# 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.IDLength of Stay (x)Total Charge (y)156275397.241085129.46157.9

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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 6 2 7 5 3 9 7.2 4 10 8 5 12 9.4 6 15 7.9
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Step 1

Given data and calculation is shown below help_outlineImage TranscriptioncloseLength of stay(x) Total Charge(y) х*x у"у ху 5 6 25 36 30 7 5 49 25 35 51.84 9 7.2 81 64.8 10 100 64 80 12 9.4 144 88.36 112.8 7.9 15 225 62.41 118.5 Sum 441.1 Sum 58 Sum 43.5 Sum 624 Sum 327.61 fullscreen
Step 2

Correlation formula help_outlineImage TranscriptioncloseΣΕxy - Σx Σν r = (1y) ( η 2-1- n Σ (Σx. i=1 6 x 441.1 58 x 43.5 = 0.74 r = (6 x 624 (58)2) (6 x 327.61 - (43.5)2) fullscreen
Step 3

Pearson correlation coefficient (r) = 0.74
Sample size (n) =6
Formulation of Hypothesis
Null Hypothesis H0 :- ρ = 0
Alt... help_outlineImage TranscriptioncloseTo test the significance, test statistic is given by 6 2 n - 2 1.78 = 0.74 x t = r 1 r2 1 -0.742 fullscreen

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