Question

Question Help The data found below measure the amounts of greenhouse gas emissions from three types of vehicles. The measurements are in tons per year expressed as CO2 equivalents Use a 005 significance level to test the claim that the different types of vehicle have the same mean amount of greenhouse gas emissions. Based on the results, does the type of vehicle appear to affect the amount of greenhouse gas emissions? III Click the icon to view the data What are the hypotheses for this test? O A. Ho: At least one of the means is different from the others. H₂ H hh O B. Ho H11 Hq : H H *H3 OC. HOE H2H H₂: H = =43 OD. HO = HH3 H, At least one of the means is different from the others Determine the test statistic F= (Round to two decimal places as needed.) Identify the P-value P-value = 0.000 (Round to two decimal places as needed) Click to select your answer(s).

Answer 1

Solution:

k = Number of Groups = 3

Group	Number of data values in group	Sum of Data Values	Sample Mean	Sample Variance	Sample Standard Deviation
Group 1	n1 = 10	64.3	X1 = 6.43	s12 = 0.446	s1 = 0.6684
Group 2	n2 = 8	66	X2 = 8.25	s22 = 0.179	s2 = 0.4242
Group 3	n3 = 9	76.7	X3 = 8.52	s32 = 0.0919	s3 = 0.3032
Total	N = 27	ΣX = 207

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


Step 2: Find the critical value
k = 3
N = total number of data values from all groups = 27
d.f.N. = k -1 = 2
d.f.D. = N -1 = 24
α = 0.05
Critical Value = 3.402826105
(Critical value is computed based on right-tailed area of 0.05.)


Step 3: Calculate F Test Value
ΣX = Sum of all data values = 64.3+66+76.7 = 207
N = 27
Grand Mean = X_GM = 207/27 = 7.667


Calculate between-group variance, denoted by S_B2:

SS_B = 10(6.43 - 7.667)²+ 8(8.25 - 7.667)²+ 9(8.522 - 7.667)²

SS_B = 24.60

S²_B =  SS_w / k -1

= 24.60 /2 = 12.30

Calculate within-group variance, denoted by S_W2:

SS_W = (10- 1)(0.44677)+ (8- 1)(0.1799)+ (9- 1)(0.09194)

SS_W = 6.0165

N - k = 27 - 3 = 24

S_W2 = SS_W/(N - k) = 6.0165/24 = 0.2507



Calculate F test value:

F = 12.30 / 0.25 = 49.07

Step 4: Make Decision:
Compare F test value = 49.0714 with Critical Value = 3.4028
The decision is to reject the null hypothesis since 49.07148 > 3.4028



Analysis of Variance Summary Table

Source	Sum of Squares	Degrees of Freedom (df)	Mean Square	F test value
Between	SSB = 24.6034	k - 1 = 2	MSB = 12.3017	F = 49.07148
Within (error)	SSW = 6.0165	N - k = 24	MSW = 0.2506
Total	30.620	26