Question

The following data refers to yield of tomatoes (kg/plot) for four different levels of salinity. Salinity level here refers to electrical conductivity (EC), where the chosen levels were EC = 1.6, 3.8, 6.0, and 10.2 nmhos/cm. (Use i = 1, 2, 3, and 4 respectively.) 1.6: 59.7 53.3 56.1 63.3 58.9 3.8: 55.1 59.7 52.8 54.6 6.0: 51.9 48.8 53.8 49.4 10.2: 44.2 48.6 40.9 47.6 46.3 Use the F test at level α = 0.05 to test for any differences in true average yield due to the different salinity levels. State the appropriate hypotheses. H0: μ1 ≠ μ2 ≠ μ3 ≠ μ4 Ha: all four μi's are equal H0: μ1 = μ2 = μ3 = μ4 Ha: all four μi's are unequal H0: μ1 = μ2 = μ3 = μ4 Ha: at least two μi's are unequal H0: μ1 ≠ μ2 ≠ μ3 ≠ μ4 Ha: at least two μi's are equal Correct: Your answer is correct. Calculate the test statistic. (Round your answer to two decimal places.) f = 3.34 Incorrect: Your answer is incorrect. What can be said about the P-value for the test? P-value > 0.100 0.050 < P-value < 0.100 0.010 < P-value < 0.050 0.001 < P-value < 0.010 P-value < 0.001 Incorrect: Your answer is incorrect. State the conclusion in the problem context. Reject H0. There is a difference in true average yield of tomatoes for the four different levels of salinity. Reject H0. There is no difference in true average yield of tomatoes for the four different levels of salinity. Fail to reject H0. There is no difference in true average yield of tomatoes for the four different levels of salinity. Fail to reject H0. There is a difference in true average yield of tomatoes for the four different levels of salinity. Incorrect: Your answer is incorrect. You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

k = Number of Groups = 4

Group	Number of data values in group	Sum of Data Values	Sample Mean	Sample Variance	Sample Standard Deviation
Group 1	n1 = 5	291.3	X1 = 58.260	s12 = 14.288	s1 = 3.78
Group 2	n2 = 4	222.2	X2 = 55.55	s22 = 8.630	s2 = 2.94
Group 3	n3 = 4	203.90	X3 = 50.975	s32 = 5.349	s3 = 2.31
Group 4	n4 = 5	227.60	X4 = 45.52	s42 = 9.377	s4 = 3.065
Total	N = 18	ΣX = 945

Step 1: State null and alternative hypotheses
H_o: μ₁= μ₂= μ₃= μ₄
H₁: At least one mean is different from the others

Step 2: Find the critical value
k = 4
N = total number of data values from all groups = 18
d.f.N. = k -1 = 3
d.f.D. = N -1 = 14
α = 0.05
Critical Value = 3.4 (Critical value is computed based on right-tailed area of 0.05.)

Step 3: Calculate F Test Value
ΣX = Sum of all data values = 291.3+222.2+203.90 +227.60 = 945
N = 18
Grand Mean = X_GM = 945/18 = 52.5
Calculate between-group variance, denoted by S_B2:
SS_B = 5(58.26 - 52.5)²+ 4(55.55 - 52.5)²+ 4(50.975 - 52.5)²+ 5(45.52 - 52.5)²
SS_B = 456.0025
S_B2 = SS_W/(k-1)

S_B2 = 456.0025/3 = 152.0008

Calculate within-group variance, denoted by S_W2:
SS_W = (5- 1)(14.288)+ (4- 1)(8.63)+ (4- 1)(5.349)+ (5- 1)(9.377)
SS_W = 136.5975
N - k = 18 - 4 = 14
S_W2 = SS_W/(N - k) = 136.5975/14 = 9.757

Calculate F test value:

F = S_B2/S_W2

F = 152.0008/9.757 = 15.5787

Step 4: Make Decision:
Compare F test value = 15.5787 with Critical Value = 3.344
The decision is to reject the null hypothesis since 15.5787 > 3.344

We can also check p value:

P value corresponding to F(3,14) = 15.5787 is 0.000097 (Obtaind using online p value calculator)

Now since p value (0.000097) < Level of significance (0.05), we reject null hypothesis.

Analysis of Variance Summary Table

Source	Sum of Squares	Degrees of Freedom (df)	Mean Square	F test value
Between	SSB = 456.0025	k - 1 = 3	MSB = 152.0008	F = 15.5787
Within (error)	SSW = 136.5975	N - k = 14	MSW = 9.757
Total	592.60	17

Based on above analysis following are answers for each part:

Appropriate hypotheses

H_o: μ₁= μ₂= μ₃= μ₄
H₁: At least one mean is different from the others

Test statistic (F) = 15.5787 = 15.58 (Rounded to 2 decimal places

P-value for the test = 0.000097. Thus, P-value < 0.001

Conclusion in the problem context:

Reject H0. There is a difference in true average yield of tomatoes for the four different levels of salinity.