MST-005 Dec 2023

MST-005 Dec 2023

Answer:

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:

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.

Answer:

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:

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.

Answer:

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}_{total}$$d{f}_{total}$df_(total)df_{total}$d{f}_{total}$): This is the total number of observations minus one.$$d{f}_{total}=N-1$$$$d{f}_{total}=N-1$$df_(total)=N-1df_{total} = N – 1$$d{f}_{total}=N-1$$where $N$$N$NN$N$ is the total number of observations.
   • Between-Groups Degrees of Freedom ($d{f}_{between}$$d{f}_{between}$df_(between)df_{between}$d{f}_{between}$): This is the number of groups minus one.$$d{f}_{between}=k-1$$$$d{f}_{between}=k-1$$df_(between)=k-1df_{between} = k – 1$$d{f}_{between}=k-1$$where $k$$k$kk$k$ is the number of groups or levels.
   • Within-Groups (Error) Degrees of Freedom ($d{f}_{within}$$d{f}_{within}$df_(within)df_{within}$d{f}_{within}$): This is the total degrees of freedom minus the between-groups degrees of freedom.$$d{f}_{within}=d{f}_{total}-d{f}_{between}$$$$d{f}_{within}=d{f}_{total}-d{f}_{between}$$df_(within)=df_(total)-df_(between)df_{within} = df_{total} – df_{between}$$d{f}_{within}=d{f}_{total}-d{f}_{between}$$

Given Data:

• Levels of the factor ($k$$k$kk$k$): 4
• Total observations ($N$$N$NN$N$): 20

Calculation:

1. Total Degrees of Freedom:
   $$d{f}_{total}=N-1=20-1=19$$$$d{f}_{total}=N-1=20-1=19$$df_(total)=N-1=20-1=19df_{total} = N – 1 = 20 – 1 = 19$$d{f}_{total}=N-1=20-1=19$$
2. Between-Groups Degrees of Freedom:
   $$d{f}_{between}=k-1=4-1=3$$$$d{f}_{between}=k-1=4-1=3$$df_(between)=k-1=4-1=3df_{between} = k – 1 = 4 – 1 = 3$$d{f}_{between}=k-1=4-1=3$$
3. Within-Groups (Error) Degrees of Freedom:
   $$d{f}_{within}=d{f}_{total}-d{f}_{between}=19-3=16$$$$d{f}_{within}=d{f}_{total}-d{f}_{between}=19-3=16$$df_(within)=df_(total)-df_(between)=19-3=16df_{within} = df_{total} – df_{between} = 19 – 3 = 16$$d{f}_{within}=d{f}_{total}-d{f}_{between}=19-3=16$$

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.

Answer:

Latin Square Design:

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

Given Data:

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

Calculation:

1. Total Degrees of Freedom:
   $$d{f}_{total}=n^2-1=4^2-1=16-1=15$$$$d{f}_{total}=n^2-1=4^2-1=16-1=15$$df_(total)=n^(2)-1=4^(2)-1=16-1=15df_{total} = n^2 – 1 = 4^2 – 1 = 16 – 1 = 15$$d{f}_{total}=n^2-1=4^2-1=16-1=15$$
2. Degrees of Freedom for Rows:
   $$d{f}_{rows}=n-1=4-1=3$$$$d{f}_{rows}=n-1=4-1=3$$df_(rows)=n-1=4-1=3df_{rows} = n – 1 = 4 – 1 = 3$$d{f}_{rows}=n-1=4-1=3$$
3. Degrees of Freedom for Columns:
   $$d{f}_{columns}=n-1=4-1=3$$$$d{f}_{columns}=n-1=4-1=3$$df_(columns)=n-1=4-1=3df_{columns} = n – 1 = 4 – 1 = 3$$d{f}_{columns}=n-1=4-1=3$$
4. Degrees of Freedom for Treatments:
   $$d{f}_{treatments}=n-1=4-1=3$$$$d{f}_{treatments}=n-1=4-1=3$$df_(treatments)=n-1=4-1=3df_{treatments} = n – 1 = 4 – 1 = 3$$d{f}_{treatments}=n-1=4-1=3$$
5. Error Degrees of Freedom (without missing value):
   $$d{f}_{error}=(n-1)(n-2)=(4-1)(4-2)=3\times 2=6$$$$d{f}_{error}=(n-1)(n-2)=(4-1)(4-2)=3\times 2=6$$df_(error)=(n-1)(n-2)=(4-1)(4-2)=3xx2=6df_{error} = (n-1)(n-2) = (4-1)(4-2) = 3 \times 2 = 6$$d{f}_{error}=(n-1)(n-2)=(4-1)(4-2)=3\times 2=6$$

Effect of Missing Value:

6. Adjusted Error Degrees of Freedom:$$d{f}_{error}=6-1=5$$$$d{f}_{error}=6-1=5$$df_(error)=6-1=5df_{error} = 6 – 1 = 5$$d{f}_{error}=6-1=5$$

Conclusion:

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

Answer:

Middle Square Method:

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:

1. Square the Seed:
   $${15}^2=225$$$${15}^2=225$$15^(2)=22515^2 = 225$${15}^2=225$$
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).

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).

Answer:

Step-by-Step Process:

1. Determine the Sampling Interval ($k$$k$kk$k$):
   • The population size ($N$$N$NN$N$) is 30.
   • The sample size ($n$$n$nn$n$) is 10.
   • The sampling interval ($k$$k$kk$k$) is calculated as:$$k=\frac{N}{n}=\frac{30}{10}=3$$$$k=\frac{N}{n}=\frac{30}{10}=3$$k=(N)/(n)=(30)/(10)=3k = \frac{N}{n} = \frac{30}{10} = 3$$k=\frac{N}{n}=\frac{30}{10}=3$$So, we will select every 3rd book.
2. Identify the Possible Starting Points:
   • The starting point must be within the first $k$$k$kk$k$ 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:

Answer:

Part (i) Proportional Allocation Without Replacement:

• Total number of plots, $N=1000$$N=1000$N=1000N = 1000$N=1000$
• Sample size, $n=100$$n=100$n=100n = 100$n=100$

• $N_1=300$$N_1=300$N_(1)=300N_1 = 300$N_1=300$
• $N_2=500$$N_2=500$N_(2)=500N_2 = 500$N_2=500$
• $N_3=200$$N_3=200$N_(3)=200N_3 = 200$N_3=200$

Calculations:

• Stratum 1: $n_1=30$$n_1=30$n_(1)=30n_1 = 30$n_1=30$
• Stratum 2: $n_2=50$$n_2=50$n_(2)=50n_2 = 50$n_2=50$
• Stratum 3: $n_3=20$$n_3=20$n_(3)=20n_3 = 20$n_3=20$

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

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

• ${\overline{Y}}_1=15$${\overline{Y}}_1=15$bar(Y)_(1)=15\overline{Y}_1 = 15${\overline{Y}}_1=15$
• ${\overline{Y}}_2=16$${\overline{Y}}_2=16$bar(Y)_(2)=16\overline{Y}_2 = 16${\overline{Y}}_2=16$
• ${\overline{Y}}_3=24$${\overline{Y}}_3=24$bar(Y)_(3)=24\overline{Y}_3 = 24${\overline{Y}}_3=24$

Calculations:

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

• $S_1=8$$S_1=8$S_(1)=8S_1 = 8$S_1=8$
• $S_2=12$$S_2=12$S_(2)=12S_2 = 12$S_2=12$
• $S_3=18$$S_3=18$S_(3)=18S_3 = 18$S_3=18$

Calculations:

1. For Stratum 1:
   $${300}^2(\frac{64}{30}-\frac{64}{300})=90000(2.1333-0.2133)=90000\times 1.92=172800$$$${300}^2\left(\frac{64}{30}-\frac{64}{300}\right)=90000\left(2.1333-0.2133\right)=90000\times 1.92=172800$$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 = 172800$${300}^2(\frac{64}{30}-\frac{64}{300})=90000(2.1333-0.2133)=90000\times 1.92=172800$$
2. For Stratum 2:
   $${500}^2(\frac{144}{50}-\frac{144}{500})=250000(2.88-0.288)=250000\times 2.592=648000$$$${500}^2\left(\frac{144}{50}-\frac{144}{500}\right)=250000\left(2.88-0.288\right)=250000\times 2.592=648000$$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 = 648000$${500}^2(\frac{144}{50}-\frac{144}{500})=250000(2.88-0.288)=250000\times 2.592=648000$$
3. For Stratum 3:
   $${200}^2(\frac{324}{20}-\frac{324}{200})=40000(16.2-1.62)=40000\times 14.58=583200$$$${200}^2\left(\frac{324}{20}-\frac{324}{200}\right)=40000\left(16.2-1.62\right)=40000\times 14.58=583200$$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 = 583200$${200}^2(\frac{324}{20}-\frac{324}{200})=40000(16.2-1.62)=40000\times 14.58=583200$$

Summary:

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

Answer:

Two-Stage Sampling: Explanation and Example

Definition:

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:

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.

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:

• 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

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

• $\left(\frac{10}{500}\right)\times \left(\frac{20}{100}\right)=\frac{1}{50}\times \frac{1}{5}=\frac{1}{250}$$\left(\frac{10}{500}\right)\times \left(\frac{20}{100}\right)=\frac{1}{50}\times \frac{1}{5}=\frac{1}{250}$((10)/(500))xx((20)/(100))=(1)/(50)xx(1)/(5)=(1)/(250)\left(\frac{10}{500}\right) \times \left(\frac{20}{100}\right) = \frac{1}{50} \times \frac{1}{5} = \frac{1}{250}$\left(\frac{10}{500}\right)\times \left(\frac{20}{100}\right)=\frac{1}{50}\times \frac{1}{5}=\frac{1}{250}$

Answer:

Step-by-Step Process:

1. State the Hypotheses:
   • $H_0$$H_0$H_(0)H_0$H_0$: The mean number of items produced is the same across all three units (${\mu }_1={\mu }_2={\mu }_3$${\mu }_1={\mu }_2={\mu }_3$mu_(1)=mu_(2)=mu_(3)\mu_1 = \mu_2 = \mu_3${\mu }_1={\mu }_2={\mu }_3$).
   • $H_1$$H_1$H_(1)H_1$H_1$: At least one unit’s mean number of items produced is different from the others.
2. Collect the Data:

3. Calculate the Group Means and the Grand Mean:

4. Calculate the Sum of Squares Between Groups (SSB):

5. Calculate the Sum of Squares Within Groups (SSW):

6. Calculate the Degrees of Freedom:
   • Between groups (df${}_{\text{between}}$${}_{\text{between}}$_(“between”)_\text{between}${}_{\text{between}}$): $k-1=3-1=2$$k-1=3-1=2$k-1=3-1=2k – 1 = 3 – 1 = 2$k-1=3-1=2$
   • Within groups (df${}_{\text{within}}$${}_{\text{within}}$_(“within”)_\text{within}${}_{\text{within}}$): $N-k=15-3=12$$N-k=15-3=12$N-k=15-3=12N – k = 15 – 3 = 12$N-k=15-3=12$
   • Total degrees of freedom: $N-1=15-1=14$$N-1=15-1=14$N-1=15-1=14N – 1 = 15 – 1 = 14$N-1=15-1=14$
7. Calculate the Mean Squares:
   • Mean Square Between Groups (MSB): $\frac{SSB}{d{f}_{\text{between}}}=\frac{130.25}{2}=65.125$$\frac{SSB}{d{f}_{\text{between}}}=\frac{130.25}{2}=65.125$(SSB)/(df_(“between”))=(130.25)/(2)=65.125\frac{SSB}{df_\text{between}} = \frac{130.25}{2} = 65.125$\frac{SSB}{d{f}_{\text{between}}}=\frac{130.25}{2}=65.125$
   • Mean Square Within Groups (MSW): $\frac{SSW}{d{f}_{\text{within}}}=\frac{19.6}{12}=1.6333$$\frac{SSW}{d{f}_{\text{within}}}=\frac{19.6}{12}=1.6333$(SSW)/(df_(“within”))=(19.6)/(12)=1.6333\frac{SSW}{df_\text{within}} = \frac{19.6}{12} = 1.6333$\frac{SSW}{d{f}_{\text{within}}}=\frac{19.6}{12}=1.6333$
8. Calculate the F-Statistic:

9. Determine the Critical Value:
   • Using an F-distribution table, find the critical value for $d{f}_1=2$$d{f}_1=2$df_(1)=2df_1 = 2$d{f}_1=2$ and $d{f}_2=12$$d{f}_2=12$df_(2)=12df_2 = 12$d{f}_2=12$ at the 5% significance level. The critical value $F_{0.05,2,12}$$F_{0.05,2,12}$F_(0.05,2,12)F_{0.05, 2, 12}$F_{0.05,2,12}$ is approximately 3.885.
10. Decision:
    • Compare the calculated F-statistic (39.86) with the critical value (3.885).
    • Since 39.86 > 3.885, we reject the null hypothesis $H_0$$H_0$H_(0)H_0$H_0$.

Conclusion:

Answer:

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

Data:

Calculations for ANOVA:

1. Group Means and Grand Mean:
   • ${\overline{X}}_A=\frac{236}{4}=59$${\overline{X}}_A=\frac{236}{4}=59$bar(X)_(A)=(236)/(4)=59\overline{X}_A = \frac{236}{4} = 59${\overline{X}}_A=\frac{236}{4}=59$
   • ${\overline{X}}_B=\frac{257}{4}=64.25$${\overline{X}}_B=\frac{257}{4}=64.25$bar(X)_(B)=(257)/(4)=64.25\overline{X}_B = \frac{257}{4} = 64.25${\overline{X}}_B=\frac{257}{4}=64.25$
   • ${\overline{X}}_C=\frac{201}{4}=50.25$${\overline{X}}_C=\frac{201}{4}=50.25$bar(X)_(C)=(201)/(4)=50.25\overline{X}_C = \frac{201}{4} = 50.25${\overline{X}}_C=\frac{201}{4}=50.25$
   • ${\overline{X}}_D=\frac{241}{4}=60.25$${\overline{X}}_D=\frac{241}{4}=60.25$bar(X)_(D)=(241)/(4)=60.25\overline{X}_D = \frac{241}{4} = 60.25${\overline{X}}_D=\frac{241}{4}=60.25$
   • Grand Mean $\overline{X}=\frac{935}{16}=58.4375$$\overline{X}=\frac{935}{16}=58.4375$bar(X)=(935)/(16)=58.4375\overline{X} = \frac{935}{16} = 58.4375$\overline{X}=\frac{935}{16}=58.4375$
2. Sum of Squares:
   • $\text{Total Sum of Squares (SST)}=\sum (X_{ij}-\overline{X}{)}^2$$\text{Total Sum of Squares (SST)}=\sum (X_{ij}-\overline{X}{)}^2$“Total Sum of Squares (SST)”=sum(X_(ij)- bar(X))^(2)\text{Total Sum of Squares (SST)} = \sum (X_{ij} – \overline{X})^2$\text{Total Sum of Squares (SST)}=\sum (X_{ij}-\overline{X}{)}^2$
   • $\text{Sum of Squares Between Groups (SSB)}=\sum n_i({\overline{X}}_i-\overline{X}{)}^2$$\text{Sum of Squares Between Groups (SSB)}=\sum n_i({\overline{X}}_i-\overline{X}{)}^2$“Sum of Squares Between Groups (SSB)”=sumn_(i)( bar(X)_(i)- bar(X))^(2)\text{Sum of Squares Between Groups (SSB)} = \sum n_i (\overline{X}_i – \overline{X})^2$\text{Sum of Squares Between Groups (SSB)}=\sum n_i({\overline{X}}_i-\overline{X}{)}^2$
   • $\text{Sum of Squares Within Groups (SSW)}=\sum (X_{ij}-{\overline{X}}_i{)}^2$$\text{Sum of Squares Within Groups (SSW)}=\sum (X_{ij}-{\overline{X}}_i{)}^2$“Sum of Squares Within Groups (SSW)”=sum(X_(ij)- bar(X)_(i))^(2)\text{Sum of Squares Within Groups (SSW)} = \sum (X_{ij} – \overline{X}_i)^2$\text{Sum of Squares Within Groups (SSW)}=\sum (X_{ij}-{\overline{X}}_i{)}^2$
   Calculations:
   $$\begin{array}{rl}& \sum X=T_1+T_2+T_3+T_4=236+257+201+241=935\\ & \frac{(\sum X{)}^2}{n}=\frac{{935}^2}{16}=54639.0625\\ & \sum \frac{T_i^2}{n_i}=(\frac{{236}^2}{4}+\frac{{257}^2}{4}+\frac{{201}^2}{4}+\frac{{241}^2}{4})=55056.75\\ & \sum X^2=14418+16711+10691+15319=57139\end{array}$$$$\begin{array}{rl}& \sum X=T_1+T_2+T_3+T_4=236+257+201+241=935\\ & \frac{(\sum X{)}^2}{n}=\frac{{935}^2}{16}=54639.0625\\ & \sum \frac{T_i^2}{n_i}=\left(\frac{{236}^2}{4}+\frac{{257}^2}{4}+\frac{{201}^2}{4}+\frac{{241}^2}{4}\right)=55056.75\\ & \sum X^2=14418+16711+10691+15319=57139\end{array}$${:[ 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}$$\begin{array}{rl}& \sum X=T_1+T_2+T_3+T_4=236+257+201+241=935\\ & \frac{(\sum X{)}^2}{n}=\frac{{935}^2}{16}=54639.0625\\ & \sum \frac{T_i^2}{n_i}=(\frac{{236}^2}{4}+\frac{{257}^2}{4}+\frac{{201}^2}{4}+\frac{{241}^2}{4})=55056.75\\ & \sum X^2=14418+16711+10691+15319=57139\end{array}$$
   $$\text{SSB}=55056.75-54639.0625=417.6875$$$$\text{SSB}=55056.75-54639.0625=417.6875$$“SSB”=55056.75-54639.0625=417.6875\text{SSB} = 55056.75 – 54639.0625 = 417.6875$$\text{SSB}=55056.75-54639.0625=417.6875$$
   $$\text{SSW}=57139-55056.75=2082.25$$$$\text{SSW}=57139-55056.75=2082.25$$“SSW”=57139-55056.75=2082.25\text{SSW} = 57139 – 55056.75 = 2082.25$$\text{SSW}=57139-55056.75=2082.25$$
   $$\text{SST}=\text{SSB}+\text{SSW}=417.6875+2082.25=2499.9375$$$$\text{SST}=\text{SSB}+\text{SSW}=417.6875+2082.25=2499.9375$$“SST”=”SSB”+”SSW”=417.6875+2082.25=2499.9375\text{SST} = \text{SSB} + \text{SSW} = 417.6875 + 2082.25 = 2499.9375$$\text{SST}=\text{SSB}+\text{SSW}=417.6875+2082.25=2499.9375$$
3. Mean Squares:
   • $\text{MSB}=\frac{\text{SSB}}{k-1}=\frac{417.6875}{3}=139.2292$$\text{MSB}=\frac{\text{SSB}}{k-1}=\frac{417.6875}{3}=139.2292$“MSB”=(“SSB”)/(k-1)=(417.6875)/(3)=139.2292\text{MSB} = \frac{\text{SSB}}{k-1} = \frac{417.6875}{3} = 139.2292$\text{MSB}=\frac{\text{SSB}}{k-1}=\frac{417.6875}{3}=139.2292$
   • $\text{MSW}=\frac{\text{SSW}}{n-k}=\frac{2082.25}{12}=173.5208$$\text{MSW}=\frac{\text{SSW}}{n-k}=\frac{2082.25}{12}=173.5208$“MSW”=(“SSW”)/(n-k)=(2082.25)/(12)=173.5208\text{MSW} = \frac{\text{SSW}}{n-k} = \frac{2082.25}{12} = 173.5208$\text{MSW}=\frac{\text{SSW}}{n-k}=\frac{2082.25}{12}=173.5208$
4. F-Statistic:
   $$F=\frac{\text{MSB}}{\text{MSW}}=\frac{139.2292}{173.5208}=0.8024$$$$F=\frac{\text{MSB}}{\text{MSW}}=\frac{139.2292}{173.5208}=0.8024$$F=(“MSB”)/(“MSW”)=(139.2292)/(173.5208)=0.8024F = \frac{\text{MSB}}{\text{MSW}} = \frac{139.2292}{173.5208} = 0.8024$$F=\frac{\text{MSB}}{\text{MSW}}=\frac{139.2292}{173.5208}=0.8024$$
5. Degrees of Freedom:
   • Between Groups: $k-1=3$$k-1=3$k-1=3k-1 = 3$k-1=3$
   • Within Groups: $n-k=12$$n-k=12$n-k=12n-k = 12$n-k=12$
6. P-value:
   $$p=F\text{Dist}(0.8024,3,12)=0.5162$$$$p=F\text{Dist}(0.8024,3,12)=0.5162$$p=F”Dist”(0.8024,3,12)=0.5162p = F\text{Dist}(0.8024, 3, 12) = 0.5162$$p=F\text{Dist}(0.8024,3,12)=0.5162$$