Consider the following data on production volume (x) and total cost (y) for a manufacturing operation.Production Volume (units)Total Cost ($)400400045050005505400600590070064007507000The estimated regression equation is y = 1246.67 + 7.6x.Use a = .05 to test whether the production volume is significantly related to the total cost.Complete the ANOVA table. Enter all values with nearest whole number, except the F test statistic (to 2 decimals). Use Table 1 ofAppendix B.Source of Variation Sum of Squares Degrees of Freedom Mean SquareFRegressionErrorTotalWhat is your conclusion?Select

ANOVA is used to measure whether there is any mean significant difference between three or more independent variables. It is used to measure the relationship between the categorical and the numerical variable. It was assumes that, one way anova has equality of variance. Given : The production volume (x) and the total cost (y) was, S No x y x-x¯ x-x¯2 y-y¯ y-y¯2 x-x¯y-y¯ 1 400 4000 -175 30625 -1616.67 2613611.111 282916.667 2 450 5000 -125 15625 -616.667 380277.7778 77083.3333 3 550 5400 -25 625 -216.667 46944.44444 5416.66667 4 600 5900 25 625 283.3333 80277.77778 7083.33333 5 700 6400 125 15625 783.3333 613611.1111 97916.6667 6 750 7000 175 30625 1383.333 1913611.111 242083.333 Total 3450 33700 93750 5648333.333 712500 Average 575 5616.667 From the table given, Mean is calculated as : x¯=∑xn=34506=575 y¯=∑yn=337006=5616.667 Thus, total sum of square, SST=SSy=5648333.333 SSx=93750Sxy=712500 Regression sum of square, SSR=Sxy2SSx=712500293750=5415000 Error sum of square, SST=SSE+SSRSSE=SST-SSR=5648333.333-5415000=233333.333Degree of freedom : Regression degree of freedom : There are two manufacturing operations, i.e. df=k-1=2-1=1 Error degree of freedom : There number of data collected is 6 df=n-k=6-2=4 Total degree of freedom : df=n-1=6-1=5 Mean sum of square for regression : MSR=SSRdf=54150001=5415000 Mean sum of square for error : MSE=SSEdf=233333.3334=58333.333 F-value : F-value=MSRMSE=541500058333.333=92.83Using the calculated values, the anova table was : Source of variation Sum of square Degree of freedom Mean square F Regression 5415000 1 5415000 92.83 Error 233333.333 4 58333.333 Total 5648333.333 5 From the anova table, the p-value at 1,4 degree of freedom at 0.05 level of significance is 0.00065 Since the p value 0.00065 is less than 0.05, the null hypothesis is rejected. Conclusion : Hence it is concluded that production volume and the total cost for a manufacturing company is related.