New representatives for the John Ford Metal and Wheel Company attend a brief trainingprogram before being assigned to a regional office. At the end of such a program, eachrepresentative was ranked with respect to future sales potential. At the end of the firstsales year, their rankings were paired with their annual sales:2.3.4.6.7.8.9.10.a.b.C.RepresentativeKitchenBondGrossArbuckleGreeneArdenCraneArthurKeeneKnopfAnnual Sales($000)319150175460348300280200190300Ranking inTraining Program49613B1527810Compute Spearman's rank correlation coefficient.Interpret.Do a hypothesis test at the .01 level of significance.

Correlation is used to measure the association between the two or more variable. It can be calculated using the two methods, they are Pearson and Spearman. Pearson correlation is used for numerical variable association calculation. Spearman rank correlation is used to find correlation between the ranked variables. a) Given : Annual sales and the ranking in training program. The data given as : Annual Sales Ranking in training program Rank X Rank Y d=X-Y d2 319 4 3 4 -1 1 150 9 10 9 1 1 175 6 9 6 3 9 460 1 1 1 0 0 348 3 2 3 -1 1 300 10 4.5 10 -5.5 30.25 280 5 6 5 1 1 200 2 7 2 5 25 190 7 8 7 1 1 300 8 4.5 8 -3.5 12.25 Sum, 81.5 From the data given, ρ=1-6∑d2+23-212nn2-1=1-681.5+0.510102-1=1-0.497=0.503 Spearman rank correlation is 0.503b) Given : From the table, Spearman rank correlation is 0.503 Interpretation : Here, the value is positive shows positive association between the annual sales and the ranking in the training. It lies above the 0.5 shows positive moderate association between the annual sales and the ranking in the training. c) Given : Let the null and alternate hypothesis : Null hypothesis : There exist correlation is zero i.e. H0 : ρ=0 Alternate hypothesis : There exist correlation is not zero i.e. Ha : ρ≠0 Level of significance is 0.05 Degree of freedom : df=n-2=10-2=8 The t critical value with 8 degree of freedom at 0.05 level of significance is 3.36 Test statistic : t=r1-r2n-2=0.5031-0.503210-2=1.646 Here, the critical value is greater than critical value and the null hypothesis failed to be rejected. i.e. There exist no correlation between the annual sales and the rank of the training program.