KAy10-Refer to the accompanying scatterplot. a. Examine the pattern of all 10 points and subjectively determine whether there appears to be astrong correlation between x and y. b. Find the value of the correlation coefficient r and determine whether there is a linear correlation.c. Remove the point with coordinates (10,10) and find the correlation coefficient r and determine whether there is a linear correlation. d.What do you conclude about the possible effect from a single pair of values?Click here to view a table of critical values for the correlation coefficient.a. Do the data points appear to have a strong linear correlation?YesNob. What is the value of the correlation coefficient for all 10 data points?(Simplify your answer. Round to three decimal places as needed.)r=View an exampleGet more help -C(8.9(….&17Table of Critical Valuesn45678 SOIEM91011121314a = .05.950.878.811.754.707.666.6327.60257655353243°F&0-0a = .01990959917875834.798765735708.68466110- X12:56 AM11/19/2022

“Since you have posted a question with multiple sub parts, we will provide the solution only to the first three sub parts as per our Q&A guidelines. Please repost the remaining sub parts separately.” The correlation coefficient (r) is calculated by using the formula : r = Sxy/√(Sxx × Syy) Where, Sxy = Σ(xi - x̄)(yi - ȳ) and Sxx = Σ(xi - x̄)2 Syy = Σ(yi–ȳ)2a) Looking at the points of the scatter plot, it can be easily seen that there exist a strong linear correlation between the data points. Part (b) : From the given scatter plot, it can be seen that x-axis and y-axis are both divided into 10 divisions. Therefore, we have From the above plot, the first data points of the type (x, y) are (1,1) i.e., when x = 1, y = 1. Similarly, we can get the other data points also. (1,2) i.e., when x = 1, y = 2 (1,3) i.e., when x = 1 and y = 3 (2,1) i.e.,when x = 2, y = 3 (2,2) i.e., when x = 2, y = 2 (2,3) i.e., when x = 2, y = 3 (3,1) i.e., when x = 3, y = 1 In a similar manner we get other two pair of points as (3,2) and (3,3). And the last pair points are as (10,10) i.e., when x = 10 and y = 10.Now writing the points that we got in step 3 in a tabular form we have, x y 1 1 1 2 1 3 2 1 2 2 2 3 3 1 3 2 3 3 10 10 Mean of x values (x̄) : x̄ = Σxi/n Substituting the values, x̄ = Σxi/n x̄ = (1+1+1+2+2+2+3+3+3+10)/10 x̄ = 28/10 = 2.8 Mean of y values (ȳ) : ȳ = Σyi/n ȳ = (1+2+3+1+2+3+1+2+3+10)/10 ȳ = 2.8 Using the the above observations, calculation table is as follows : x y (xi–x̄) (yi–ȳ) (xi–x̄)2 (yi–ȳ)2 (xi–x̄)(yi–ȳ) 11122233310 12312312310 -1.800-1.800-1.800-0.800-0.800-0.8000.2000.2000.2007.200 -1.800-0.8000.200-1.800-0.8000.200-1.800-0.8000.2007.200 3.2403.2403.2400.6400.6400.6400.0400.0400.04051.840 3.2400.6400.0403.2400.6400.0403.2400.6400.04051.840 3.2401.440-0.3601.4400.640-0.160-0.360-0.1600.04051.840 Σ(xi–x̄)2 =63.600 Σ(yi–ȳ)2 =63.600 Σ(xi –x̄)(yi –ȳ)=57.600From the calculation table given in step 4, we have Sxy = Σ(xi –x̄)(yi –ȳ) = 57.600 Sxx = Σ(xi–x̄)2 = 63.600 Syy = Σ(yi–ȳ)2 = 63.600 Substituting the values in the formula of correlation coefficient (r) provided in step 1 we get, r = 57.600/√(63.600 × 63.600) r = 57.600/√ 4044.96 r = 57.600/63.600 r = 0.905660 On rounding off to 3 decimal places we get, r = 0.906 (ANSWER) Part (c) : On removing the coordinates (10,10), the correlation coefficient is calculated as, Mean of x values : x̄ = Σxi/n x̄ = (1+1+1+2+2+2+3+3+3) x̄ = 18/9 x̄ = 2 Mean of y values : ȳ = Σyi/n ȳ = (1+2+3+1+2+3+1+2+3)/9 ȳ = 2 x y (xi –x̄) (yi –ȳ) (xi –x̄)2 (yi –ȳ)2 (xi –x̄)(yi –ȳ) 111222333 123123123 -1.000-1.000-1.0000.0000.0000.0001.0001.0001.000 -1.0000.0001.000-1.0000.0001.000-1.0000.0001.000 1.0001.0001.0000.0000.0000.0001.0001.0001.000 1.0000.0001.0001.0000.0001.0001.0000.0001.000 1.0000.000-1.0000.0000.0000.000-1.0000.0001.000 Σ(xi –x̄)2= 6.000 Σ(yi –ȳ)2= 6.000 Σ(xi –x̄)(yi –ȳ)= 0.000From the table provided in step 6, we have, Sxy = Σ(xi –x̄)(yi –ȳ) = 0.000 Sxx = Σ(xi–x̄)2 = 6.000 Syy = Σ(yi–ȳ)2 = 6.000 Substituting the values in the formula of correlation coefficient (r) provided in step 1 we get, r = 0.000/√(6.000× 6.000) r = 0.000 (ANSWER) Since, r = 0. Hence, on removing coordinate (10,10) there exist no linear correlation between the data points.