Calculate the Pearson Correlation Coefficient for these two variables and b) test whether the correlation is significantly different from 0. Run the test at a 5% level of significance. Give each of the following for part b to receive full credit: 1) the appropriate null and alternative hypotheses; 2) the appropriate test; 3) the decision rule; 4) the calculation of the test statistic; and 5) your conclusion including a comparison to alpha or the critical value. You MUST show your work to receive full credit. Partial credit is available.ID Length of Stay (x) Total Charge (y)1)5 62)7 53)9 7.24)10 85)12 9.46)15 7.9

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Calculate the Pearson Correlation Coefficient for these two variables and b) test whether the correlation is significantly different from 0. Run the test at a 5% level of significance. Give each of the following for part b to receive full credit: 1) the appropriate null and alternative hypotheses; 2) the appropriate test; 3) the decision rule; 4) the calculation of the test statistic; and 5) your conclusion including a comparison to alpha or the critical value. You MUST show your work to receive full credit. Partial credit is available.ID Length of Stay (x)  Total Charge (y)1)5                          62)7                          53)9                          7.24)10                        85)12                        9.46)15                        7.9

Rebekah Zimmerman · Accepted Answer

Length of stay(x          )                          Total charge(y)                          x          ∗          x                          y          ∗          y                          x          y                                      5                          6                          25                          36                          30                                      7                          5                          49                          25                          35                                      9                          7.2                          81                          51.84                          64.8                                      10                          8                          100                          64                          80                                      12                          9.4                          144                          88.36                          112.8                                      15                          7.9                          225                          62.41                          118.5                                      Sum - 58                          Sum - 43.5                          Sum - 624                          Sum - 327.61                          Sum - 441.1                      r  =            n              ∑        i        6                    =        1                    x        i                    y        i            =      −              ∑        i        6                    x        i                    ∑        i        6                    y        i                    n              ∑        i        6                    =        1                    x        i        2            −      (              ∑        i        6                    =        1                    y        1        2            −      (              ∑        i        6                    =        1                    y        i                    )        2            )        r  =            6      ×      441.1      −      58      ×      43.5              6      ×      624      −      (      58              )        2            )      (      6      ×      327.61      −      (      43.5              )        2            )        =  0.74Pearson correlation coefficient (r) = 0.74Sample size (n) =6Formulation of HypothesisNull Hypothesis H0 :- ρ = 0Alternate Hypothesis H1 :- ρ ≠ 0To test the significance, test statistic is given byTo test the significance, test statistic is geven by  t  =  r                    n        −        2                    1        −                  r          2                      =  0.74  ×                    6        −        2                    1        −                  0.74          2                      =  1.78Finding T-criticalDegree Of Freedom (D.O.F) = 6 – 2 = 4t-critical = 2.7764 (From Excel  =T.INV.2T(0.05,4)Finding the value of pp-value = 0.149 (From Excel =T.INV.2T(0.05,4))Decision ruleIf p &amp;gt; 0.05 we accept the null hypothesisIf p &amp;lt; 0.05 we reject the null hypothesisConclusionp-value(=0.149) &amp;gt; 0.05, So we failed to reject  null hypothesis and conclude that correlation is not different from 0

Calculate the Pearson Correlation Coefficient for these two variables and b) test whether the correl

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