tat Identify the P-value.P-value= (Round to two decimal places as needed.)What is the conclusion of the test?the null hypothesis. Conclude that the type of vehicleappear to affect the amount of greenhouse gas emissions for these three types. 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 0.025 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 ofvehicle appear to affect the amount of greenhouse gas emissions?What are the hypotheses for this test?OA. Ho: H₁ H₂ H3H₁: At least one of the means is different from the others.OB. Ho: H₁ H₂ #13H₁: H1 H₂ H3OC. Ho: H₁ H₂ H3H₁: H1 H₂ H3O D. Ho: At least one of the means is different from the others.H₁ H₁ H₂ H3Determine the test statistic.F= (Round to two decimal places as needed.)C...Type A5.75.95.57.55.76.47.16.55.86.9Type B9.27.78.58.78.18.18.57.7Type C9.99.49.99.28.88.696D8.88.7

The sample observation of the amount of greenhouse gas emission for three types of vehicles are as follows:Type AType BType C5.79.29.95.97.79.45.58.59.97.58.79.25.78.18.86.48.18.67.18.59.66.57.78.85.88.76.9Considering μ1, μ2, and μ3 as the average amount of greenhouse gas emission for vehicle types of A, B, and C, it is set up as follows:H0:μ1=μ2=μ3; the average amount of greenhouse emissions for three types of vehicles are hypothesized to be indifferent.H1:μ1≠μ2≠μ3; the average amount of greenhouse emissions for three types of vehicles are hypothesized to be not indifferent.Calculation Table:TypeSample observation; yijSample sizeTotalType A5.75.95.57.55.76.47.16.55.86.910 (=n1)63 (=T1.)3969 (=T1.2)Type B9.27.78.58.78.18.18.57.78 (=n2)66.5 (=T2.)4422.25 (=T2.2)Type C9.99.49.99.28.88.69.68.88.79 (=n3)82.9 (=T3.)6872.41 (=T3.2)Total212.4 (=G)The number of vehicle types, k=3.The total number of sampled vehicles, n=n1+n2+n3=27.The Raw Sum of Squares:RSS=∑i∑jyij2=5.72+5.92+...+9.22+7.72+...+9.92+9.42+...+8.72=1721.5Correction Factor:CF=G2n=212.4227=1670.88Total Sum of Squares:TSS=RSS−CF=1721.5−1670.88=50.62Treatment Sum of Squares:SST=T1.2n1+T2.2n2+T3.2n3−CF=396910+4422.258+6872.419−1670.88=42.40236111Error Sum of Squares,SSE=TSS−SST=50.62−42.40236111=8.217638889The test statistic is:F=SST(k−1)SSE(n−k)=42.40236111(3−1)8.217638889(27−3)=61.91904271≈61.92The p-value for the F=61.92 with the distribution degrees of freedom df1=k−1=3−1=2 and df2=n−k=27−3=24 is obtained using the excel function "F.DIST.RT(F,df1,df2)" as follows:The p-value is 0.00.The decision rule is: reject H0, if p-value≤α.For α=0.025, p-value (=0.025)<α. Thus H0 is rejected.The average amount of greenhouse emissions for three types of vehicles is concluded to be not indifferent.The test statistic, F=61.92.The p-value is 0.00.Reject the null hypothesis. Conclude that the type of vehicles does appear to affect the amount of greenhouse gas emissions.