Mean of API= 165.49/13= 12.73Variance= 15,818.46- 15563.08= 255.38/12= 21.28Standard error= 0.035Z-score of 95% confidence = 1.96 but Z= (x-mean)/standard deviationPrice= 0.976×0.035+12.73= 12.8Question 295% confidence= Z0.05= 1.96Mean oil price= 165.49/13 = 12.73Variance= 1/n-1(x-mean) 21/12(2109.1937-2106.6877) = 0.2088s.t = 0.20881/2/13= 0.035z= (x-u)/s.t where s.t is the standard error.Lower limit= 12.73- 1.96×0.035= 12.6614Upper limit = 12.73+1.96×0.035= 12.7986Confidence region for the mean= 12.66≤x≤12.7986Question 3The interval for x is narrower than the interval for y since the standard error is smallQuestion4Predicted values of y given the regression equation y= 9.4349+0.095235x, replacing values for x and solving for y we get the predicted values of y to bePredicted values residual values 10.58 16.4210.58 17.9210.61 20.1910.60 20.710.62 21.2810.64 23.8610.65 23.3510.67 24.0310.67 26.3310.69 30.3110.69 30.3110.69 28.1110.7 28.6Question 6A graph of residual values against the predicted valuesQuestion 7Heterosdacity is a problem since there is a wider variability between the two variables in the scatter diagram. I.e. as the independent variable increases the gap, hence, makes it difficult to analyze the data.Question 9Standardized residuals are values divided by their standard errorsStandard erro= 0.35 and dividing we get the following values46.9151.259.1460.868.1766.7168.6575.2386.686.680.3181.71Question 10Possible outliers30.31, 30.31, 28.11, 28.6

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