Find the following sums of squares (SS):SP or SSXY = ["1125", "30", "27", "35", "28", "34"]SSX or SSXX = ["1125", "30", "27", "35", "28", "34"]SSY or SSYY = ["1125", "30", "27", "35", "28", "34"]Find the correlation (r) between X and Y: ["0.88", "0.03", "5.10", "1.14", "-0.88", "-0.03"]What is the value of rho for the null distribution? ["3", "27", "28", "34", "0"]What is the df for this test? ["2", "3", "4", "9", "8"]What is the critical value for this test? ["3.18", "4.18", "2.35", "2.78"]What is the obtained test statistic for this test? ["11.21", "3.13", "4.18", "2.78"]Do you decide to reject or retain the null hypothesis? For the data listed below, do a hypothesis test for the significance of the Pearson's correlationcoefficient (alpha = .05, 2-tailed) and use your calculations to fill in the following answers.For the data listed below, do a hypothesis test for the significance of the Pearson's correlationcoefficient (alpha = .05, 2-tailed) and use your calculations to fill in the following answers.X683108Y78186Find the following sums of squares (SS):Find the following sums of squares (SS):

From the information, given thatThat is,The sum of squares of XY is,Thus, the sum of squares of XY is 27.Correct option: 27The sum of squares of XX is,Thus, the sum of squares of XX is 28.Correct option: 28The sum of squares of YY is,Thus, the sum of squares of YY is 34.Correct option: 34 The correlation value is,Thus, the value of the correlation is 0.88..The correct option: 0.88.The null hypothesis is stated below,Null hypothesis: That is, there is no correlation between the two variables.Thus, the value of the rh0 for the null distribution is 0.Correct option: 0. Alternative hypothesis: That is, there is no correlation between the two variables.The df is,Thus, the df is 3.Use EXCEL Procedure to find the critical value of t.Follow the instructions to obtain the critical value of t: 1. Open Excel2. Go to Formula bar.3. In formula bar enter the function as“=TINV”4. Enter the probability as 0.055. Enter the degrees of freedom as 36. Click enterFrom the Excel sheet the critical value of t at 0.05 level of significance with the 4 degrees of freedom is 3.18 Thus, the critical value is 3.18..Correct option: 3.18The value of the test statistic is,Thus, the value of the test statistic is 3.13.Conclusion: Use the level of significance is 0.05The value of test statistic is 3.13 and the critical value is 3.18.That is, By, the rejection rule , fails to reject the null hypothesis.Correct option: Retain the null hypothesis. Thus, the sum of squares of XY is 27.Thus, the sum of squares of XX is 28.Thus, the sum of squares of YY is 34.Thus, the value of the correlation is 0.88..Thus, the value of the rh0 for the null distribution is 0.Thus, the df is 3.Thus, the critical value is 3.18..Thus, the value of the test statistic is 3.13..Correct option: Retain the null hypothesis.