V prop = ∑ N i ( x i ¯ − x ¯ ) 2 n N ≥ 0 V ran − V prop = ∑ N i x i ¯ − x ¯ 2 n N ≥ 0 V_(“ran”)-V_(“prop”)=(sumN_(i)( bar(x_(i))-( bar(x)))^(2))/(nN) >= 0V_{\text{ran}} – V_{\text{prop}} = \frac{\sum N_i\left(\overline{x_i}-\bar{x}\right)^2}{n N} \geq 0Vran−Vprop=∑Ni(xi¯−x¯)2nN≥0
    This implies that V ran ≥ V prop V ran ≥ V prop V_(“ran”) >= V_(“prop”)V_{\text{ran}} \geq V_{\text{prop}}Vran≥Vprop.
  2. For V prop − V opt V prop − V opt V_(“prop”)-V_(“opt”)V_{\text{prop}} – V_{\text{opt}}Vprop−Vopt:
    V prop − V opt = 1 n N [ ∑ i = 1 k N i ( S i − S ¯ ) 2 ] ≥ 0 V prop − V opt = 1 n N ∑ i = 1 k   N i S i − S ¯ 2 ≥ 0 V_(“prop”)-V_(“opt”)=(1)/(nN)[sum_(i=1)^(k)N_(i)(S_(i)-( bar(S)))^(2)] >= 0V_{\text{prop}} – V_{\text{opt}} = \frac{1}{n N}\left[\sum_{i=1}^k N_i\left(S_i-\bar{S}\right)^2\right] \geq 0Vprop−Vopt=1nN[∑i=1kNi(Si−S¯)2]≥0
    This implies that V prop ≥ V opt V prop ≥ V opt V_(“prop”) >= V_(“opt”)V_{\text{prop}} \geq V_{\text{opt}}Vprop≥Vopt.

Conclusion

Combining the above two results, we can conclude that:
V opt ( x ¯ st ) ≤ V prop ( x ¯ st ) ≤ V ran ( x ¯ st ) V opt x ¯ st ≤ V prop x ¯ st ≤ V ran x ¯ st V_(“opt”)( bar(x)_(“st”)) <= V_(“prop”)( bar(x)_(“st”)) <= V_(“ran”)( bar(x)_(“st”))V_{\text{opt}}\left(\bar{x}_{\text{st}}\right) \leq V_{\text{prop}}\left(\bar{x}_{\text{st}}\right) \leq V_{\text{ran}}\left(\bar{x}_{\text{st}}\right)Vopt(x¯st)≤Vprop(x¯st)≤Vran(x¯st)
This validates the given statement as False, with V opt V opt V_(“opt”)V_{\text{opt}}Vopt being the lower bound, V ran V ran V_(“ran”)V_{\text{ran}}Vran being the upper bound, and V prop V prop V_(“prop”)V_{\text{prop}}Vprop lying between them.
\(cos\left(\theta +\phi \right)=cos\:\theta \:cos\:\phi -sin\:\theta \:sin\:\phi \)
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\(sin\left(\theta +\phi \right)=sin\:\theta \:cos\:\phi +cos\:\theta \:sin\:\phi \)

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