MST-005 Dec 2023

MST-005 Dec 2023

Question:-01

1.State whether the following statements are True or False. Give reasons in support of your answer :
(a) Sampling error occurs in both census and sample survey.

Answer:

To determine whether the statement "Sampling error occurs in both census and sample survey" is true or false, let’s analyze the concepts of sampling error, census, and sample survey.

Definitions:

  1. Sampling Error:
    • Sampling error is the error that occurs when a sample from a population is used to estimate some characteristics of the population. This error arises because the sample is only a part of the population, and there might be differences between the sample and the population.
  2. Census:
    • A census is a survey that attempts to collect data from every member of a population. Because it includes every member, there should, in theory, be no sampling error, as there is no sampling involved.
  3. Sample Survey:
    • A sample survey collects data from a subset (sample) of a population. Since not all members of the population are included, there is an inherent risk of sampling error.

Statement Analysis:

Statement: Sampling error occurs in both census and sample survey.

Evaluation:

  • Census:
    • Since a census involves collecting data from the entire population, there is no sampling process involved. Therefore, there is no sampling error in a census. Any errors in a census would be due to non-sampling errors such as measurement errors, data processing errors, or nonresponse errors.
  • Sample Survey:
    • A sample survey involves collecting data from a sample of the population. Because a sample is only a part of the whole population, there is a possibility that the sample may not perfectly represent the population. This discrepancy leads to sampling error.

Conclusion:

  • True or False: The statement is false.
  • Justification:
    • False for Census: Sampling error does not occur in a census because a census includes the entire population. There is no sampling process, so there cannot be any sampling error.
    • True for Sample Survey: Sampling error does occur in a sample survey because it relies on a sample, which might not perfectly represent the entire population.
Therefore, the statement "Sampling error occurs in both census and sample survey" is false because sampling error does not occur in a census but does occur in a sample survey.

(b) In a cluster sampling, the elements within a cluster should be as homogeneous as possible.

Answer:

To determine whether the statement "In a cluster sampling, the elements within a cluster should be as homogeneous as possible" is true or false, let’s first review the concept of cluster sampling and the requirements for effective clustering.

Definitions and Concepts:

  1. Cluster Sampling:
    • Cluster sampling is a method where the population is divided into clusters, which are groups of elements. A random sample of clusters is then selected, and all elements within the chosen clusters are studied. This method is often used when a population is large and spread out, making it impractical to conduct a simple random sample.
  2. Homogeneity within Clusters:
    • Homogeneity within clusters means that the elements within each cluster are similar to each other.
  3. Heterogeneity within Clusters:
    • Heterogeneity within clusters means that the elements within each cluster are diverse or different from each other.

Analysis of the Statement:

Statement: In a cluster sampling, the elements within a cluster should be as homogeneous as possible.

Evaluation:

  • Goal of Cluster Sampling:
    • The primary goal of cluster sampling is to achieve practical and cost-effective sampling by grouping elements into clusters. For the method to be efficient and provide accurate results, the clusters themselves should be as similar to each other as possible (i.e., homogeneous between clusters) to ensure that the sample is representative of the entire population.
  • Homogeneity Within Clusters:
    • If elements within a cluster are too homogeneous, it may reduce the variability captured by the sample, leading to less efficient estimates. For cluster sampling to be effective, it is generally preferable that clusters be internally heterogeneous. This way, the variability within each cluster mirrors the variability of the entire population.
  • Heterogeneity Within Clusters:
    • Heterogeneous clusters ensure that each cluster reflects the diversity of the population, which helps in achieving a more representative sample when only a few clusters are selected.

Conclusion:

  • True or False: The statement is false.
  • Justification:
    • In cluster sampling, the goal is to have clusters that are heterogeneous internally but homogeneous with respect to each other. This means that the elements within a cluster should be as diverse as possible to reflect the overall population variability, while the clusters themselves should be similar to each other to ensure that the sample represents the population accurately.
    • Homogeneous clusters would reduce the effectiveness of cluster sampling because they wouldn’t capture the population’s variability well, leading to less accurate estimates.
Therefore, the statement "In a cluster sampling, the elements within a cluster should be as homogeneous as possible" is false. In cluster sampling, it is preferable for the elements within each cluster to be as heterogeneous as possible.

(c) The error degrees of freedom in an one-way analysis of variance of population means at 4 levels of a factor with total 20 observations will be 16 .

Answer:

To determine whether the statement "The error degrees of freedom in a one-way analysis of variance of population means at 4 levels of a factor with a total of 20 observations will be 16" is true or false, let’s first review the calculation of degrees of freedom in a one-way ANOVA.

Definitions and Concepts:

  1. One-Way ANOVA:
    • One-way ANOVA is used to compare the means of three or more independent groups to determine if there is a statistically significant difference between them.
  2. Degrees of Freedom:
    • Total Degrees of Freedom ( d f t o t a l d f t o t a l df_(total)df_{total}dftotal): This is the total number of observations minus one. d f t o t a l = N 1 d f t o t a l = N 1 df_(total)=N-1df_{total} = N – 1dftotal=N1where N N NNN is the total number of observations.
    • Between-Groups Degrees of Freedom ( d f b e t w e e n d f b e t w e e n df_(between)df_{between}dfbetween): This is the number of groups minus one. d f b e t w e e n = k 1 d f b e t w e e n = k 1 df_(between)=k-1df_{between} = k – 1dfbetween=k1where k k kkk is the number of groups or levels.
    • Within-Groups (Error) Degrees of Freedom ( d f w i t h i n d f w i t h i n df_(within)df_{within}dfwithin): This is the total degrees of freedom minus the between-groups degrees of freedom. d f w i t h i n = d f t o t a l d f b e t w e e n d f w i t h i n = d f t o t a l d f b e t w e e n df_(within)=df_(total)-df_(between)df_{within} = df_{total} – df_{between}dfwithin=dftotaldfbetween

Given Data:

  • Levels of the factor ( k k kkk): 4
  • Total observations ( N N NNN): 20

Calculation:

  1. Total Degrees of Freedom:
    d f t o t a l = N 1 = 20 1 = 19 d f t o t a l = N 1 = 20 1 = 19 df_(total)=N-1=20-1=19df_{total} = N – 1 = 20 – 1 = 19dftotal=N1=201=19
  2. Between-Groups Degrees of Freedom:
    d f b e t w e e n = k 1 = 4 1 = 3 d f b e t w e e n = k 1 = 4 1 = 3 df_(between)=k-1=4-1=3df_{between} = k – 1 = 4 – 1 = 3dfbetween=k1=41=3
  3. Within-Groups (Error) Degrees of Freedom:
    d f w i t h i n = d f t o t a l d f b e t w e e n = 19 3 = 16 d f w i t h i n = d f t o t a l d f b e t w e e n = 19 3 = 16 df_(within)=df_(total)-df_(between)=19-3=16df_{within} = df_{total} – df_{between} = 19 – 3 = 16dfwithin=dftotaldfbetween=193=16

Conclusion:

  • True or False: The statement is true.
  • Justification:
    • The error degrees of freedom (within-groups degrees of freedom) in a one-way analysis of variance with 4 levels of a factor and a total of 20 observations is indeed calculated as 16. The calculations align with the formula for within-groups degrees of freedom in ANOVA.
Therefore, the statement "The error degrees of freedom in a one-way analysis of variance of population means at 4 levels of a factor with a total of 20 observations will be 16" is true.

(d) If there is one missing value in a Latin Square Design with 4 treatments, the error degrees of freedom will be 5 .

Answer:

To determine whether the statement "If there is one missing value in a Latin Square Design with 4 treatments, the error degrees of freedom will be 5" is true or false, we need to understand the degrees of freedom in a Latin Square Design and how missing values affect them.

Latin Square Design:

  1. Latin Square Design (LSD):
    • It is an experimental design used to control for two blocking factors.
    • The design involves n n nnn treatments, n n nnn rows, and n n nnn columns, where each treatment appears exactly once in each row and each column.
  2. Degrees of Freedom in LSD:
    • Total degrees of freedom: n 2 1 n 2 1 n^(2)-1n^2 – 1n21
    • Degrees of freedom for rows: n 1 n 1 n-1n – 1n1
    • Degrees of freedom for columns: n 1 n 1 n-1n – 1n1
    • Degrees of freedom for treatments: n 1 n 1 n-1n – 1n1
    • Error degrees of freedom: ( n 1 ) ( n 2 ) ( n 1 ) ( n 2 ) (n-1)(n-2)(n-1)(n-2)(n1)(n2)

Given Data:

  • Number of treatments ( n n nnn): 4
  • One missing value.

Calculation:

  1. Total Degrees of Freedom:
    d f t o t a l = n 2 1 = 4 2 1 = 16 1 = 15 d f t o t a l = n 2 1 = 4 2 1 = 16 1 = 15 df_(total)=n^(2)-1=4^(2)-1=16-1=15df_{total} = n^2 – 1 = 4^2 – 1 = 16 – 1 = 15dftotal=n21=421=161=15
  2. Degrees of Freedom for Rows:
    d f r o w s = n 1 = 4 1 = 3 d f r o w s = n 1 = 4 1 = 3 df_(rows)=n-1=4-1=3df_{rows} = n – 1 = 4 – 1 = 3dfrows=n1=41=3
  3. Degrees of Freedom for Columns:
    d f c o l u m n s = n 1 = 4 1 = 3 d f c o l u m n s = n 1 = 4 1 = 3 df_(columns)=n-1=4-1=3df_{columns} = n – 1 = 4 – 1 = 3dfcolumns=n1=41=3
  4. Degrees of Freedom for Treatments:
    d f t r e a t m e n t s = n 1 = 4 1 = 3 d f t r e a t m e n t s = n 1 = 4 1 = 3 df_(treatments)=n-1=4-1=3df_{treatments} = n – 1 = 4 – 1 = 3dftreatments=n1=41=3
  5. Error Degrees of Freedom (without missing value):
    d f e r r o r = ( n 1 ) ( n 2 ) = ( 4 1 ) ( 4 2 ) = 3 × 2 = 6 d f e r r o r = ( n 1 ) ( n 2 ) = ( 4 1 ) ( 4 2 ) = 3 × 2 = 6 df_(error)=(n-1)(n-2)=(4-1)(4-2)=3xx2=6df_{error} = (n-1)(n-2) = (4-1)(4-2) = 3 \times 2 = 6dferror=(n1)(n2)=(41)(42)=3×2=6

Effect of Missing Value:

When there is one missing value in a Latin Square Design, the error degrees of freedom reduce by 1.
  1. Adjusted Error Degrees of Freedom: d f e r r o r = 6 1 = 5 d f e r r o r = 6 1 = 5 df_(error)=6-1=5df_{error} = 6 – 1 = 5dferror=61=5

Conclusion:

  • True or False: The statement is true.
  • Justification:
    • For a Latin Square Design with 4 treatments (i.e., n = 4 n = 4 n=4n = 4n=4), the error degrees of freedom are calculated as ( n 1 ) ( n 2 ) ( n 1 ) ( n 2 ) (n-1)(n-2)(n-1)(n-2)(n1)(n2). Without any missing values, this is 6. With one missing value, it is reduced by 1, making it 5.
Therefore, the statement "If there is one missing value in a Latin Square Design with 4 treatments, the error degrees of freedom will be 5" is true.

(e) In a middle square method, the next generated random number using random number 15 , will be 22 .

Answer:

To determine whether the statement "In a middle square method, the next generated random number using random number 15, will be 22" is true or false, we need to understand the middle square method and apply it to the given number.

Middle Square Method:

The middle square method is a simple pseudo-random number generator (PRNG) method. Here’s how it works:
  1. Start with an initial seed (the random number).
  2. Square the seed to get a new number.
  3. Extract the middle digits of the squared number to form the next random number.
  4. Repeat the process using the newly generated number.

Steps to Apply the Middle Square Method:

Given the initial random number: 15
  1. Square the Seed:
    15 2 = 225 15 2 = 225 15^(2)=22515^2 = 225152=225
  2. Extract the Middle Digits:
    • Since the middle square method typically uses a fixed number of digits, we need to decide how many digits to keep.
    • If we assume a 2-digit middle square method (since the seed is a 2-digit number), we extract the middle 2 digits from 225.
    • The middle digits of 225 are 22 (since we take the two digits from the center).
Thus, following the middle square method with the initial number 15, the next generated random number is indeed 22.

Conclusion:

  • True or False: The statement is true.
  • Justification:
    • Using the middle square method with the initial random number 15, the next generated random number is 22, as derived from squaring 15 to get 225 and extracting the middle digits (22).
Therefore, the statement "In a middle square method, the next generated random number using random number 15, will be 22" is true.

Question:-02

2.(a) 30 books of Statistics are arranged in serial numbers 1 to 30 in a library. Select all possible systematic random samples of 10 books.

Answer:

To select all possible systematic random samples of 10 books from 30 books arranged in serial numbers 1 to 30, we need to follow a systematic sampling procedure. Systematic sampling involves selecting every k k kkk-th item from the population. Here’s the step-by-step process:

Step-by-Step Process:

  1. Determine the Sampling Interval ( k k kkk):
    • The population size ( N N NNN) is 30.
    • The sample size ( n n nnn) is 10.
    • The sampling interval ( k k kkk) is calculated as: k = N n = 30 10 = 3 k = N n = 30 10 = 3 k=(N)/(n)=(30)/(10)=3k = \frac{N}{n} = \frac{30}{10} = 3k=Nn=3010=3So, we will select every 3rd book.
  2. Identify the Possible Starting Points:
    • The starting point must be within the first k k kkk elements. Therefore, the possible starting points are 1, 2, and 3.
  3. Generate Samples Based on Starting Points:
    • From each starting point, select every 3rd book until you have 10 books.

Generating the Samples:

  • Starting Point 1:
    • Books selected: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28
    • Sample: [ 1 , 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , 28 ] [ 1 , 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , 28 ] [1,4,7,10,13,16,19,22,25,28][1, 4, 7, 10, 13, 16, 19, 22, 25, 28][1,4,7,10,13,16,19,22,25,28]
  • Starting Point 2:
    • Books selected: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29
    • Sample: [ 2 , 5 , 8 , 11 , 14 , 17 , 20 , 23 , 26 , 29 ] [ 2 , 5 , 8 , 11 , 14 , 17 , 20 , 23 , 26 , 29 ] [2,5,8,11,14,17,20,23,26,29][2, 5, 8, 11, 14, 17, 20, 23, 26, 29][2,5,8,11,14,17,20,23,26,29]
  • Starting Point 3:
    • Books selected: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
    • Sample: [ 3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 , 30 ] [ 3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 , 30 ] [3,6,9,12,15,18,21,24,27,30][3, 6, 9, 12, 15, 18, 21, 24, 27, 30][3,6,9,12,15,18,21,24,27,30]

All Possible Systematic Random Samples:

  1. [ 1 , 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , 28 ] [ 1 , 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , 28 ] [1,4,7,10,13,16,19,22,25,28][1, 4, 7, 10, 13, 16, 19, 22, 25, 28][1,4,7,10,13,16,19,22,25,28]
  2. [ 2 , 5 , 8 , 11 , 14 , 17 , 20 , 23 , 26 , 29 ] [ 2 , 5 , 8 , 11 , 14 , 17 , 20 , 23 , 26 , 29 ] [2,5,8,11,14,17,20,23,26,29][2, 5, 8, 11, 14, 17, 20, 23, 26, 29][2,5,8,11,14,17,20,23,26,29]
  3. [ 3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 , 30 ] [ 3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 , 30 ] [3,6,9,12,15,18,21,24,27,30][3, 6, 9, 12, 15, 18, 21, 24, 27, 30][3,6,9,12,15,18,21,24,27,30]

Conclusion:

These are the three possible systematic random samples of 10 books from the 30 books arranged in serial numbers 1 to 30.

(b) One thousand plots in a state of India were stratified according to their sizes. The number of plots ( N i ) N i (N_(i))\left(\mathrm{N}_i\right)(Ni), mean production of wheat per plot ( Y ¯ i ) Y ¯ i ( bar(Y)_(i))\left(\overline{\mathrm{Y}}_i\right)(Y¯i) and standard deviation of production of wheat per plot ( S i ) S i (S_(i))\left(\mathrm{S}_i\right)(Si) for each stratum are given as follows :
Stratum N i N i N_(i)\mathrm{N}_iNi Y ¯ i Y ¯ i bar(Y)_(i)\overline{\mathrm{Y}}_iY¯i S i S i S_(i)\mathrm{~S}_i Si
1 300 15 8
2 500 16 12
3 200 24 18
Stratum N_(i) bar(Y)_(i) S_(i) 1 300 15 8 2 500 16 12 3 200 24 18| Stratum | $\mathrm{N}_i$ | $\overline{\mathrm{Y}}_i$ | $\mathrm{~S}_i$ | | :—: | :—: | :—: | :—: | | 1 | 300 | 15 | 8 | | 2 | 500 | 16 | 12 | | 3 | 200 | 24 | 18 |
(i) Determine the sample size drawn from each stratum for drawing a sample of 100 plots under proportional allocation without replacement.
(ii) Also, estimate the sample mean and the variance of sample mean under given sampling scheme.

Answer:

To determine the sample size drawn from each stratum for drawing a sample of 100 plots under proportional allocation without replacement, and to estimate the sample mean and the variance of the sample mean under the given sampling scheme, we follow these steps:

Part (i) Proportional Allocation Without Replacement:

In proportional allocation, the sample size drawn from each stratum is proportional to the size of the stratum. The total sample size is denoted as n n nnn, and the total number of plots is N N NNN.
  • Total number of plots, N = 1000 N = 1000 N=1000N = 1000N=1000
  • Sample size, n = 100 n = 100 n=100n = 100n=100
The sample size for each stratum ( n i n i n_(i)n_ini) is calculated as:
n i = N i N × n n i = N i N × n n_(i)=(N_(i))/(N)xx nn_i = \frac{N_i}{N} \times nni=NiN×n
Given:
  • N 1 = 300 N 1 = 300 N_(1)=300N_1 = 300N1=300
  • N 2 = 500 N 2 = 500 N_(2)=500N_2 = 500N2=500
  • N 3 = 200 N 3 = 200 N_(3)=200N_3 = 200N3=200

Calculations:

n 1 = 300 1000 × 100 = 30 n 1 = 300 1000 × 100 = 30 n_(1)=(300)/(1000)xx100=30n_1 = \frac{300}{1000} \times 100 = 30n1=3001000×100=30
n 2 = 500 1000 × 100 = 50 n 2 = 500 1000 × 100 = 50 n_(2)=(500)/(1000)xx100=50n_2 = \frac{500}{1000} \times 100 = 50n2=5001000×100=50
n 3 = 200 1000 × 100 = 20 n 3 = 200 1000 × 100 = 20 n_(3)=(200)/(1000)xx100=20n_3 = \frac{200}{1000} \times 100 = 20n3=2001000×100=20
So, the sample sizes for each stratum under proportional allocation are:
  • Stratum 1: n 1 = 30 n 1 = 30 n_(1)=30n_1 = 30n1=30
  • Stratum 2: n 2 = 50 n 2 = 50 n_(2)=50n_2 = 50n2=50
  • Stratum 3: n 3 = 20 n 3 = 20 n_(3)=20n_3 = 20n3=20

Part (ii) Estimating the Sample Mean and the Variance of the Sample Mean:

Sample Mean ( Y ¯ Y ¯ bar(Y)\overline{Y}Y¯):

The sample mean under proportional allocation is the weighted mean of the stratum means. It is given by:
Y ¯ = i = 1 L ( N i N Y ¯ i ) Y ¯ = i = 1 L N i N Y ¯ i bar(Y)=sum_(i=1)^(L)((N_(i))/(N) bar(Y)_(i))\overline{Y} = \sum_{i=1}^{L} \left(\frac{N_i}{N} \overline{Y}_i \right)Y¯=i=1L(NiNY¯i)
Given:
  • Y ¯ 1 = 15 Y ¯ 1 = 15 bar(Y)_(1)=15\overline{Y}_1 = 15Y¯1=15
  • Y ¯ 2 = 16 Y ¯ 2 = 16 bar(Y)_(2)=16\overline{Y}_2 = 16Y¯2=16
  • Y ¯ 3 = 24 Y ¯ 3 = 24 bar(Y)_(3)=24\overline{Y}_3 = 24Y¯3=24

Calculations:

Y ¯ = 300 1000 × 15 + 500 1000 × 16 + 200 1000 × 24 Y ¯ = 300 1000 × 15 + 500 1000 × 16 + 200 1000 × 24 bar(Y)=(300)/(1000)xx15+(500)/(1000)xx16+(200)/(1000)xx24\overline{Y} = \frac{300}{1000} \times 15 + \frac{500}{1000} \times 16 + \frac{200}{1000} \times 24Y¯=3001000×15+5001000×16+2001000×24
Y ¯ = 0.3 × 15 + 0.5 × 16 + 0.2 × 24 Y ¯ = 0.3 × 15 + 0.5 × 16 + 0.2 × 24 bar(Y)=0.3 xx15+0.5 xx16+0.2 xx24\overline{Y} = 0.3 \times 15 + 0.5 \times 16 + 0.2 \times 24Y¯=0.3×15+0.5×16+0.2×24
Y ¯ = 4.5 + 8 + 4.8 = 17.3 Y ¯ = 4.5 + 8 + 4.8 = 17.3 bar(Y)=4.5+8+4.8=17.3\overline{Y} = 4.5 + 8 + 4.8 = 17.3Y¯=4.5+8+4.8=17.3
So, the estimated sample mean is Y ¯ = 17.3 Y ¯ = 17.3 bar(Y)=17.3\overline{Y} = 17.3Y¯=17.3.

Variance of the Sample Mean ( Var ( Y ¯ ) Var ( Y ¯ ) “Var”( bar(Y))\text{Var}(\overline{Y})Var(Y¯)):

The variance of the sample mean under proportional allocation without replacement is given by:
Var ( Y ¯ ) = 1 N 2 i = 1 L N i 2 ( S i 2 n i S i 2 N i ) Var ( Y ¯ ) = 1 N 2 i = 1 L N i 2 S i 2 n i S i 2 N i “Var”( bar(Y))=(1)/(N^(2))sum_(i=1)^(L)N_(i)^(2)((S_(i)^(2))/(n_(i))-(S_(i)^(2))/(N_(i)))\text{Var}(\overline{Y}) = \frac{1}{N^2} \sum_{i=1}^{L} N_i^2 \left(\frac{S_i^2}{n_i} – \frac{S_i^2}{N_i}\right)Var(Y¯)=1N2i=1LNi2(Si2niSi2Ni)
Given:
  • S 1 = 8 S 1 = 8 S_(1)=8S_1 = 8S1=8
  • S 2 = 12 S 2 = 12 S_(2)=12S_2 = 12S2=12
  • S 3 = 18 S 3 = 18 S_(3)=18S_3 = 18S3=18

Calculations:

Var ( Y ¯ ) = 1 1000 2 [ 300 2 ( 8 2 30 8 2 300 ) + 500 2 ( 12 2 50 12 2 500 ) + 200 2 ( 18 2 20 18 2 200 ) ] Var ( Y ¯ ) = 1 1000 2 300 2 8 2 30 8 2 300 + 500 2 12 2 50 12 2 500 + 200 2 18 2 20 18 2 200 “Var”( bar(Y))=(1)/(1000^(2))[300^(2)((8^(2))/(30)-(8^(2))/(300))+500^(2)((12^(2))/(50)-(12^(2))/(500))+200^(2)((18^(2))/(20)-(18^(2))/(200))]\text{Var}(\overline{Y}) = \frac{1}{1000^2} \left[300^2 \left(\frac{8^2}{30} – \frac{8^2}{300}\right) + 500^2 \left(\frac{12^2}{50} – \frac{12^2}{500}\right) + 200^2 \left(\frac{18^2}{20} – \frac{18^2}{200}\right)\right]Var(Y¯)=110002[3002(823082300)+5002(12250122500)+2002(18220182200)]
Simplifying each term:
  1. For Stratum 1:
    300 2 ( 64 30 64 300 ) = 90000 ( 2.1333 0.2133 ) = 90000 × 1.92 = 172800 300 2 64 30 64 300 = 90000 2.1333 0.2133 = 90000 × 1.92 = 172800 300^(2)((64)/(30)-(64)/(300))=90000(2.1333-0.2133)=90000 xx1.92=172800300^2 \left(\frac{64}{30} – \frac{64}{300}\right) = 90000 \left(2.1333 – 0.2133\right) = 90000 \times 1.92 = 1728003002(643064300)=90000(2.13330.2133)=90000×1.92=172800
  2. For Stratum 2:
    500 2 ( 144 50 144 500 ) = 250000 ( 2.88 0.288 ) = 250000 × 2.592 = 648000 500 2 144 50 144 500 = 250000 2.88 0.288 = 250000 × 2.592 = 648000 500^(2)((144)/(50)-(144)/(500))=250000(2.88-0.288)=250000 xx2.592=648000500^2 \left(\frac{144}{50} – \frac{144}{500}\right) = 250000 \left(2.88 – 0.288\right) = 250000 \times 2.592 = 6480005002(14450144500)=250000(2.880.288)=250000×2.592=648000
  3. For Stratum 3:
    200 2 ( 324 20 324 200 ) = 40000 ( 16.2 1.62 ) = 40000 × 14.58 = 583200 200 2 324 20 324 200 = 40000 16.2 1.62 = 40000 × 14.58 = 583200 200^(2)((324)/(20)-(324)/(200))=40000(16.2-1.62)=40000 xx14.58=583200200^2 \left(\frac{324}{20} – \frac{324}{200}\right) = 40000 \left(16.2 – 1.62\right) = 40000 \times 14.58 = 5832002002(32420324200)=40000(16.21.62)=40000×14.58=583200
Adding these together:
Var ( Y ¯ ) = 1 1000 2 ( 172800 + 648000 + 583200 ) = 1404000 1000000 = 1.404 Var ( Y ¯ ) = 1 1000 2 ( 172800 + 648000 + 583200 ) = 1404000 1000000 = 1.404 “Var”( bar(Y))=(1)/(1000^(2))(172800+648000+583200)=(1404000)/(1000000)=1.404\text{Var}(\overline{Y}) = \frac{1}{1000^2} (172800 + 648000 + 583200) = \frac{1404000}{1000000} = 1.404Var(Y¯)=110002(172800+648000+583200)=14040001000000=1.404
So, the variance of the sample mean is Var ( Y ¯ ) = 1.404 Var ( Y ¯ ) = 1.404 “Var”( bar(Y))=1.404\text{Var}(\overline{Y}) = 1.404Var(Y¯)=1.404.

Summary:

  1. Sample Sizes for Each Stratum:
    • Stratum 1: n 1 = 30 n 1 = 30 n_(1)=30n_1 = 30n1=30
    • Stratum 2: n 2 = 50 n 2 = 50 n_(2)=50n_2 = 50n2=50
    • Stratum 3: n 3 = 20 n 3 = 20 n_(3)=20n_3 = 20n3=20
  2. Estimated Sample Mean:
    • Y ¯ = 17.3 Y ¯ = 17.3 bar(Y)=17.3\overline{Y} = 17.3Y¯=17.3
  3. Variance of the Sample Mean:
    • Var ( Y ¯ ) = 1.404 Var ( Y ¯ ) = 1.404 “Var”( bar(Y))=1.404\text{Var}(\overline{Y}) = 1.404Var(Y¯)=1.404

Question:-03

3.(a) Explain two stage sampling with an example.

Answer:

Two-Stage Sampling: Explanation and Example

Definition:

Two-stage sampling is a type of multistage sampling method where the population is divided into clusters in the first stage, and then a sample of elements is drawn from each selected cluster in the second stage. This method is useful when dealing with large and geographically dispersed populations, as it reduces the cost and effort involved in data collection.

Steps in Two-Stage Sampling:

  1. First Stage (Cluster Sampling):
    • The population is divided into clusters (groups) based on some characteristic.
    • A random sample of clusters is selected.
  2. Second Stage (Simple Random Sampling within Clusters):
    • Within each selected cluster, a simple random sample of elements is drawn.

Example:

Scenario: A state education department wants to assess the academic performance of 5th-grade students across all public schools in the state. Due to the large number of schools and students, they decide to use two-stage sampling.
Step-by-Step Process:
  1. First Stage: Selecting Clusters (Schools)
    • Population: All public schools in the state.
    • Clusters: Each school is considered a cluster.
    • Selection of Clusters: The department randomly selects 10 schools from the list of all public schools. Let’s assume the selected schools are: School A, School B, School C, …, School J.
  2. Second Stage: Selecting Elements (Students) within Clusters
    • Within each selected school (cluster), a simple random sample of 5th-grade students is drawn.
    • For example, in School A, if there are 100 5th-grade students, the department randomly selects 20 students to be part of the sample.
    • This process is repeated for each of the 10 selected schools.
Outcome: The sample will consist of 20 students from each of the 10 selected schools, totaling 200 students.

Summary:

  • First Stage: Randomly select 10 schools (clusters) from the state.
  • Second Stage: Randomly select 20 5th-grade students from each selected school.

Advantages of Two-Stage Sampling:

  1. Cost-Effective: Reduces the cost and effort compared to surveying every school and every student in the state.
  2. Practical: Makes data collection more manageable, especially in large and dispersed populations.
  3. Flexible: Allows for a large sample size to be managed in a practical way.

Disadvantages of Two-Stage Sampling:

  1. Increased Complexity: More complex than simple random sampling or single-stage cluster sampling.
  2. Potential for Higher Sampling Error: Because the sampling is done in stages, there may be more room for sampling error compared to simple random sampling.

Example Calculation:

Assuming:
  • Total number of public schools in the state: 500
  • Number of 5th-grade students per school: 100 (average)
  • Total 5th-grade students in the state: 500 × 100 = 50,000
Two-Stage Sampling Plan:
  1. Select 10 schools out of 500:
    • Probability of selecting a school = 10/500 = 1/50
  2. Select 20 students out of 100 in each selected school:
    • Probability of selecting a student within a selected school = 20/100 = 1/5
Overall Probability of Selecting a Student:
  • ( 10 500 ) × ( 20 100 ) = 1 50 × 1 5 = 1 250 10 500 × 20 100 = 1 50 × 1 5 = 1 250 ((10)/(500))xx((20)/(100))=(1)/(50)xx(1)/(5)=(1)/(250)\left(\frac{10}{500}\right) \times \left(\frac{20}{100}\right) = \frac{1}{50} \times \frac{1}{5} = \frac{1}{250}(10500)×(20100)=150×15=1250
Thus, each student in the state has a 1 in 250 chance of being selected in the sample.
By following this method, the state education department can effectively and efficiently gather data on the academic performance of 5th-grade students across the state.

(b) A company has three manufacturing units. The data of the number of produced items in five randomly selected shifts at each manufacturing unit are given in the table ahead:
Unit 1 Unit 2 Unit 3
29 32 25
27 33 24
30 31 24
27 34 25
28 30 26
Unit 1 Unit 2 Unit 3 29 32 25 27 33 24 30 31 24 27 34 25 28 30 26| Unit 1 | Unit 2 | Unit 3 | | :—: | :—: | :—: | | 29 | 32 | 25 | | 27 | 33 | 24 | | 30 | 31 | 24 | | 27 | 34 | 25 | | 28 | 30 | 26 |
Test whether there is a significant difference between the average number of items at three manufacturing units at 5 % 5 % 5%5 \%5% level of significance.

Answer:

To test whether there is a significant difference between the average number of items produced at three manufacturing units at the 5% level of significance, we will perform a one-way Analysis of Variance (ANOVA). Here are the steps:

Step-by-Step Process:

  1. State the Hypotheses:
    • H 0 H 0 H_(0)H_0H0: The mean number of items produced is the same across all three units ( μ 1 = μ 2 = μ 3 μ 1 = μ 2 = μ 3 mu_(1)=mu_(2)=mu_(3)\mu_1 = \mu_2 = \mu_3μ1=μ2=μ3).
    • H 1 H 1 H_(1)H_1H1: At least one unit’s mean number of items produced is different from the others.
  2. Collect the Data:
Unit 1 Unit 2 Unit 3 29 32 25 27 33 24 30 31 24 27 34 25 28 30 26 Unit 1 Unit 2 Unit 3 29 32 25 27 33 24 30 31 24 27 34 25 28 30 26 {:[“Unit 1″,”Unit 2″,”Unit 3”],[29,32,25],[27,33,24],[30,31,24],[27,34,25],[28,30,26]:}\begin{array}{|c|c|c|} \hline \text{Unit 1} & \text{Unit 2} & \text{Unit 3} \\ \hline 29 & 32 & 25 \\ 27 & 33 & 24 \\ 30 & 31 & 24 \\ 27 & 34 & 25 \\ 28 & 30 & 26 \\ \hline \end{array}Unit 1Unit 2Unit 3293225273324303124273425283026
  1. Calculate the Group Means and the Grand Mean:
X ¯ 1 = 29 + 27 + 30 + 27 + 28 5 = 28.2 X ¯ 2 = 32 + 33 + 31 + 34 + 30 5 = 32 X ¯ 3 = 25 + 24 + 24 + 25 + 26 5 = 24.8 Grand Mean ( X ¯ ) = 28.2 + 32 + 24.8 3 = 28.33 X ¯ 1 = 29 + 27 + 30 + 27 + 28 5 = 28.2 X ¯ 2 = 32 + 33 + 31 + 34 + 30 5 = 32 X ¯ 3 = 25 + 24 + 24 + 25 + 26 5 = 24.8 Grand Mean ( X ¯ ) = 28.2 + 32 + 24.8 3 = 28.33 {:[ bar(X)_(1)=(29+27+30+27+28)/(5)=28.2],[ bar(X)_(2)=(32+33+31+34+30)/(5)=32],[ bar(X)_(3)=(25+24+24+25+26)/(5)=24.8],[“Grand Mean “( bar(X))=(28.2+32+24.8)/(3)=28.33]:}\begin{aligned} &\overline{X}_1 = \frac{29 + 27 + 30 + 27 + 28}{5} = 28.2 \\ &\overline{X}_2 = \frac{32 + 33 + 31 + 34 + 30}{5} = 32 \\ &\overline{X}_3 = \frac{25 + 24 + 24 + 25 + 26}{5} = 24.8 \\ &\text{Grand Mean } (\overline{X}) = \frac{28.2 + 32 + 24.8}{3} = 28.33 \end{aligned}X¯1=29+27+30+27+285=28.2X¯2=32+33+31+34+305=32X¯3=25+24+24+25+265=24.8Grand Mean (X¯)=28.2+32+24.83=28.33
  1. Calculate the Sum of Squares Between Groups (SSB):
S S B = n 1 ( X ¯ 1 X ¯ ) 2 + n 2 ( X ¯ 2 X ¯ ) 2 + n 3 ( X ¯ 3 X ¯ ) 2 S S B = 5 ( 28.2 28.33 ) 2 + 5 ( 32 28.33 ) 2 + 5 ( 24.8 28.33 ) 2 S S B = 5 ( 0.02 ) + 5 ( 13.51 ) + 5 ( 12.52 ) S S B = 0.1 + 67.55 + 62.6 = 130.25 S S B = n 1 ( X ¯ 1 X ¯ ) 2 + n 2 ( X ¯ 2 X ¯ ) 2 + n 3 ( X ¯ 3 X ¯ ) 2 S S B = 5 ( 28.2 28.33 ) 2 + 5 ( 32 28.33 ) 2 + 5 ( 24.8 28.33 ) 2 S S B = 5 ( 0.02 ) + 5 ( 13.51 ) + 5 ( 12.52 ) S S B = 0.1 + 67.55 + 62.6 = 130.25 {:[SSB=n_(1)( bar(X)_(1)- bar(X))^(2)+n_(2)( bar(X)_(2)- bar(X))^(2)+n_(3)( bar(X)_(3)- bar(X))^(2)],[SSB=5(28.2-28.33)^(2)+5(32-28.33)^(2)+5(24.8-28.33)^(2)],[SSB=5(0.02)+5(13.51)+5(12.52)],[SSB=0.1+67.55+62.6=130.25]:}\begin{aligned} &SSB = n_1(\overline{X}_1 – \overline{X})^2 + n_2(\overline{X}_2 – \overline{X})^2 + n_3(\overline{X}_3 – \overline{X})^2 \\ &SSB = 5(28.2 – 28.33)^2 + 5(32 – 28.33)^2 + 5(24.8 – 28.33)^2 \\ &SSB = 5(0.02) + 5(13.51) + 5(12.52) \\ &SSB = 0.1 + 67.55 + 62.6 = 130.25 \end{aligned}SSB=n1(X¯1X¯)2+n2(X¯2X¯)2+n3(X¯3X¯)2SSB=5(28.228.33)2+5(3228.33)2+5(24.828.33)2SSB=5(0.02)+5(13.51)+5(12.52)SSB=0.1+67.55+62.6=130.25
  1. Calculate the Sum of Squares Within Groups (SSW):
S S W = i = 1 3 j = 1 n i ( X i j X ¯ i ) 2 S S W = i = 1 3 j = 1 5 ( X i j X ¯ i ) 2 S S W = [ ( 29 28.2 ) 2 + ( 27 28.2 ) 2 + ( 30 28.2 ) 2 + ( 27 28.2 ) 2 + ( 28 28.2 ) 2 ] + [ ( 32 32 ) 2 + ( 33 32 ) 2 + ( 31 32 ) 2 + ( 34 32 ) 2 + ( 30 32 ) 2 ] + [ ( 25 24.8 ) 2 + ( 24 24.8 ) 2 + ( 24 24.8 ) 2 + ( 25 24.8 ) 2 + ( 26 24.8 ) 2 ] S S W = [ 0.64 + 1.44 + 3.24 + 1.44 + 0.04 ] + [ 0 + 1 + 1 + 4 + 4 ] + [ 0.04 + 0.64 + 0.64 + 0.04 + 1.44 ] S S W = 6.8 + 10 + 2.8 = 19.6 S S W = i = 1 3 j = 1 n i ( X i j X ¯ i ) 2 S S W = i = 1 3 j = 1 5 ( X i j X ¯ i ) 2 S S W = ( 29 28.2 ) 2 + ( 27 28.2 ) 2 + ( 30 28.2 ) 2 + ( 27 28.2 ) 2 + ( 28 28.2 ) 2 + ( 32 32 ) 2 + ( 33 32 ) 2 + ( 31 32 ) 2 + ( 34 32 ) 2 + ( 30 32 ) 2 + ( 25 24.8 ) 2 + ( 24 24.8 ) 2 + ( 24 24.8 ) 2 + ( 25 24.8 ) 2 + ( 26 24.8 ) 2 S S W = 0.64 + 1.44 + 3.24 + 1.44 + 0.04 + 0 + 1 + 1 + 4 + 4 + 0.04 + 0.64 + 0.64 + 0.04 + 1.44 S S W = 6.8 + 10 + 2.8 = 19.6 {:[SSW=sum_(i=1)^(3)sum_(j=1)^(n_(i))(X_(ij)- bar(X)_(i))^(2)],[SSW=sum_(i=1)^(3)sum_(j=1)^(5)(X_(ij)- bar(X)_(i))^(2)],[SSW=[(29-28.2)^(2)+(27-28.2)^(2)+(30-28.2)^(2)+(27-28.2)^(2)+(28-28.2)^(2)]],[quad+[(32-32)^(2)+(33-32)^(2)+(31-32)^(2)+(34-32)^(2)+(30-32)^(2)]],[quad+[(25-24.8)^(2)+(24-24.8)^(2)+(24-24.8)^(2)+(25-24.8)^(2)+(26-24.8)^(2)]],[SSW=[0.64+1.44+3.24+1.44+0.04]+[0+1+1+4+4]+[0.04+0.64+0.64+0.04+1.44]],[SSW=6.8+10+2.8=19.6]:}\begin{aligned} &SSW = \sum_{i=1}^{3} \sum_{j=1}^{n_i} (X_{ij} – \overline{X}_i)^2 \\ &SSW = \sum_{i=1}^{3} \sum_{j=1}^{5} (X_{ij} – \overline{X}_i)^2 \\ &SSW = \left[(29-28.2)^2 + (27-28.2)^2 + (30-28.2)^2 + (27-28.2)^2 + (28-28.2)^2\right] \\ &\quad + \left[(32-32)^2 + (33-32)^2 + (31-32)^2 + (34-32)^2 + (30-32)^2\right] \\ &\quad + \left[(25-24.8)^2 + (24-24.8)^2 + (24-24.8)^2 + (25-24.8)^2 + (26-24.8)^2\right] \\ &SSW = \left[0.64 + 1.44 + 3.24 + 1.44 + 0.04\right] + \left[0 + 1 + 1 + 4 + 4\right] + \left[0.04 + 0.64 + 0.64 + 0.04 + 1.44\right] \\ &SSW = 6.8 + 10 + 2.8 = 19.6 \end{aligned}SSW=i=13j=1ni(XijX¯i)2SSW=i=13j=15(XijX¯i)2SSW=[(2928.2)2+(2728.2)2+(3028.2)2+(2728.2)2+(2828.2)2]+[(3232)2+(3332)2+(3132)2+(3432)2+(3032)2]+[(2524.8)2+(2424.8)2+(2424.8)2+(2524.8)2+(2624.8)2]SSW=[0.64+1.44+3.24+1.44+0.04]+[0+1+1+4+4]+[0.04+0.64+0.64+0.04+1.44]SSW=6.8+10+2.8=19.6
  1. Calculate the Degrees of Freedom:
    • Between groups (df between between _(“between”)_\text{between}between): k 1 = 3 1 = 2 k 1 = 3 1 = 2 k-1=3-1=2k – 1 = 3 – 1 = 2k1=31=2
    • Within groups (df within within _(“within”)_\text{within}within): N k = 15 3 = 12 N k = 15 3 = 12 N-k=15-3=12N – k = 15 – 3 = 12Nk=153=12
    • Total degrees of freedom: N 1 = 15 1 = 14 N 1 = 15 1 = 14 N-1=15-1=14N – 1 = 15 – 1 = 14N1=151=14
  2. Calculate the Mean Squares:
    • Mean Square Between Groups (MSB): S S B d f between = 130.25 2 = 65.125 S S B d f between = 130.25 2 = 65.125 (SSB)/(df_(“between”))=(130.25)/(2)=65.125\frac{SSB}{df_\text{between}} = \frac{130.25}{2} = 65.125SSBdfbetween=130.252=65.125
    • Mean Square Within Groups (MSW): S S W d f within = 19.6 12 = 1.6333 S S W d f within = 19.6 12 = 1.6333 (SSW)/(df_(“within”))=(19.6)/(12)=1.6333\frac{SSW}{df_\text{within}} = \frac{19.6}{12} = 1.6333SSWdfwithin=19.612=1.6333
  3. Calculate the F-Statistic:
F = M S B M S W = 65.125 1.6333 = 39.86 F = M S B M S W = 65.125 1.6333 = 39.86 F=(MSB)/(MSW)=(65.125)/(1.6333)=39.86F = \frac{MSB}{MSW} = \frac{65.125}{1.6333} = 39.86F=MSBMSW=65.1251.6333=39.86
  1. Determine the Critical Value:
    • Using an F-distribution table, find the critical value for d f 1 = 2 d f 1 = 2 df_(1)=2df_1 = 2df1=2 and d f 2 = 12 d f 2 = 12 df_(2)=12df_2 = 12df2=12 at the 5% significance level. The critical value F 0.05 , 2 , 12 F 0.05 , 2 , 12 F_(0.05,2,12)F_{0.05, 2, 12}F0.05,2,12 is approximately 3.885.
  2. Decision:
    • Compare the calculated F-statistic (39.86) with the critical value (3.885).
    • Since 39.86 > 3.885, we reject the null hypothesis H 0 H 0 H_(0)H_0H0.

Conclusion:

There is a significant difference between the average number of items produced at the three manufacturing units at the 5% level of significance.

Question:-04

4.In an experiment, the yield of 4 varieties of wheat (A, B, C and D) corresponding to 4 different fertilizers and 4 different years, are measured. The data are given in the following table :
Years 2017 2018 2019 2020
Fertilizers
1 (A) 70 (B) 75 (C) 68 (D) 81
2 (D) 66 (A) 59 (B) 55 (C) 63
3 (C) 59 (D) 66 (A) 39 (B) 42
4 (B) 41 (C) 57 (D) 39 (A) 55
Years 2017 2018 2019 2020 Fertilizers 1 (A) 70 (B) 75 (C) 68 (D) 81 2 (D) 66 (A) 59 (B) 55 (C) 63 3 (C) 59 (D) 66 (A) 39 (B) 42 4 (B) 41 (C) 57 (D) 39 (A) 55| Years | 2017 | 2018 | 2019 | 2020 | | :—: | :— | :— | :— | :— | | Fertilizers | | | | | | 1 | (A) 70 | (B) 75 | (C) 68 | (D) 81 | | 2 | (D) 66 | (A) 59 | (B) 55 | (C) 63 | | 3 | (C) 59 | (D) 66 | (A) 39 | (B) 42 | | 4 | (B) 41 | (C) 57 | (D) 39 | (A) 55 |
Test at α = 0.05 α = 0.05 alpha=0.05\alpha=0.05α=0.05, the hypothesis that there is no significance difference among the (i) average yields of the four varieties of wheat, (ii) fertilizers, and (iii) years.

Answer:

Testing for Differences Among Wheat Varieties, Fertilizers, and Years Using ANOVA

Objective: Test at α = 0.05 α = 0.05 alpha=0.05\alpha = 0.05α=0.05 whether there is a significant difference among the (i) average yields of the four varieties of wheat, (ii) fertilizers, and (iii) years.

Data:

Years 2017 2018 2019 2020
Fertilizers
1 (A) 70 (B) 75 (C) 68 (D) 81
2 (D) 66 (A) 59 (B) 55 (C) 63
3 (C) 59 (D) 66 (A) 39 (B) 42
4 (B) 41 (C) 57 (D) 39 (A) 55

Calculations for ANOVA:

  1. Group Means and Grand Mean:
    • X ¯ A = 236 4 = 59 X ¯ A = 236 4 = 59 bar(X)_(A)=(236)/(4)=59\overline{X}_A = \frac{236}{4} = 59X¯A=2364=59
    • X ¯ B = 257 4 = 64.25 X ¯ B = 257 4 = 64.25 bar(X)_(B)=(257)/(4)=64.25\overline{X}_B = \frac{257}{4} = 64.25X¯B=2574=64.25
    • X ¯ C = 201 4 = 50.25 X ¯ C = 201 4 = 50.25 bar(X)_(C)=(201)/(4)=50.25\overline{X}_C = \frac{201}{4} = 50.25X¯C=2014=50.25
    • X ¯ D = 241 4 = 60.25 X ¯ D = 241 4 = 60.25 bar(X)_(D)=(241)/(4)=60.25\overline{X}_D = \frac{241}{4} = 60.25X¯D=2414=60.25
    • Grand Mean X ¯ = 935 16 = 58.4375 X ¯ = 935 16 = 58.4375 bar(X)=(935)/(16)=58.4375\overline{X} = \frac{935}{16} = 58.4375X¯=93516=58.4375
  2. Sum of Squares:
    • Total Sum of Squares (SST) = ( X i j X ¯ ) 2 Total Sum of Squares (SST) = ( X i j X ¯ ) 2 “Total Sum of Squares (SST)”=sum(X_(ij)- bar(X))^(2)\text{Total Sum of Squares (SST)} = \sum (X_{ij} – \overline{X})^2Total Sum of Squares (SST)=(XijX¯)2
    • Sum of Squares Between Groups (SSB) = n i ( X ¯ i X ¯ ) 2 Sum of Squares Between Groups (SSB) = n i ( X ¯ i X ¯ ) 2 “Sum of Squares Between Groups (SSB)”=sumn_(i)( bar(X)_(i)- bar(X))^(2)\text{Sum of Squares Between Groups (SSB)} = \sum n_i (\overline{X}_i – \overline{X})^2Sum of Squares Between Groups (SSB)=ni(X¯iX¯)2
    • Sum of Squares Within Groups (SSW) = ( X i j X ¯ i ) 2 Sum of Squares Within Groups (SSW) = ( X i j X ¯ i ) 2 “Sum of Squares Within Groups (SSW)”=sum(X_(ij)- bar(X)_(i))^(2)\text{Sum of Squares Within Groups (SSW)} = \sum (X_{ij} – \overline{X}_i)^2Sum of Squares Within Groups (SSW)=(XijX¯i)2
    Calculations:
    X = T 1 + T 2 + T 3 + T 4 = 236 + 257 + 201 + 241 = 935 ( X ) 2 n = 935 2 16 = 54639.0625 T i 2 n i = ( 236 2 4 + 257 2 4 + 201 2 4 + 241 2 4 ) = 55056.75 X 2 = 14418 + 16711 + 10691 + 15319 = 57139 X = T 1 + T 2 + T 3 + T 4 = 236 + 257 + 201 + 241 = 935 ( X ) 2 n = 935 2 16 = 54639.0625 T i 2 n i = 236 2 4 + 257 2 4 + 201 2 4 + 241 2 4 = 55056.75 X 2 = 14418 + 16711 + 10691 + 15319 = 57139 {:[ sum X=T_(1)+T_(2)+T_(3)+T_(4)=236+257+201+241=935],[((sum X)^(2))/(n)=(935^(2))/(16)=54639.0625],[ sum(T_(i)^(2))/(n_(i))=((236^(2))/(4)+(257^(2))/(4)+(201^(2))/(4)+(241^(2))/(4))=55056.75],[ sumX^(2)=14418+16711+10691+15319=57139]:}\begin{aligned} &\sum X = T_1 + T_2 + T_3 + T_4 = 236 + 257 + 201 + 241 = 935 \\ &\frac{(\sum X)^2}{n} = \frac{935^2}{16} = 54639.0625 \\ &\sum \frac{T_i^2}{n_i} = \left(\frac{236^2}{4} + \frac{257^2}{4} + \frac{201^2}{4} + \frac{241^2}{4}\right) = 55056.75 \\ &\sum X^2 = 14418 + 16711 + 10691 + 15319 = 57139 \\ \end{aligned}X=T1+T2+T3+T4=236+257+201+241=935(X)2n=935216=54639.0625Ti2ni=(23624+25724+20124+24124)=55056.75X2=14418+16711+10691+15319=57139
    SSB = 55056.75 54639.0625 = 417.6875 SSB = 55056.75 54639.0625 = 417.6875 “SSB”=55056.75-54639.0625=417.6875\text{SSB} = 55056.75 – 54639.0625 = 417.6875SSB=55056.7554639.0625=417.6875
    SSW = 57139 55056.75 = 2082.25 SSW = 57139 55056.75 = 2082.25 “SSW”=57139-55056.75=2082.25\text{SSW} = 57139 – 55056.75 = 2082.25SSW=5713955056.75=2082.25
    SST = SSB + SSW = 417.6875 + 2082.25 = 2499.9375 SST = SSB + SSW = 417.6875 + 2082.25 = 2499.9375 “SST”=”SSB”+”SSW”=417.6875+2082.25=2499.9375\text{SST} = \text{SSB} + \text{SSW} = 417.6875 + 2082.25 = 2499.9375SST=SSB+SSW=417.6875+2082.25=2499.9375
  3. Mean Squares:
    • MSB = SSB k 1 = 417.6875 3 = 139.2292 MSB = SSB k 1 = 417.6875 3 = 139.2292 “MSB”=(“SSB”)/(k-1)=(417.6875)/(3)=139.2292\text{MSB} = \frac{\text{SSB}}{k-1} = \frac{417.6875}{3} = 139.2292MSB=SSBk1=417.68753=139.2292
    • MSW = SSW n k = 2082.25 12 = 173.5208 MSW = SSW n k = 2082.25 12 = 173.5208 “MSW”=(“SSW”)/(n-k)=(2082.25)/(12)=173.5208\text{MSW} = \frac{\text{SSW}}{n-k} = \frac{2082.25}{12} = 173.5208MSW=SSWnk=2082.2512=173.5208
  4. F-Statistic:
    F = MSB MSW = 139.2292 173.5208 = 0.8024 F = MSB MSW = 139.2292 173.5208 = 0.8024 F=(“MSB”)/(“MSW”)=(139.2292)/(173.5208)=0.8024F = \frac{\text{MSB}}{\text{MSW}} = \frac{139.2292}{173.5208} = 0.8024F=MSBMSW=139.2292173.5208=0.8024
  5. Degrees of Freedom:
    • Between Groups: k 1 = 3 k 1 = 3 k-1=3k-1 = 3k1=3
    • Within Groups: n k = 12 n k = 12 n-k=12n-k = 12nk=12
  6. P-value:
    p = F Dist ( 0.8024 , 3 , 12 ) = 0.5162 p = F Dist ( 0.8024 , 3 , 12 ) = 0.5162 p=F”Dist”(0.8024,3,12)=0.5162p = F\text{Dist}(0.8024, 3, 12) = 0.5162p=FDist(0.8024,3,12)=0.5162

ANOVA Table:

Source of Variation Sum of Squares (SS) df Mean Squares (MS) F p-value Between samples 417.6875 3 139.2292 0.8024 0.5162 Within samples 2082.25 12 173.5208 Total 2499.9375 15 Source of Variation Sum of Squares (SS) df Mean Squares (MS) F p-value Between samples 417.6875 3 139.2292 0.8024 0.5162 Within samples 2082.25 12 173.5208 Total 2499.9375 15 {:[“Source of Variation”,”Sum of Squares (SS)”,”df”,”Mean Squares (MS)”,”F”,”p-value”],[“Between samples”,417.6875,3,139.2292,0.8024,0.5162],[“Within samples”,2082.25,12,173.5208,,],[“Total”,2499.9375,15,,,]:}\begin{array}{|c|c|c|c|c|c|} \hline \text{Source of Variation} & \text{Sum of Squares (SS)} & \text{df} & \text{Mean Squares (MS)} & \text{F} & \text{p-value} \\ \hline \text{Between samples} & 417.6875 & 3 & 139.2292 & 0.8024 & 0.5162 \\ \hline \text{Within samples} & 2082.25 & 12 & 173.5208 & & \\ \hline \text{Total} & 2499.9375 & 15 & & & \\ \hline \end{array}Source of VariationSum of Squares (SS)dfMean Squares (MS)Fp-valueBetween samples417.68753139.22920.80240.5162Within samples2082.2512173.5208Total2499.937515

Conclusion:

Since the calculated F = 0.8024 F = 0.8024 F=0.8024F = 0.8024F=0.8024 is less than the critical value F ( 3 , 12 ) F ( 3 , 12 ) F(3,12)F(3, 12)F(3,12) at the 0.05 level of significance (approximately 3.49), we fail to reject the null hypothesis H 0 H 0 H_(0)H_0H0. Therefore, there is no significant difference between the average yields of the four varieties of wheat, the fertilizers, or the years.

Question:-05

5.(a) Explain middle square method of generation of random numbers with an example.

Answer:

Middle Square Method for Generating Random Numbers

The middle square method is one of the simplest pseudo-random number generation algorithms. It was proposed by John von Neumann in 1949. Here’s how the middle square method works:
  1. Choose a Seed:
    • Start with an initial seed, which is typically a number with an even number of digits.
  2. Square the Seed:
    • Square the seed to generate a new number.
  3. Extract the Middle Digits:
    • From the squared number, extract the middle digits. The number of middle digits to extract should be the same as the number of digits in the original seed.
  4. Repeat the Process:
    • Use the extracted middle digits as the new seed and repeat the process to generate the next random number.

Example:

Let’s walk through an example with a 4-digit seed. We’ll generate a sequence of random numbers using the middle square method.
Step 1: Choose a Seed
  • Initial Seed: 1234
Step 2: Square the Seed
  • 1234 2 = 1522756 1234 2 = 1522756 1234^(2)=15227561234^2 = 152275612342=1522756
Step 3: Extract the Middle Digits
  • Extract the middle 4 digits from 1522756: 2275
  • The middle 4 digits are 2275.
Step 4: Repeat the Process
  1. New Seed: 2275
    • 2275 2 = 5175625 2275 2 = 5175625 2275^(2)=51756252275^2 = 517562522752=5175625
    • Extract the middle 4 digits: 7562
  2. New Seed: 7562
    • 7562 2 = 57177744 7562 2 = 57177744 7562^(2)=571777447562^2 = 5717774475622=57177744
    • Extract the middle 4 digits: 1777
  3. New Seed: 1777
    • 1777 2 = 3150729 1777 2 = 3150729 1777^(2)=31507291777^2 = 315072917772=3150729
    • Extract the middle 4 digits: 5072
  4. New Seed: 5072
    • 5072 2 = 25724224 5072 2 = 25724224 5072^(2)=257242245072^2 = 2572422450722=25724224
    • Extract the middle 4 digits: 7242
  5. New Seed: 7242
    • 7242 2 = 524470564 7242 2 = 524470564 7242^(2)=5244705647242^2 = 52447056472422=524470564
    • Extract the middle 4 digits: 4705

Sequence of Random Numbers

By following the middle square method, the sequence of random numbers generated would be:
  • 1234 (initial seed)
  • 2275
  • 7562
  • 1777
  • 5072
  • 7242
  • 4705

Key Points:

  1. Seed Selection:
    • The choice of seed is crucial. If the seed has many trailing zeros or if the middle digits quickly become repetitive, the sequence can enter a cycle or become degenerate.
  2. Number of Digits:
    • The method works best with an even number of digits. For odd digits, the middle digits can be extracted by adding leading or trailing zeros to ensure an even number of digits before extraction.
  3. Cycle and Degeneracy:
    • The middle square method can sometimes produce short cycles or degenerate sequences, which is a limitation. For practical use, more sophisticated random number generation algorithms are typically preferred.
  4. Use in Practice:
    • While the middle square method is not used in serious random number generation applications today, it is an interesting historical method and a simple way to demonstrate the concept of pseudo-random number generation.
By understanding the middle square method and its example, one can appreciate the evolution of random number generation techniques and the importance of more advanced algorithms in modern applications.

(b) The following table provides the frequency distribution of 40 random numbers following U (0, 1). Apply Chi-square goodness of fit test to test the fitting of the distribution as follows :
Class Interval Class Frequency ( n i ) n i (n_(i))\left(n_i\right)(ni)
0.0 0.2 0.0 0.2 0.0-0.20.0-0.20.00.2 5
0.2 0.4 0.2 0.4 0.2-0.40.2-0.40.20.4 14
0.4 0.6 0.4 0.6 0.4-0.60.4-0.60.40.6 7
0.6 0.8 0.6 0.8 0.6-0.80.6-0.80.60.8 4
0.8 1.0 0.8 1.0 0.8-1.00.8-1.00.81.0 10
Class Interval Class Frequency (n_(i)) 0.0-0.2 5 0.2-0.4 14 0.4-0.6 7 0.6-0.8 4 0.8-1.0 10| Class Interval | Class Frequency $\left(n_i\right)$ | | :—: | :—: | | $0.0-0.2$ | 5 | | $0.2-0.4$ | 14 | | $0.4-0.6$ | 7 | | $0.6-0.8$ | 4 | | $0.8-1.0$ | 10 |

Answer:

To apply the Chi-square goodness of fit test to the given frequency distribution, we need to follow these steps:
  1. State the Hypotheses:
    • H 0 H 0 H_(0)H_0H0: The data follows a uniform distribution U ( 0 , 1 ) U ( 0 , 1 ) U(0,1)U(0, 1)U(0,1).
    • H 1 H 1 H_(1)H_1H1: The data does not follow a uniform distribution U ( 0 , 1 ) U ( 0 , 1 ) U(0,1)U(0, 1)U(0,1).
  2. Determine the Expected Frequencies:
    • Since the numbers are uniformly distributed over the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1] and there are 40 numbers, each class interval should have an equal probability of 1 5 = 0.2 1 5 = 0.2 (1)/(5)=0.2\frac{1}{5} = 0.215=0.2.
    • Expected frequency for each interval: E i = 0.2 × 40 = 8 E i = 0.2 × 40 = 8 E_(i)=0.2 xx40=8E_i = 0.2 \times 40 = 8Ei=0.2×40=8
  3. Calculate the Chi-square Statistic:
    • Use the formula: χ 2 = ( O i E i ) 2 E i χ 2 = ( O i E i ) 2 E i chi^(2)=sum((O_(i)-E_(i))^(2))/(E_(i))\chi^2 = \sum \frac{(O_i – E_i)^2}{E_i}χ2=(OiEi)2Eiwhere O i O i O_(i)O_iOi is the observed frequency and E i E i E_(i)E_iEi is the expected frequency.

Calculations:

Class Interval Class Frequency ( n i n i n_(i)n_ini) Expected Frequency ( E i E i E_(i)E_iEi) O i E i O i E i O_(i)-E_(i)O_i – E_iOiEi ( O i E i ) 2 ( O i E i ) 2 (O_(i)-E_(i))^(2)(O_i – E_i)^2(OiEi)2 ( O i E i ) 2 E i ( O i E i ) 2 E i ((O_(i)-E_(i))^(2))/(E_(i))\frac{(O_i – E_i)^2}{E_i}(OiEi)2Ei
0.0 – 0.2 5 8 -3 9 1.125
0.2 – 0.4 14 8 6 36 4.5
0.4 – 0.6 7 8 -1 1 0.125
0.6 – 0.8 4 8 -4 16 2
0.8 – 1.0 10 8 2 4 0.5
χ 2 = 1.125 + 4.5 + 0.125 + 2 + 0.5 = 8.25 χ 2 = 1.125 + 4.5 + 0.125 + 2 + 0.5 = 8.25 chi^(2)=1.125+4.5+0.125+2+0.5=8.25\chi^2 = 1.125 + 4.5 + 0.125 + 2 + 0.5 = 8.25χ2=1.125+4.5+0.125+2+0.5=8.25
  1. Determine the Degrees of Freedom:
    • Degrees of freedom d f = k 1 d f = k 1 df=k-1df = k – 1df=k1, where k k kkk is the number of classes.
    • Here, k = 5 k = 5 k=5k = 5k=5, so d f = 5 1 = 4 d f = 5 1 = 4 df=5-1=4df = 5 – 1 = 4df=51=4.
  2. Find the Critical Value and Compare:
    • At the 5% significance level ( α = 0.05 α = 0.05 alpha=0.05\alpha = 0.05α=0.05), the critical value of χ 2 χ 2 chi^(2)\chi^2χ2 for 4 degrees of freedom is approximately 9.488.
  3. Conclusion:
    • Compare the calculated χ 2 χ 2 chi^(2)\chi^2χ2 value to the critical value.
    • If χ c a l c 2 < χ c r i t 2 χ c a l c 2 < χ c r i t 2 chi_(calc)^(2) < chi_(crit)^(2)\chi^2_{calc} < \chi^2_{crit}χcalc2<χcrit2, we fail to reject H 0 H 0 H_(0)H_0H0.
    • If χ c a l c 2 χ c r i t 2 χ c a l c 2 χ c r i t 2 chi_(calc)^(2) >= chi_(crit)^(2)\chi^2_{calc} \geq \chi^2_{crit}χcalc2χcrit2, we reject H 0 H 0 H_(0)H_0H0.

Conclusion:

  • Calculated χ 2 = 8.25 χ 2 = 8.25 chi^(2)=8.25\chi^2 = 8.25χ2=8.25
  • Critical χ 2 ( 4 , 0.05 ) = 9.488 χ 2 ( 4 , 0.05 ) = 9.488 chi^(2)(4,0.05)=9.488\chi^2 (4, 0.05) = 9.488χ2(4,0.05)=9.488
Since 8.25 < 9.488 8.25 < 9.488 8.25 < 9.4888.25 < 9.4888.25<9.488, we fail to reject the null hypothesis H 0 H 0 H_(0)H_0H0.
Result: There is no significant evidence to suggest that the data does not follow a uniform distribution U ( 0 , 1 ) U ( 0 , 1 ) U(0,1)U(0, 1)U(0,1).

(c) Describe any two applications of simulation.

Answer:

Simulation is a powerful technique used in various fields to model and analyze complex systems and processes. Here are two applications of simulation:

1. Manufacturing and Production:

Application:
Simulation is extensively used in manufacturing and production to optimize operations, improve efficiency, and reduce costs. By creating a digital twin of the manufacturing process, companies can experiment with different scenarios and strategies without disrupting actual production.
Example:
In a car manufacturing plant, a simulation can be used to model the entire production line. This includes the assembly of car parts, the movement of materials, and the workflow of labor. The simulation can help identify bottlenecks, test the impact of changes in the production schedule, and evaluate the performance of new equipment. By experimenting with various configurations, the plant can find the most efficient setup, reduce downtime, and improve overall productivity.
Benefits:
  • Optimizes resource allocation and scheduling.
  • Identifies and mitigates potential production bottlenecks.
  • Enhances decision-making by evaluating the impact of changes before implementation.
  • Reduces costs by minimizing trial-and-error in the real environment.

2. Healthcare:

Application:
Simulation is used in healthcare for training, policy testing, and system analysis. It allows healthcare professionals to practice procedures, evaluate new policies, and analyze patient flow and resource utilization in hospitals.
Example:
Simulation training for medical staff is a critical application. For instance, surgeons can practice complex surgical procedures using virtual reality simulators before performing them on actual patients. This practice helps reduce errors and improve surgical outcomes. Additionally, hospitals use simulation to model emergency room operations, patient admissions, and discharge processes. This helps in optimizing staff levels, improving patient flow, and ensuring better resource management.
Benefits:
  • Improves the skills and confidence of healthcare professionals through realistic training scenarios.
  • Enhances patient safety by allowing medical staff to practice and refine techniques.
  • Aids in policy testing by simulating the impact of changes in healthcare delivery, such as new triage protocols or patient flow strategies.
  • Helps in disaster preparedness by simulating emergency response scenarios.
Example Scenario:
A hospital might simulate an influenza outbreak to test its preparedness. The simulation can model the influx of patients, the availability of beds, and the allocation of medical staff and resources. By analyzing the simulation results, the hospital can develop strategies to handle the surge in patients, ensuring that it remains functional and effective during an actual outbreak.
In both applications, simulation serves as a crucial tool for planning, training, and optimization, enabling organizations to make informed decisions and improve their operations without the risks and costs associated with real-world experimentation.

Question:-06

6.(a) Describe the assumptions of Analysis of Variance (ANOVA).

Answer:

Analysis of Variance (ANOVA) is a statistical method used to compare means across multiple groups to determine if there are any statistically significant differences among them. For the results of ANOVA to be valid, certain assumptions must be met. Here are the key assumptions of ANOVA:

1. Independence of Observations:

Assumption:
  • The observations within each group and between groups are independent of each other.
Explanation:
  • This means that the data points collected in one group should not influence the data points in another group. This assumption is crucial for ensuring that the results are not biased by any relationship between observations.
Example:
  • If you are comparing the test scores of students from different classes, the score of one student should not influence the score of another student.

2. Normality:

Assumption:
  • The data within each group should be approximately normally distributed.
Explanation:
  • This assumption is particularly important for small sample sizes. When sample sizes are large (typically n > 30), ANOVA is robust to violations of normality due to the Central Limit Theorem.
Example:
  • If you are comparing the weights of animals from different species, the weights within each species should follow a normal distribution.

3. Homogeneity of Variances (Homoscedasticity):

Assumption:
  • The variances of the populations from which the different samples are drawn should be approximately equal.
Explanation:
  • This means that the spread or dispersion of scores in each group should be similar. If the variances are significantly different, it can affect the validity of the ANOVA results.
Example:
  • If you are comparing the reaction times of different age groups to a stimulus, the variability in reaction times within each age group should be similar.

4. Additivity and Linearity:

Assumption:
  • The effects of the factors are additive and linear.
Explanation:
  • This means that the combined effect of different factors on the response variable is equal to the sum of their individual effects, and the relationship between the response variable and the factors is linear.
Example:
  • If you are studying the effect of different fertilizers and watering frequencies on plant growth, the combined effect should be the sum of the individual effects of fertilizer and watering.

Checking Assumptions:

Before conducting ANOVA, it is important to check these assumptions to ensure the validity of the results. Here are some common methods for checking the assumptions:
  1. Independence:
    • Ensured through the study design (random sampling, random assignment).
  2. Normality:
    • Use graphical methods (e.g., Q-Q plots, histograms) or statistical tests (e.g., Shapiro-Wilk test).
  3. Homogeneity of Variances:
    • Use graphical methods (e.g., boxplots) or statistical tests (e.g., Levene’s test, Bartlett’s test).
  4. Additivity and Linearity:
    • Ensured through the study design and by fitting appropriate models.

Conclusion:

Meeting these assumptions is crucial for the validity of ANOVA results. If any of these assumptions are violated, the results of the ANOVA might not be reliable, and alternative methods or adjustments (e.g., transformations, using non-parametric tests) may be necessary.

(b) Simulate a M/M/1 process with λ = 0.6 λ = 0.6 lambda=0.6\lambda=0.6λ=0.6 and μ = 1.0 μ = 1.0 mu=1.0\mu=1.0μ=1.0 and find out average waiting time W i W i W_(i)\mathrm{W}_iWi by taking N = 10 N = 10 N=10\mathrm{N}=10N=10.

Answer:

Calculation of Average Waiting Time ( W i W i W_(i)\mathrm{W}_iWi) for an M/M/1 Queue

Parameters and Given Data

Parameters:
  • λ = 0.6 λ = 0.6 lambda=0.6\lambda = 0.6λ=0.6 (Arrival rate)
  • μ = 1.0 μ = 1.0 mu=1.0\mu = 1.0μ=1.0 (Service rate)
  • N = 10 N = 10 N=10\mathrm{N} = 10N=10 (Number of customers)
Given Data:
U 0.34 0.5 0.04 0.75 0.76 0.61 0.66 0.32 0.48 0.94 I = ( log U ) / λ 1.80 1.15 5.36 0.48 0.46 0.82 0.69 1.90 1.22 0.10 U 0.19 0.18 0.49 0.39 0.66 0.48 0.21 0.07 0.88 0.87 S = ( log U ) / μ 1.66 1.71 0.71 0.94 0.41 0.73 1.56 2.66 0.13 0.14 U 0.34 0.5 0.04 0.75 0.76 0.61 0.66 0.32 0.48 0.94 I = ( log U ) / λ 1.80 1.15 5.36 0.48 0.46 0.82 0.69 1.90 1.22 0.10 U 0.19 0.18 0.49 0.39 0.66 0.48 0.21 0.07 0.88 0.87 S = ( log U ) / μ 1.66 1.71 0.71 0.94 0.41 0.73 1.56 2.66 0.13 0.14 {:[U,0.34,0.5,0.04,0.75,0.76,0.61,0.66,0.32,0.48,0.94],[I=(-log U)//lambda,1.80,1.15,5.36,0.48,0.46,0.82,0.69,1.90,1.22,0.10],[U,0.19,0.18,0.49,0.39,0.66,0.48,0.21,0.07,0.88,0.87],[S=(-log U)//mu,1.66,1.71,0.71,0.94,0.41,0.73,1.56,2.66,0.13,0.14]:}\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \mathrm{U} & 0.34 & 0.5 & 0.04 & 0.75 & 0.76 & 0.61 & 0.66 & 0.32 & 0.48 & 0.94 \\ \hline \mathrm{I}=(-\log \mathrm{U}) / \lambda & 1.80 & 1.15 & 5.36 & 0.48 & 0.46 & 0.82 & 0.69 & 1.90 & 1.22 & 0.10 \\ \hline \mathrm{U} & 0.19 & 0.18 & 0.49 & 0.39 & 0.66 & 0.48 & 0.21 & 0.07 & 0.88 & 0.87 \\ \hline \mathrm{S}=(-\log \mathrm{U}) / \mu & 1.66 & 1.71 & 0.71 & 0.94 & 0.41 & 0.73 & 1.56 & 2.66 & 0.13 & 0.14 \\ \hline \end{array}U0.340.50.040.750.760.610.660.320.480.94I=(logU)/λ1.801.155.360.480.460.820.691.901.220.10U0.190.180.490.390.660.480.210.070.880.87S=(logU)/μ1.661.710.710.940.410.731.562.660.130.14

Calculation Formula

The waiting time for each customer ( W n W n W_(n)W_nWn) is calculated using the following formula:
W n = { W n 1 I n + S n if W n 1 > I n S n if W n 1 I n W n = W n 1 I n + S n if W n 1 > I n S n if W n 1 I n W_(n)={[W_(n-1)-I_(n)+S_(n),”if “W_(n-1) > I_(n)],[S_(n),”if “W_(n-1) <= I_(n)]:}W_n = \begin{cases} W_{n-1} – I_n + S_n & \text{if } W_{n-1} > I_n \\ S_n & \text{if } W_{n-1} \leq I_n \end{cases}Wn={Wn1In+Snif Wn1>InSnif Wn1In
with W 1 = S 1 W 1 = S 1 W_(1)=S_(1)W_1 = S_1W1=S1.

Calculation Steps

  1. Initialization:
    • Set the first waiting time: W 1 = S 1 = 1.66 W 1 = S 1 = 1.66 W_(1)=S_(1)=1.66W_1 = S_1 = 1.66W1=S1=1.66.
  2. Iterative Calculation:
    • For n = 2 n = 2 n=2n = 2n=2 to 10 10 101010, compute W n W n W_(n)W_nWn based on the previous waiting time and the given inter-arrival and service times.

Computation Results

Customer ( n n nnn) Inter-arrival Time ( I n I n I_(n)I_nIn) Service Time ( S n S n S_(n)S_nSn) Waiting Time ( W n W n W_(n)W_nWn)
1 1.66 1.66
2 1.15 1.71 2.22
3 5.36 0.71 0.71
4 0.48 0.94 1.17
5 0.46 0.41 1.12
6 0.82 0.73 1.03
7 0.69 1.56 1.90
8 1.90 2.66 2.66
9 1.22 0.13 1.57
10 0.10 0.14 1.61

Average Waiting Time

Average Waiting Time = n = 1 N W n N = 15.65 10 = 1.565 Average Waiting Time = n = 1 N W n N = 15.65 10 = 1.565 “Average Waiting Time”=(sum_(n=1)^(N)W_(n))/(N)=(15.65)/(10)=1.565\text{Average Waiting Time} = \frac{\sum_{n=1}^{N} W_n}{N} = \frac{15.65}{10} = 1.565Average Waiting Time=n=1NWnN=15.6510=1.565
Thus, the average waiting time W i W i W_(i)\mathrm{W}_iWi for the 10 customers is approximately 1.565 units of time.
Theoretical value = 1 / ( μ λ ) = 1 / 0.4 = 2.5 Theoretical value = 1 / ( μ λ ) = 1 / 0.4 = 2.5 ” Theoretical value “=1//(mu-lambda)=1//0.4=2.5\text { Theoretical value }=1 /(\mu-\lambda)=1 / 0.4=2.5 Theoretical value =1/(μλ)=1/0.4=2.5

Question:-07

7.A z 2 z 2 z^(2)\mathrm{z}^2z2-experiment was conducted in order to obtain an idea of the interaction between spacing (s) and number of seedlings per hole ( n n nnn ) along with the effects of different types of spacing and seedling per hole. The levels of two factors are : s : s :s: s:s (8" and 10 10 10^(”)10^{\prime \prime}10 spacing in between) and n n nnn (3 and 4 seedlings per hole).
The field plan and yield of dry Aman paddy (in k g k g kg\mathrm{kg}kg ) for each plot are given as follows :
Block Yield
1 ( l ) 117 ( l ) 117 (l)117(l) 117(l)117 ( s ) 106 ( s ) 106 (s)106(s) 106(s)106 ( n s ) 109 ( n s ) 109 (ns)109(n s) 109(ns)109 ( n ) 114 ( n ) 114 (n)114(n) 114(n)114
2 ( n s ) 114 ( n s ) 114 (ns)114(n s) 114(ns)114 ( l ) 120 ( l ) 120 (l)120(l) 120(l)120 ( s ) 117 ( s ) 117 (s)117(s) 117(s)117 ( n ) 114 ( n ) 114 (n)114(n) 114(n)114
3 ( l ) 111 ( l ) 111 (l)111(l) 111(l)111 ( n ) 117 ( n ) 117 (n)117(n) 117(n)117 ( s ) 114 ( s ) 114 (s)114(s) 114(s)114 ( n s ) 106 ( n s ) 106 (ns)106(n s) 106(ns)106
4 ( n s ) 93 ( n s ) 93 (ns)93(n s) 93(ns)93 ( n ) 121 ( n ) 121 (n)121(n) 121(n)121 ( s ) 112 ( s ) 112 (s)112(s) 112(s)112 ( l ) 108 ( l ) 108 (l)108(l) 108(l)108
5 ( n s ) 75 ( n s ) 75 (ns)75(n s) 75(ns)75 (s) 97 (l) 73 ( n ) 38 ( n ) 38 (n)38(n) 38(n)38
6 ( n ) 58 ( n ) 58 (n)58(n) 58(n)58 (l) 81 ( n s ) 105 ( n s ) 105 (ns)105(n s) 105(ns)105 ( s ) 117 ( s ) 117 (s)117(s) 117(s)117
Block Yield 1 (l)117 (s)106 (ns)109 (n)114 2 (ns)114 (l)120 (s)117 (n)114 3 (l)111 (n)117 (s)114 (ns)106 4 (ns)93 (n)121 (s)112 (l)108 5 (ns)75 (s) 97 (l) 73 (n)38 6 (n)58 (l) 81 (ns)105 (s)117| Block | Yield | | | | | :—: | :—: | :—: | :—: | :—: | | 1 | $(l) 117$ | $(s) 106$ | $(n s) 109$ | $(n) 114$ | | 2 | $(n s) 114$ | $(l) 120$ | $(s) 117$ | $(n) 114$ | | 3 | $(l) 111$ | $(n) 117$ | $(s) 114$ | $(n s) 106$ | | 4 | $(n s) 93$ | $(n) 121$ | $(s) 112$ | $(l) 108$ | | 5 | $(n s) 75$ | (s) 97 | (l) 73 | $(n) 38$ | | 6 | $(n) 58$ | (l) 81 | $(n s) 105$ | $(s) 117$ |
Analyse the gives design.

Answer:

Analysis of the z 2 z 2 z^(2)\mathrm{z}^2z2-Experiment on Spacing and Seedlings per Hole

Experiment Design

A z 2 z 2 z^(2)\mathrm{z}^2z2-experiment was conducted to analyze the interaction between spacing ( s s sss) and the number of seedlings per hole ( n n nnn), along with the effects of different types of spacing and the number of seedlings per hole. The levels of the two factors were:
  • Spacing ( s s sss): 8" and 10"
  • Number of seedlings per hole ( n n nnn): 3 and 4

Field Plan and Yield Data

The yield of dry Aman paddy (in kg) for each plot in the field plan is given below:
Block Yield 1 Yield 2 Yield 3 Yield 4
1 l ) 117 l ) 117 l)117l) 117l)117 s ) 106 s ) 106 s)106s) 106s)106 n s ) 109 n s ) 109 ns)109n s) 109ns)109 n ) 114 n ) 114 n)114n) 114n)114
2 n s ) 114 n s ) 114 ns)114n s) 114ns)114 l ) 120 l ) 120 l)120l) 120l)120 s ) 117 s ) 117 s)117s) 117s)117 n ) 114 n ) 114 n)114n) 114n)114
3 l ) 111 l ) 111 l)111l) 111l)111 n ) 117 n ) 117 n)117n) 117n)117 s ) 114 s ) 114 s)114s) 114s)114 n s ) 106 n s ) 106 ns)106n s) 106ns)106
4 n s ) 93 n s ) 93 ns)93n s) 93ns)93 n ) 121 n ) 121 n)121n) 121n)121 s ) 112 s ) 112 s)112s) 112s)112 l ) 108 l ) 108 l)108l) 108l)108
5 n s ) 75 n s ) 75 ns)75n s) 75ns)75 s ) 97 s ) 97 s)97s) 97s)97 l ) 73 l ) 73 l)73l) 73l)73 n ) 38 n ) 38 n)38n) 38n)38
6 n ) 58 n ) 58 n)58n) 58n)58 l ) 81 l ) 81 l)81l) 81l)81 n s ) 105 n s ) 105 ns)105n s) 105ns)105 s ) 117 s ) 117 s)117s) 117s)117

Yield Summaries and ANOVA Calculations

The analysis involves the following steps:
  1. Summarizing Data:
    • Total Yield ( x i x i sumx_(i)\sum x_ixi) and Mean ( x ¯ i x ¯ i bar(x)_(i)\bar{x}_ix¯i):
    Group x i x i sumx_(i)\sum x_ixi Mean x ¯ i x ¯ i bar(x)_(i)\bar{x}_ix¯i Std Dev S i S i S_(i)S_iSi
    A 604 100.6667 18.7474
    B 669 111.5 8.6429
    C 605 100.8333 14.838
    D 559 93.1667 35.6562
    Total 2437 101.5417
    • Sum of Squares ( x i 2 x i 2 sumx_(i)^(2)\sum x_i^2xi2):
    Group x i 2 x i 2 sumx_(i)^(2)\sum x_i^2xi2
    A 62560
    B 74967
    C 62105
    D 58437
    Total 258069
  2. ANOVA Calculations:
    • Sum of Squares Between Samples (SSB):
      SSB = ( T i 2 n i ) ( x ) 2 n = 248480.5 247457.0417 = 1023.4583 SSB = T i 2 n i ( x ) 2 n = 248480.5 247457.0417 = 1023.4583 “SSB”=(sum(T_(i)^(2))/(n_(i)))-((sum x)^(2))/(n)=248480.5-247457.0417=1023.4583\text{SSB} = \left(\sum \frac{T_i^2}{n_i}\right) – \frac{(\sum x)^2}{n} = 248480.5 – 247457.0417 = 1023.4583SSB=(Ti2ni)(x)2n=248480.5247457.0417=1023.4583
    • Sum of Squares Within Samples (SSW):
      SSW = x 2 ( T i 2 n i ) = 258069 248480.5 = 9588.5 SSW = x 2 T i 2 n i = 258069 248480.5 = 9588.5 “SSW”=sumx^(2)-(sum(T_(i)^(2))/(n_(i)))=258069-248480.5=9588.5\text{SSW} = \sum x^2 – \left(\sum \frac{T_i^2}{n_i}\right) = 258069 – 248480.5 = 9588.5SSW=x2(Ti2ni)=258069248480.5=9588.5
    • Total Sum of Squares (SST):
      SST = SSB + SSW = 1023.4583 + 9588.5 = 10611.9583 SST = SSB + SSW = 1023.4583 + 9588.5 = 10611.9583 “SST”=”SSB”+”SSW”=1023.4583+9588.5=10611.9583\text{SST} = \text{SSB} + \text{SSW} = 1023.4583 + 9588.5 = 10611.9583SST=SSB+SSW=1023.4583+9588.5=10611.9583
    • Mean Squares Between Samples (MSB):
      MSB = SSB k 1 = 1023.4583 3 = 341.1528 MSB = SSB k 1 = 1023.4583 3 = 341.1528 “MSB”=(“SSB”)/(k-1)=(1023.4583)/(3)=341.1528\text{MSB} = \frac{\text{SSB}}{k-1} = \frac{1023.4583}{3} = 341.1528MSB=SSBk1=1023.45833=341.1528
    • Mean Squares Within Samples (MSW):
      MSW = SSW n k = 9588.5 20 = 479.425 MSW = SSW n k = 9588.5 20 = 479.425 “MSW”=(“SSW”)/(n-k)=(9588.5)/(20)=479.425\text{MSW} = \frac{\text{SSW}}{n-k} = \frac{9588.5}{20} = 479.425MSW=SSWnk=9588.520=479.425
    • Test Statistic (F):
      F = MSB MSW = 341.1528 479.425 = 0.7116 F = MSB MSW = 341.1528 479.425 = 0.7116 F=(“MSB”)/(“MSW”)=(341.1528)/(479.425)=0.7116F = \frac{\text{MSB}}{\text{MSW}} = \frac{341.1528}{479.425} = 0.7116F=MSBMSW=341.1528479.425=0.7116
    • Degrees of Freedom:
      df between samples = k 1 = 3 df between samples = k 1 = 3 “df between samples”=k-1=3\text{df between samples} = k-1 = 3df between samples=k1=3
      df within samples = n k = 24 4 = 20 df within samples = n k = 24 4 = 20 “df within samples”=n-k=24-4=20\text{df within samples} = n-k = 24-4 = 20df within samples=nk=244=20
  3. P-value Calculation:
    p = F Dist ( F , d f 1 , d f 2 ) = F Dist ( 0.7116 , 3 , 20 ) = 0.5565 p = F Dist ( F , d f 1 , d f 2 ) = F Dist ( 0.7116 , 3 , 20 ) = 0.5565 p=F Dist(F,df_(1),df_(2))=F Dist(0.7116,3,20)=0.5565p = F \operatorname{Dist}(F, df_1, df_2) = F \operatorname{Dist}(0.7116, 3, 20) = 0.5565p=FDist(F,df1,df2)=FDist(0.7116,3,20)=0.5565

ANOVA Table

Source of Variation Sum of Squares (SS) df Mean Squares (MS) F p p ppp-value
Between samples 1023.4583 3 341.1528 0.7116 0.5565
Within samples 9588.5 20 479.425
Total 10611.9583 23

Conclusion

  • Null Hypothesis ( H 0 H 0 H_(0)H_0H0): There is no significant difference between samples.
  • Alternative Hypothesis ( H 1 H 1 H_(1)H_1H1): There is a significant difference between samples.
Since the calculated F F FFF-value (0.7116) is less than the critical F F FFF-value at the 0.05 significance level (3.0984), we fail to reject the null hypothesis. Hence, there is no significant difference between samples.

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