BECS-184 Solved Assignment 2024 | DATA ANALYSIS | IGNOU

1. The following table presents the number of hours a group of school students played video games during the weekends and the test scores attained by each of them in a test, the following Monday.


\begin{equation}
\begin{array}{|c|c|}
\hline \text { Time (in hours) } & \text { Test score } \\
\hline 0 & 96 \\
\hline 1 & 85 \\
\hline 2 & 82 \\
\hline 3 & 74 \\
\hline 3 & 95 \\
\hline 5 & 68 \\
\hline 5 & 76 \\
\hline 5 & 84 \\
\hline 6 & 58 \\
\hline 7 & 65 \\
\hline 7 & 75 \\
\hline 10 & 50 \\
\hline
\end{array}
\end{equation}

 

(a.) It is believed that a linear relationship exists between the time spent on playing video games and test score attained. Find out the strength of this linear relationship.
(b.) Estimate the line of best fit in the scenario. Use this line to find the expected test score for a student who plays video games for 9 hours.
2. A study involves testing whether or not the amount of caffeine consumed affected memory. Fifteen volunteers took part in this study. They were given three types of drink (type A,B and C) containing different levels of caffeine \((50 \mathrm{mg}, 100 \mathrm{mg}\), and \(150 \mathrm{mg}\),

respectively). Volunteers were divided into three groups of five each and were assigned the drink groupwise. They were then given a memory test (In terms of number of words remembered from a list). The results are given in the following table:
\begin{equation}
\begin{array}{|c|c|c|}
\hline \text { Group A (50 mg) } & \text { Group B }(100 \mathrm{mg}) & \text { Group C }(150 \mathrm{mg}) \\
\hline 7 & 11 & 14 \\
\hline 8 & 14 & 12 \\
\hline 10 & 14 & 10 \\
\hline 12 & 12 & 16 \\
\hline 7 & 10 & 13 \\
\hline
\end{array}
\end{equation}At significance level of \(5 \%\), check whether the mean number of words remembered from the list by the participants belonging to the three groups are significantly different.
Assignment Two

Answer the following questions. Each question carries 12 marks.

3. a.) What could be a structured approach to multivariate model building?
b.) What are the various assumptions on which the multivariate regression analysis rests?
4. Explain the following:
a. ANOVA and MANOVA
a. Normal distribution curve
b. Snowball sampling techniques
c. Degrees of freedom
5. What is the difference between census and survey? Explain the various stages involved in planning and organizing the censuses and surveys.
6. Differentiate between quantitative and qualitative research in the context of data analysis. Discuss tools of data collection used in qualitative research.
7. Differentiate between:
a. Type I and type II errors
b. Phenomenology and Ethnography
c. \(t\) test and \(f\) test
d. discrete and continuous variable

Expert Answer
becs-184-solved-assignment-2024-4881b57e-e851-4d6d-bbd7-c9d76e6db204

becs-184-solved-assignment-2024-4881b57e-e851-4d6d-bbd7-c9d76e6db204

BECS-184 Solved Assignment 2024
Answer the following questions. Each question carries 20 marks
  1. The following table presents the number of hours a group of school students played video games during the weekends and the test scores attained by each of them in a test, the following Monday.
Time (in hours) Test score
0 96
1 85
2 82
3 74
3 95
5 68
5 76
5 84
6 58
7 65
7 75
10 50
Time (in hours) Test score 0 96 1 85 2 82 3 74 3 95 5 68 5 76 5 84 6 58 7 65 7 75 10 50| Time (in hours) | Test score | | :—: | :—: | | 0 | 96 | | 1 | 85 | | 2 | 82 | | 3 | 74 | | 3 | 95 | | 5 | 68 | | 5 | 76 | | 5 | 84 | | 6 | 58 | | 7 | 65 | | 7 | 75 | | 10 | 50 |
(a.) It is believed that a linear relationship exists between the time spent on playing video games and test score attained. Find out the strength of this linear relationship.
(b.) Estimate the line of best fit in the scenario. Use this line to find the expected test score for a student who plays video games for 9 hours.
Answer:

Part (a): Finding the Strength of the Linear Relationship

Step 1: Calculating the Means

First, we calculate the mean of the hours played ( x ¯ x ¯ bar(x)\bar{x}x¯) and the mean of the test scores ( y ¯ y ¯ bar(y)\bar{y}y¯).
  • Mean of hours played ( x ¯ x ¯ bar(x)\bar{x}x¯):
    x ¯ = ∑ x i n = 0 + 1 + 2 + 3 + 3 + 5 + 5 + 5 + 6 + 7 + 7 + 10 12 = 54 12 = 4.5 x ¯ = ∑ x i n = 0 + 1 + 2 + 3 + 3 + 5 + 5 + 5 + 6 + 7 + 7 + 10 12 = 54 12 = 4.5 bar(x)=(sumx_(i))/(n)=(0+1+2+3+3+5+5+5+6+7+7+10)/(12)=(54)/(12)=4.5\bar{x} = \frac{\sum x_i}{n} = \frac{0 + 1 + 2 + 3 + 3 + 5 + 5 + 5 + 6 + 7 + 7 + 10}{12} = \frac{54}{12} = 4.5x¯=∑xin=0+1+2+3+3+5+5+5+6+7+7+1012=5412=4.5
  • Mean of test scores ( y ¯ y ¯ bar(y)\bar{y}y¯):
    y ¯ = ∑ y i n = 96 + 85 + 82 + 74 + 95 + 68 + 76 + 84 + 58 + 65 + 75 + 50 12 = 908 12 = 75.6667 y ¯ = ∑ y i n = 96 + 85 + 82 + 74 + 95 + 68 + 76 + 84 + 58 + 65 + 75 + 50 12 = 908 12 = 75.6667 bar(y)=(sumy_(i))/(n)=(96+85+82+74+95+68+76+84+58+65+75+50)/(12)=(908)/(12)=75.6667\bar{y} = \frac{\sum y_i}{n} = \frac{96 + 85 + 82 + 74 + 95 + 68 + 76 + 84 + 58 + 65 + 75 + 50}{12} = \frac{908}{12} = 75.6667y¯=∑yin=96+85+82+74+95+68+76+84+58+65+75+5012=90812=75.6667

Step 2: Using Assumed Means

Since x ¯ = 4.5 x ¯ = 4.5 bar(x)=4.5\bar{x} = 4.5x¯=4.5 and y ¯ = 75.6667 y ¯ = 75.6667 bar(y)=75.6667\bar{y} = 75.6667y¯=75.6667 are not integers, we use assumed means A = 5 A = 5 A=5A = 5A=5 and B = 76 B = 76 B=76B = 76B=76, respectively.

Step 3: Calculating Deviations

We calculate the deviations from the assumed means ( d x = x − A d x = x − A dx=x-Ad x = x – Adx=x−A and d y = y − B d y = y − B dy=y-Bd y = y – Bdy=y−B) and their products and squares.
  • The table of deviations and their products and squares is as follows:
x y d x = x − A = x − 5 d y = y − B = y − 76 d x 2 d y 2 d x ⋅ d y 0 96 − 5 20 25 400 − 100 1 85 − 4 9 16 81 − 36 2 82 − 3 6 9 36 − 18 3 74 − 2 − 2 4 4 4 3 95 − 2 19 4 361 − 38 5 68 0 − 8 0 64 0 5 76 0 0 0 0 0 5 84 0 8 0 64 0 6 58 1 − 18 1 324 − 18 7 65 2 − 11 4 121 − 22 7 75 2 − 1 4 1 − 2 10 50 5 − 26 25 676 − 130 — — — − − — — — 54 908 ∑ d x = − 6 ∑ d y = − 4 ∑ d x 2 = 92 ∑ d y 2 = 2132 ∑ d x ⋅ d y = − 360 x y d x = x − A = x − 5 d y = y − B = y − 76 d x 2 d y 2 d x ⋅ d y 0 96 − 5 20 25 400 − 100 1 85 − 4 9 16 81 − 36 2 82 − 3 6 9 36 − 18 3 74 − 2 − 2 4 4 4 3 95 − 2 19 4 361 − 38 5 68 0 − 8 0 64 0 5 76 0 0 0 0 0 5 84 0 8 0 64 0 6 58 1 − 18 1 324 − 18 7 65 2 − 11 4 121 − 22 7 75 2 − 1 4 1 − 2 10 50 5 − 26 25 676 − 130 — — — − − — — — 54 908 ∑ d x = − 6 ∑ d y = − 4 ∑ d x 2 = 92 ∑ d y 2 = 2132 ∑ d x ⋅ d y = − 360 {:[x,y,dx=x-A=x-5,dy=y-B=y-76,dx^(2),dy^(2),dx*dy],[0,96,-5,20,25,400,-100],[1,85,-4,9,16,81,-36],[2,82,-3,6,9,36,-18],[3,74,-2,-2,4,4,4],[3,95,-2,19,4,361,-38],[5,68,0,-8,0,64,0],[5,76,0,0,0,0,0],[5,84,0,8,0,64,0],[6,58,1,-18,1,324,-18],[7,65,2,-11,4,121,-22],[7,75,2,-1,4,1,-2],[10,50,5,-26,25,676,-130],[“— “,”— “,”— “,–,”— “,”— “,”— “],[54,908,sum dx=-6,sum dy=-4,sum dx^(2)=92,sum dy^(2)=2132,sum dx*dy=-360]:}\begin{array}{|c|c|c|c|c|c|c|} \hline x & y & d x=x-A=x-5 & d y=y-B=y-76 & d x^2 & d y^2 & d x \cdot d y \\ \hline 0 & 96 & -5 & 20 & 25 & 400 & -100 \\ \hline 1 & 85 & -4 & 9 & 16 & 81 & -36 \\ \hline 2 & 82 & -3 & 6 & 9 & 36 & -18 \\ \hline 3 & 74 & -2 & -2 & 4 & 4 & 4 \\ \hline 3 & 95 & -2 & 19 & 4 & 361 & -38 \\ \hline 5 & 68 & 0 & -8 & 0 & 64 & 0 \\ \hline 5 & 76 & 0 & 0 & 0 & 0 & 0 \\ \hline 5 & 84 & 0 & 8 & 0 & 64 & 0 \\ \hline 6 & 58 & 1 & -18 & 1 & 324 & -18 \\ \hline 7 & 65 & 2 & -11 & 4 & 121 & -22 \\ \hline 7 & 75 & 2 & -1 & 4 & 1 & -2 \\ \hline 10 & 50 & 5 & -26 & 25 & 676 & -130 \\ \hline \text {— } & \text {— } & \text {— } & — & \text {— } & \text {— } & \text {— } \\ \hline 54 & 908 & \sum d x=-6 & \sum d y=-4 & \sum d x^2=92 & \sum d y^2=2132 & \sum d x \cdot d y=-360 \\ \hline \end{array}xydx=x−A=x−5dy=y−B=y−76dx2dy2dx⋅dy096−52025400−100185−491681−36282−36936−18374−2−2444395−2194361−385680−80640576000005840806406581−181324−187652−114121−227752−141−210505−2625676−130— — — −−— — — 54908∑dx=−6∑dy=−4∑dx2=92∑dy2=2132∑dx⋅dy=−360
After calculating, we get:
∑ d x = − 6 , ∑ d y = − 4 , ∑ d x 2 = 92 , ∑ d y 2 = 2132 , ∑ d x ⋅ d y = − 360 ∑ d x = − 6 , ∑ d y = − 4 , ∑ d x 2 = 92 , ∑ d y 2 = 2132 , ∑ d x ⋅ d y = − 360 sum dx=-6,quad sum dy=-4,quad sum dx^(2)=92,quad sum dy^(2)=2132,quad sum dx*dy=-360\sum d x = -6, \quad \sum d y = -4, \quad \sum d x^2 = 92, \quad \sum d y^2 = 2132, \quad \sum d x \cdot d y = -360∑dx=−6,∑dy=−4,∑dx2=92,∑dy2=2132,∑dx⋅dy=−360

Step 4: Calculating the Regression Coefficient

The regression coefficient ( b y x b y x b_(yx)b_{yx}byx) is calculated as follows:
b y x = n ∑ d x d y − ( ∑ d x ) ( ∑ d y ) n ∑ d x 2 − ( ∑ d x ) 2 b y x = n ∑ d x d y − ( ∑ d x ) ( ∑ d y ) n ∑ d x 2 − ( ∑ d x ) 2 b_(yx)=(n sum dxdy-(sum dx)(sum dy))/(n sum dx^(2)-(sum dx)^(2))b_{yx} = \frac{n \sum d x d y – (\sum d x)(\sum d y)}{n \sum d x^2 – (\sum d x)^2}byx=n∑dxdy−(∑dx)(∑dy)n∑dx2−(∑dx)2
Substituting the values:
b y x = 12 × − 360 − ( − 6 ) × − 4 12 × 92 − ( − 6 ) 2 = − 4320 − 24 1104 − 36 b y x = 12 × − 360 − ( − 6 ) × − 4 12 × 92 − ( − 6 ) 2 = − 4320 − 24 1104 − 36 b_(yx)=(12 xx-360-(-6)xx-4)/(12 xx92-(-6)^(2))=(-4320-24)/(1104-36)b_{yx} = \frac{12 \times -360 – (-6) \times -4}{12 \times 92 – (-6)^2} = \frac{-4320 – 24}{1104 – 36}byx=12×−360−(−6)×−412×92−(−6)2=−4320−241104−36
After calculating, we find:
b y x = − 4.0674 b y x = − 4.0674 b_(yx)=-4.0674b_{yx} = -4.0674byx=−4.0674

Part (b): Estimating the Line of Best Fit and Expected Test Score

Step 1: Formulating the Regression Line

The regression line of y y yyy on x x xxx is given by:
y − y ¯ = b y x ( x − x ¯ ) y − y ¯ = b y x ( x − x ¯ ) y- bar(y)=b_(yx)(x- bar(x))y – \bar{y} = b_{yx}(x – \bar{x})y−y¯=byx(x−x¯)
Substituting the means and the regression coefficient:
y − 75.6667 = − 4.0674 ( x − 4.5 ) y − 75.6667 = − 4.0674 ( x − 4.5 ) y-75.6667=-4.0674(x-4.5)y – 75.6667 = -4.0674(x – 4.5)y−75.6667=−4.0674(x−4.5)
Expanding and rearranging:
y = − 4.0674 x + 18.3034 + 75.6667 y = − 4.0674 x + 18.3034 + 75.6667 y=-4.0674 x+18.3034+75.6667y = -4.0674x + 18.3034 + 75.6667y=−4.0674x+18.3034+75.6667
Simplifying:
y = − 4.0674 x + 93.97 y = − 4.0674 x + 93.97 y=-4.0674 x+93.97y = -4.0674x + 93.97y=−4.0674x+93.97

Step 2: Estimating the Test Score for 9 Hours of Gameplay

Now, we estimate the test score ( y y yyy) for a student who plays video games for 9 hours ( x = 9 x = 9 x=9x = 9x=9):
y = − 4.0674 × 9 + 93.97 y = − 4.0674 × 9 + 93.97 y=-4.0674 xx9+93.97y = -4.0674 \times 9 + 93.97y=−4.0674×9+93.97
After calculating, we find:
y = 57.3633 y = 57.3633 y=57.3633y = 57.3633y=57.3633

Summary

  • The strength of the linear relationship between time spent on playing video games and test scores is represented by the regression coefficient b y x = − 4.0674 b y x = − 4.0674 b_(yx)=-4.0674b_{yx} = -4.0674byx=−4.0674.
  • The line of best fit is y = − 4.0674 x + 93.97 y = − 4.0674 x + 93.97 y=-4.0674 x+93.97y = -4.0674x + 93.97y=−4.0674x+93.97.
  • For a student playing video games for 9 hours, the expected test score is approximately 57.36.
  1. A study involves testing whether or not the amount of caffeine consumed affected memory. Fifteen volunteers took part in this study. They were given three types of drink (type A,B and C) containing different levels of caffeine ( 50 m g , 100 m g ( 50 m g , 100 m g (50mg,100mg(50 \mathrm{mg}, 100 \mathrm{mg}(50mg,100mg, and 150 m g 150 m g 150mg150 \mathrm{mg}150mg,respectively). Volunteers were divided into three groups of five each and were assigned the drink groupwise. They were then given a memory test (In terms of number of words remembered from a list). The results are given in the following table:
Group A (50 mg) Group B (100 mg) Group C (150 mg)
7 11 14
8 14 12
10 14 10
12 12 16
7 10 13
Group A (50 mg) Group B (100 mg) Group C (150 mg) 7 11 14 8 14 12 10 14 10 12 12 16 7 10 13| Group A (50 mg) | Group B (100 mg) | Group C (150 mg) | | :—: | :—: | :—: | | 7 | 11 | 14 | | 8 | 14 | 12 | | 10 | 14 | 10 | | 12 | 12 | 16 | | 7 | 10 | 13 |
At significance level of 5 % 5 % 5%5 \%5%, check whether the mean number of words remembered from the list by the participants belonging to the three groups are significantly different.
Answer:
To determine whether the mean number of words remembered by participants in the three groups (A, B, and C) are significantly different, we will conduct an ANOVA (Analysis of Variance) test. This test is appropriate when comparing the means of three or more groups.

Hypotheses Formulation

  • Null Hypothesis ( H 0 H 0 H_(0)H_0H0): The means of all three groups are equal ( μ A = μ B = μ C μ A = μ B = μ C mu _(A)=mu _(B)=mu _(C)\mu_A = \mu_B = \mu_CμA=μB=μC).
  • Alternative Hypothesis ( H 1 H 1 H_(1)H_1H1): At least one group mean is different.
A B C 7 11 14 8 14 12 10 14 10 12 12 16 7 10 13 ∑ A = 4 4 ∑ B = 6 1 ∑ C = 6 5 A 2 B 2 C 2 49 121 196 64 196 144 100 196 100 144 144 256 49 100 169 ∑ A 2 = 4 0 6 ∑ B 2 = 7 5 7 ∑ C 2 = 8 6 5 A B C 7 11 14 8 14 12 10 14 10 12 12 16 7 10 13 ∑ A = 4 4 ∑ B = 6 1 ∑ C = 6 5 A 2 B 2 C 2 49 121 196 64 196 144 100 196 100 144 144 256 49 100 169 ∑ A 2 = 4 0 6 ∑ B 2 = 7 5 7 ∑ C 2 = 8 6 5 {:[{:[A,B,C],[7,11,14],[8,14,12],[10,14,10],[12,12,16],[7,10,13],[sum A=44,sum B=61,sum C=65]:}],[{:[A^(2),B^(2),C^(2)],[49,121,196],[64,196,144],[100,196,100],[144,144,256],[49,100,169],[sumA^(2)=406,sumB^(2)=757,sumC^(2)=865]:}]:}\begin{aligned} &\begin{array}{|c|c|c|} \hline \boldsymbol{A} & \boldsymbol{B} & \boldsymbol{C} \\ \hline 7 & 11 & 14 \\ \hline 8 & 14 & 12 \\ \hline 10 & 14 & 10 \\ \hline 12 & 12 & 16 \\ \hline 7 & 10 & 13 \\ \hline \sum \boldsymbol{A}=\mathbf{4 4} & \sum \boldsymbol{B}=\mathbf{6 1} & \sum \boldsymbol{C}=\mathbf{6 5} \\ \hline \end{array}\\ &\begin{array}{|c|c|c|} \hline \boldsymbol{A}^{\mathbf{2}} & \boldsymbol{B}^{\mathbf{2}} & \boldsymbol{C}^{\mathbf{2}} \\ \hline 49 & 121 & 196 \\ \hline 64 & 196 & 144 \\ \hline 100 & 196 & 100 \\ \hline 144 & 144 & 256 \\ \hline 49 & 100 & 169 \\ \hline \sum \boldsymbol{A}^{\mathbf{2}}=\mathbf{4 0 6} & \sum \boldsymbol{B}^{\mathbf{2}}=\mathbf{7 5 7} & \sum \boldsymbol{C}^{\mathbf{2}}= \mathbf{8 6 5} \\ \hline \end{array} \end{aligned}ABC711148141210141012121671013∑A=44∑B=61∑C=65A2B2C2491211966419614410019610014414425649100169∑A2=406∑B2=757∑C2=865
Data table
Group A A AAA B B BBB C C CCC Total
N N N\mathrm{N}N n 1 = 5 n 1 = 5 n_(1)=5n_1=5n1=5 n 2 = 5 n 2 = 5 n_(2)=5n_2=5n2=5 n 3 = 5 n 3 = 5 n_(3)=5n_3=5n3=5 n = 15 n = 15 n=15n=15n=15
∑ x i ∑ x i sumx_(i)\sum x_i∑xi T 1 = ∑ x 1 = 44 T 1 = ∑ x 1 = 44 T_(1)=sumx_(1)=44T_1=\sum x_1=44T1=∑x1=44 T 2 = ∑ x 2 = 61 T 2 = ∑ x 2 = 61 T_(2)=sumx_(2)=61T_2=\sum x_2=61T2=∑x2=61 T 3 = ∑ x 3 = 65 T 3 = ∑ x 3 = 65 T_(3)=sumx_(3)=65T_3=\sum x_3=65T3=∑x3=65 ∑ x = 170 ∑ x = 170 sum x=170\sum x=170∑x=170
∑ x i 2 ∑ x i 2 sumx_(i)^(2)\sum x_i^2∑xi2 ∑ x 1 2 = 406 ∑ x 1 2 = 406 sumx_(1)^(2)=406\sum x_1^2=406∑x12=406 ∑ x 2 2 = 757 ∑ x 2 2 = 757 sumx_(2)^(2)=757\sum x_2^2=757∑x22=757 ∑ x 3 2 = 865 ∑ x 3 2 = 865 sumx_(3)^(2)=865\sum x_3^2=865∑x32=865 ∑ x 2 = 2028 ∑ x 2 = 2028 sumx^(2)=2028\sum x^2=2028∑x2=2028
Mean x ¯ i x ¯ i bar(x)_(i)\bar{x}_ix¯i x ¯ 1 = 8.8 x ¯ 1 = 8.8 bar(x)_(1)=8.8\bar{x}_1=8.8x¯1=8.8 x ¯ 2 = 12.2 x ¯ 2 = 12.2 bar(x)_(2)=12.2\bar{x}_2=12.2x¯2=12.2 x ¯ 3 = 13 x ¯ 3 = 13 bar(x)_(3)=13\bar{x}_3=13x¯3=13 Overall x ¯ = 11.3333 x ¯ = 11.3333 bar(x)=11.3333\bar{x}=11.3333x¯=11.3333
Std Dev S i S i S_(i)S_iSi S 1 = 2.1679 S 1 = 2.1679 S_(1)=2.1679S_1=2.1679S1=2.1679 S 2 = 1.7889 S 2 = 1.7889 S_(2)=1.7889S_2=1.7889S2=1.7889 S 3 = 2.2361 S 3 = 2.2361 S_(3)=2.2361S_3=2.2361S3=2.2361
Group A B C Total N n_(1)=5 n_(2)=5 n_(3)=5 n=15 sumx_(i) T_(1)=sumx_(1)=44 T_(2)=sumx_(2)=61 T_(3)=sumx_(3)=65 sum x=170 sumx_(i)^(2) sumx_(1)^(2)=406 sumx_(2)^(2)=757 sumx_(3)^(2)=865 sumx^(2)=2028 Mean bar(x)_(i) bar(x)_(1)=8.8 bar(x)_(2)=12.2 bar(x)_(3)=13 Overall bar(x)=11.3333 Std Dev S_(i) S_(1)=2.1679 S_(2)=1.7889 S_(3)=2.2361 | Group | $A$ | $B$ | $C$ | Total | | :—: | :—: | :—: | :—: | :—: | | $\mathrm{N}$ | $n_1=5$ | $n_2=5$ | $n_3=5$ | $n=15$ | | $\sum x_i$ | $T_1=\sum x_1=44$ | $T_2=\sum x_2=61$ | $T_3=\sum x_3=65$ | $\sum x=170$ | | $\sum x_i^2$ | $\sum x_1^2=406$ | $\sum x_2^2=757$ | $\sum x_3^2=865$ | $\sum x^2=2028$ | | Mean $\bar{x}_i$ | $\bar{x}_1=8.8$ | $\bar{x}_2=12.2$ | $\bar{x}_3=13$ | Overall $\bar{x}=11.3333$ | | Std Dev $S_i$ | $S_1=2.1679$ | $S_2=1.7889$ | $S_3=2.2361$ | |
Let k = k = k=\mathrm{k}=k= the number of different samples = 3 = 3 =3=3=3
n = n 1 + n 2 + n 3 = 5 + 5 + 5 = 15 n = n 1 + n 2 + n 3 = 5 + 5 + 5 = 15 n=n_(1)+n_(2)+n_(3)=5+5+5=15n=n_1+n_2+n_3=5+5+5=15n=n1+n2+n3=5+5+5=15
Overall x ¯ = 170 15 = 11.3333 x ¯ = 170 15 = 11.3333 bar(x)=(170)/(15)=11.3333\bar{x}=\frac{170}{15}=11.3333x¯=17015=11.3333
∑ x = T 1 + T 2 + T 3 = 44 + 61 + 65 = 170 → ( 1 ) ( ∑ x ) 2 n = 170 2 15 = 1926.6667 → ( 2 ) ∑ x = T 1 + T 2 + T 3 = 44 + 61 + 65 = 170 → ( 1 ) ∑ x 2 n = 170 2 15 = 1926.6667 → ( 2 ) {:[ sum x=T_(1)+T_(2)+T_(3)=44+61+65=170 rarr(1)],[((sum x)^(2))/(n)=(170^(2))/(15)=1926.6667 rarr(2)]:}\begin{aligned} & \sum x=T_1+T_2+T_3=44+61+65=170 \rightarrow(1) \\ & \frac{\left(\sum x\right)^2}{n}=\frac{170^2}{15}=1926.6667 \rightarrow(2) \end{aligned}∑x=T1+T2+T3=44+61+65=170→(1)(∑x)2n=170215=1926.6667→(2)
∑ T i 2 n i = ( 44 2 5 + 61 2 5 + 65 2 5 ) = 1976.4 → ( 3 ) ∑ x 2 = ∑ x 1 2 + ∑ x 2 2 + ∑ x 3 2 = 406 + 757 + 865 = 2028 → ( 4 ) ∑ T i 2 n i = 44 2 5 + 61 2 5 + 65 2 5 = 1976.4 → ( 3 ) ∑ x 2 = ∑ x 1 2 + ∑ x 2 2 + ∑ x 3 2 = 406 + 757 + 865 = 2028 → ( 4 ) {:[ sum(T_(i)^(2))/(n_(i))=((44^(2))/(5)+(61^(2))/(5)+(65^(2))/(5))=1976.4 rarr(3)],[ sumx^(2)=sumx_(1)^(2)+sumx_(2)^(2)+sumx_(3)^(2)=406+757+865=2028 rarr(4)]:}\begin{aligned} & \sum \frac{T_i^2}{n_i}=\left(\frac{44^2}{5}+\frac{61^2}{5}+\frac{65^2}{5}\right)=1976.4 \rightarrow(3) \\ & \sum x^2=\sum x_1^2+\sum x_2^2+\sum x_3^2=406+757+865=2028 \rightarrow(4)\end{aligned}∑Ti2ni=(4425+6125+6525)=1976.4→(3)∑x2=∑x12+∑x22+∑x32=406+757+865=2028→(4)
ANOVA:
Step-1: sum of squares between samples
SSB = ( ∑ T i 2 n i ) − ( ∑ x ) 2 n = ( 3 ) − ( 2 ) = 1976.4 − 1926.6667 = 49.7333 SSB = ∑ T i 2 n i − ∑ x 2 n = ( 3 ) − ( 2 ) = 1976.4 − 1926.6667 = 49.7333 {:[SSB=(sum(T_(i)^(2))/(n_(i)))-((sum x)^(2))/(n)=(3)-(2)],[=1976.4-1926.6667],[=49.7333]:}\begin{aligned} & \operatorname{SSB}=\left(\sum \frac{T_i^2}{n_i}\right)-\frac{\left(\sum x\right)^2}{n}=(3)-(2) \\ & =1976.4-1926.6667 \\ & =49.7333 \end{aligned}SSB=(∑Ti2ni)−(∑x)2n=(3)−(2)=1976.4−1926.6667=49.7333
Or
SSB = ∑ n j ⋅ ( x ¯ j − x ¯ ) 2 = 5 × ( 8.8 − 11.3333 ) 2 + 5 × ( 12.2 − 11.3333 ) 2 + 5 × ( 13 − 11.3333 ) 2 = 49.7333 SSB = ∑ n j ⋅ x ¯ j − x ¯ 2 = 5 × ( 8.8 − 11.3333 ) 2 + 5 × ( 12.2 − 11.3333 ) 2 + 5 × ( 13 − 11.3333 ) 2 = 49.7333 {:[SSB=sumn_(j)*( bar(x)_(j)-( bar(x)))^(2)],[=5xx(8.8-11.3333)^(2)+5xx(12.2-11.3333)^(2)+5xx(13-11.3333)^(2)],[=49.7333]:}\begin{aligned} & \operatorname{SSB}=\sum n_j \cdot\left(\bar{x}_j-\bar{x}\right)^2 \\ & =5 \times(8.8-11.3333)^2+5 \times(12.2-11.3333)^2+5 \times(13-11.3333)^2 \\ & =49.7333 \end{aligned}SSB=∑nj⋅(x¯j−x¯)2=5×(8.8−11.3333)2+5×(12.2−11.3333)2+5×(13−11.3333)2=49.7333
Step-2 : sum of squares within samples
SSW = ∑ x 2 − ( ∑ T i 2 n i ) = ( 4 ) − ( 3 ) = 2028 − 1976.4 = 51.6 SSW = ∑ x 2 − ∑ T i 2 n i = ( 4 ) − ( 3 ) = 2028 − 1976.4 = 51.6 {:[SSW=sumx^(2)-(sum(T_(i)^(2))/(n_(i)))=(4)-(3)],[=2028-1976.4],[=51.6]:}\begin{aligned} & \operatorname{SSW}=\sum x^2-\left(\sum \frac{T_i^2}{n_i}\right)=(4)-(3) \\ & =2028-1976.4 \\ & =51.6 \end{aligned}SSW=∑x2−(∑Ti2ni)=(4)−(3)=2028−1976.4=51.6
Step-3 : Total sum of squares