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Assig Testing nment 4 ANOVA TEST and comparing three variety of tomatoes for yield Three variety of tomatoes are being compared for yield. The experiment was carried out by assigning each variety at random to four different plots. One variety of potato in each of the 4 locations. Tomato type (3 levels) and location (4 levels). The plot is as follows Tomato Type P-1 P-2 P-3 YIELD Y-1 18 13 12 Y-2 20 23 21 Y-3 14 12 9 Y-4 11 17 10 Step 1: Define hypothesis H0: μ1 = μ2 = μ3 = μ4 the population means of the tomato types are equal. H1: μ1 ≠ μ2 ≠ μ3≠ μ At least one mean is different. WE can reject the null hypothesis and accept alternative hypothesis Step 2: Calculate means of rows and columns Tomato Type (B) P-1 P-2 P-3 Total Mean YIELD (A) Y-1 18 13 12 43 14.33 Y-2 20 23 21 64 21.33 Y-3 14 12 9 35 11.66 Y-4 11 17 10 38 12.66 Total 63 65 52 180 15 (Group Mean Mean 15.75 16.25 13 Step 3: Frame the ANOVA Summary table Step 4: Calculate the DF (degree of freedom) DF for yield Number of rows –1 = 4 –1 = 3 DF for tomato Number of columns –1 = 3 –1 = 2 DF total Total number of observations in dataset –1 = 12 –1 = 11 DF residual DF total – (DF yield + DF tomato) = 11 –5 = 6 Step 5: Calculate SS (sum of squares) SS yield = Variance (14.33, 21.33, 11.66, 12.66)2 x 12= 171.44 n = 3 (no. of columns) 15 = group mean 3(14.33-15)2+3(21.33-15)2+3(11.66-15)2+3(12.66-15)2 1.34+120.2+33.46+16.43=171.44 SS tomato = Variance (15.75, 16.25, 13)2 x 12= 24.5 n = 4 (no. of rows) 15 = group mean 4(15.75-15)2+4(16.25-15)2+4(13-15)2 2.25+6.25+16=24.5 SS total = Standard deviation2 x 11= 238 All variety of tomatoes (18,20,14,11,13,23,12,17,12,21,9,10) 15 = group mean SStotal = (18-15)2+(20-15)2+(14-15)2+(11-15)2+(13-15)2+(23-15)2+(12-15)2+(17-15)2+(12- 15)2+(21-15)2+(9-15)2+(10-15)2 = 9+25+1+16+4+64+9+4+9+36+36+25 =238 SS residual = SS total – (SS yield + SS tomato) = 238 – (171.44 + 24.5) = 42.39 Step 6: Calculate MS (mean square) The formula of MS is as follows MS = SS ÷ DF Therefore, MS for each row is calculated as follows. MS yield = SS yield ÷ DF yield = 171.44 ÷ 3 = 57.14 MS potato = SS tomato ÷ DF tomato = 24.5 ÷ 2 = 12.25 MS residual = SS residual ÷ DF residual = 42.17 ÷ 6 = 7.03 Step 7: Calculate F (F value) F yield = MS yield ÷ MS residual = 57.14 ÷ 7.03 = 8.12 F tomato = MS tomato ÷ MS residual = 12.55 ÷ 7.03 = 1.74 Step 8: Calculate F-critical values Check the critical F values in F table (p=0.05) with calculated DFs. F-critical = F 0.05, DF effect (F table column), DF residual (F table row) Therefore, F-critical yield = F0.05, 3, 6= 4.76 F-critical tomato = F0.05, 2, 6= 5.14 DF SS MS F F-critical P value Yield 3 171.44 57.14 8.12 4.76 (3,6) 0.04994 Tomato 2 24.5 12.25 1.74 5.14 (2,6) 0.05006 Residual 6 42.17 7.03 7.03 Total 11 238 F yield> F-critical yield (H1) Reject null hypothesis and accept alternative hypothesis. Yield has significant effect on the dependent variable. The result is significant at p < .05 F potato <F-critical tomato (H0) Accept null hypothesis and reject alternative hypothesis. Tomato has no significant effect on the dependent variable. The result is not significant at p < .05 Assig Testing nment 4 ANOVA TEST and comparing three variety of tomatoes for yield Three variety of tomatoes are being compared for yield. The experiment was carried out by assigning each variety at random to four different plots. One variety of potato in each of the 4 locations. Tomato type (3 levels) and location (4 levels). The plot is as follows Tomato Type P-1 P-2 P-3 YIELD Y-1 18 13 12 Y-2 20 23 21 Y-3 14 12 9 Y-4 11 17 10 Step 1: Define hypothesis H0: μ1 = μ2 = μ3 = μ4 the population means of the tomato types are equal. H1: μ1 ≠ μ2 ≠ μ3≠ μ At least one mean is different. WE can reject the null hypothesis and accept alternative hypothesis Step 2: Calculate means of rows and columns Tomato Type

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Assignment

Assignment 3 on Pearson’s correlation coefficient

TEST FOR SIGNIFICANCE OF PEARSON’S CORRELATION COEFFICIENT (r)

Using the distribution of business and mathematics scores of SS2 students of Nazarene International Secondary School below,

(i) Compute the Pearson’s Product Moment Correlation Coefficient (r),

(ii) Interpret your result, and

Business and mathematics scores of SS2 students of Rusinga Secondary School

business

9

12

13

15

11

11

10

12

13

9

17

mathematics

33

46

61

59

52

39

37

41

57

31

66

business (x)

mathematics (y)

xy

x2

y2

9

33

297

81

1089

12

46

552

144

2116

13

61

793

169

3721

15

59

885

225

3481

11

52

572

121

2704

11

39

429

121

1521

10

37

370

100

1369

12

41

492

144

1681

13

57

741

169

3249

9

31

249

81

961

17

66

1122

289

4356

Σx= 132

Σy= 522

Σxy= 6532

Σx2= 1644

Σy2= 26247

From our table:

Σx = 132

Σy = 522

Σxy = 6532

Σx2 = 1644

Σy2 = 26247

📷n is the sample size, in our case = 11

Substituting, we have r= 2948

R=0.90065

.

Second, the computed 0.96 was transformed to the t-distribution using the formula

📷

t = 0.9 📷

= 0.9 × 6.19

= 5.57

Third, the degrees of freedom (df) was found thus: df = (n – 2) = 11 – 2 = 9

A positive correlation coefficient of 0.96 implies that business and mathematics are positively related. In other words, a student who scores highly in business is likely to score highly in mathematics and vice versa

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Assignment 2 Chi square

JOB SATISFACTORY LEVEL

A survey of 200 workers was conducted regarding their education (school graduates or less, college graduates, university graduates) and the level of their job satisfaction (low, medium, and high). These are the results:

Low Medium High

School 20 35 25

College 17 33 20

University 11 18 21

We will test at a 2.5% level of significance whether the level of job satisfaction depends on the level of education.

Calculating Chi-square

∑ χ2 i-j= (O – E) 2/E

Where:

O = Observed (the actual count of cases in each cell of the table)

E = Expected value (calculated below)

χ2 = the cell Chi-square value

∑ χ2 = Formula instruction to sum all the cell Chi square values

χ2 i-j = i-j is the correct notation to represent all the cells, from the first cell (i) to the last cell (j); in this case Cell 1 (i) through Cell 9 (j).

The first step is calculating χ2, the sum of each row and sum of each column.

Low Medium High Row Sum

School 20 35 25 80

College 17 33 20 70

University 11 18 21 50

Row Column 48 86 66 200

N=200

Second step is calculating expected values.

E= MR × MC/n

Where:

E = represents the cell expected value,

MR = represents the row marginal for that cell,

MC = represents the column marginal for that cell,

n = represents the total sample size.

Specifically, for each cell, its row marginal is multiplied by its column marginal, and that product is divided by the sample size.

Cell 1 (80 x 48)/200= 19.2 Cell 5 (70 x 86)/200 = 30.1 Cell 9 (50 x 66)/200 = 16.5

Cell 2 (70 x 48)/200 = 16.8 Cell 6 (50 x 86)/200 = 21.5

Cell 3 (50 x 48)/200= 12 Cell 7 (80 x 66)/200 = 26.4

Cell 4 (80 x 86)/200= 34.4 Cell 8 (70 x 66)/200 = 23.1

Once the expected values have been calculated, the cell χ2 values are calculated with the following formula:

E= (O – E)2 /E

C1 χ2= (20-19.2)2/19.2 C4 χ2= (35-34.4)2/34.4 C7 χ2 = (25-26.4)2/26.4

= 0.03 = 0.01 = 0.07

C2 χ2= (17-16.8)2/16.8 C5 χ2 = (33-30.1)2/30.1 C8 χ2 = (20-23.1)2/23.1

= 0.0023 = 0.28 = 0.42

C3 χ2 = (11-12)2/12 C6 χ2 = (18-21.5)2/21.5 C9 χ2 = (21-16.5)2/16.5

= 0.08 = 0.57 = 1.23

Low Medium High Row Sum

School 19.2 (0.03) 34.4 (0.01) 26.4 (0.07) 80

College 16.8 (0.0023) 30.1 (0.28) 23.1 (0.42) 70

University 12 (0.08) 21.5 (0.57) 16.5 (1.23) 50

Row Column 48 86 66 200

Summed to obtain the χ2 statistic for the table

(0.03+0.0023+0.08+0.01+0.28+0.57+0.07+0.42+1.23)

χ2= 2.69

The degrees of freedom for a χ2 table are calculated with the formula:

(Number of rows -1) x (Number of columns -1).

3 x 3 table

(3-1) x (3-1)

df=4

• Given that p = 0.025 and df = 4, we see that the critical value 𝜒2 (4, 0.025) = 11.14 and thus our test statistic 𝜒 2 = 2.694 is in the region of acceptance.

• We can also see from the table that the p-value corresponding to our test statistic is between 0.5 and 0.75, and thus it is bigger than p.

• Therefore, we can state that:

Accept the null hypothesis (Ho) and

Reject the (H1) alternate hypothesis.

(Ho) The level of job satisfaction and the level of education are independent.

(H1) The level of job satisfaction and the level of education are not independent

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Assignment 2

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Assignment 1 ANOVA TEST

Testing and comparing three variety of tomatoes in the field

Three variety of tomatoes are being compared for yield. The experiment was carried out by assigning each variety at random to four different plots. One variety of tomato in each of the 4 locations. Tomato type (3 levels) and location (4 levels). The plot is as follows

Tomato Type

P-1 P-2 P-3

YIELD Y-1 18 13 12

Y-2 20 23 21

Y-3 14 12 9

Y-4 11 17 10

Step 1: Define hypothesis

H0: μ1 = μ2 = μ3 = μ4 the population means of the tomato types are equal.

H1: μ1 ≠ μ2 ≠ μ3≠ μ At least one mean is different.

WE can reject the null hypothesis and accept alternative hypothesis

Step 2: Calculate means of rows and columns

Tomato Type (B)

P-1 P-2 P-3 Total Mean

YIELD (A) Y-1 18 13 12 43 14.33

Y-2 20 23 21 64 21.33

Y-3 14 12 9 35 11.66

Y-4 11 17 10 38 12.66

Total 63 65 52 180 15 (Group Mean

Mean 15.75 16.25 13

Step 3: Frame the ANOVA Summary table

Step 4: Calculate the DF (degree of freedom)

DF for yield Number of rows –1 = 4 –1 = 3

DF for tomato Number of columns –1 = 3 –1 = 2

DF total Total number of observations in dataset –1 = 12 –1 = 11

DF residual DF total – (DF yield + DF tomato) = 11 –5 = 6

Step 5: Calculate SS (sum of squares)

SS yield = Variance (14.33, 21.33, 11.66, 12.66)2 x 12= 171.44

n = 3 (no. of columns)

15 = group mean

3(14.33-15)2+3(21.33-15)2+3(11.66-15)2+3(12.66-15)2

1.34+120.2+33.46+16.43=171.44

SS tomato = Variance (15.75, 16.25, 13)2 x 12= 24.5

n = 4 (no. of rows)

15 = group mean

4(15.75-15)2+4(16.25-15)2+4(13-15)2

2.25+6.25+16=24.5

SS total = Standard deviation2 x 11= 238

All variety of tomatoes (18,20,14,11,13,23,12,17,12,21,9,10)

15 = group mean

SStotal = (18-15)2+(20-15)2+(14-15)2+(11-15)2+(13-15)2+(23-15)2+(12-15)2+(17-15)2+(12- 15)2+(21-15)2+(9-15)2+(10-15)2

= 9+25+1+16+4+64+9+4+9+36+36+25

=238

SS residual = SS total – (SS yield + SS tomato) = 238 – (171.44 + 24.5) = 42.39

Step 6: Calculate MS (mean square)

The formula of MS is as follows

MS = SS ÷ DF

Therefore, MS for each row is calculated as follows.

MS yield = SS yield ÷ DF yield = 171.44 ÷ 3 = 57.14

MS tomato = SS tomato ÷ DF tomato = 24.5 ÷ 2 = 12.25

MS residual = SS residual ÷ DF residual = 42.17 ÷ 6 = 7.03

Step 7: Calculate F (F value)

F yield = MS yield ÷ MS residual = 57.14 ÷ 7.03 = 8.12

F potato = MS tomato ÷ MS residual = 12.55 ÷ 7.03 = 1.74

Step 8: Calculate F-critical values

Check the critical F values in F table (p=0.05) with calculated DFs.

F-critical = F 0.05, DF effect (F table column), DF residual (F table row)

Therefore,

F-critical yield = F0.05, 3, 6= 4.76

F-critical tomato = F0.05, 2, 6= 5.14

DF SS MS F F-critical P value

Yield 3 171.44 57.14 8.12 4.76 (3,6) 0.04994

Tomato 2 24.5 12.25 1.74 5.14 (2,6) 0.05006

Residual 6 42.17 7.03 7.03

Total 11 238

F yield> F-critical yield (H1)

Reject null hypothesis and accept alternative hypothesis. Yield has significant effect on the dependent variable. The result is significant at p < .05

F tomato <F-critical tomato

Accept null hypothesis and reject alternative hypothesis. Tomato has no significant effect on the dependent variable. The result is not significant at p < .05