Please anwser using SPSS!An accurate assessment of oxygen consumption provides important information for determining energy expenditure requirements for physically demanding tasks. The paper “Oxygen Consumption During Fire Suppression: Error of Heart Rate Estimation” (Ergo-nomics [1991]: 1469–1474) reported on a study in which x 5 oxygen consumption (in milliliters per kilogram per minute) during a treadmill test was determined for a sample of 10 fire fighters. Then y 5 oxygen consumption at a comparable heart rate was measured for each of the 10 individuals while they performed a fire-suppression simulation. This resulted in the following data and scatterplot: simulation. This resulted in the following data andscatterplot:FirefighteryFirefighter 635.426.3y50403020151.349.3Fire-simulationconsumption3534.129.52427835.4 38.633.9 29.43The regression equation isfirecon=-11.4 + 1.09 treadconPredictorCoefConstant -11.37treadcon1.0906S = 4.70R-sq = 59.5%41.130.6Stdev12.460.3181436.328.2a. Does the scatterplot suggest an approximate linearrelationship?Treadmill49 consumptionb. The investigators fit a least-squares line. The result-ing Minitab output is given in the following:91040.639.523.5 31.6t-ratio-0.91536.528.00.3883.43 0.009R-sq(adj) = 54.4%Predict fire-simulation consumption when treadmillconsumption is 40.c.How effectively does a straight line summarize therelationship?d. Delete the first observation, (51.3, 49.3), and calcu-late the new equation of the least-squares lineand the value of r². What do you conclude? (Hint:For the original data, Σx = 388.8, Σy = 310.3,Ex² = 15,338.54, Σ xy = 12,306.58, and2y = 10,072.41.)

Let the variable X denote the oxygen consumption during a treadmill test and Y denote the oxygen consumption during fire-suppression simulation. The data of firefighters is given below.(a)Enter the data into the SPSS as follows: Click on Graphs ---> Scatterplot--> Move y variable to y-axis and x variable to x-axis. Click "OK". The scatterplot from the SPSS is displayed below. From the scatterplot, the datapoints do not fall along the straight line but the datapoints are scattered almost in a random manner without showing any particular pattern. There is one point which is far away from the remaining data points appears to provide an impression that there may be linear relationship exists between the variables. Once, the extreme point is ignored, it is clear that no relationship exists. Thus, the scatterplot does not indicate approximate linear relationship between two variables.Enter the data into the SPSS as follows: Click on Analyze--> Regression--->Linear A new-window is displayed move y variable to dependent and x variable to independent list. Click Ok. The output of the SPSS is displayed below. The unstandardized coefficients column represents the estimates of regression equation. From the above output, the estimate of constant is, bo=-11.37. The estimate of the variable x is, b1=1.09 The regression equation is, y=bo+b1x =-11.37+(1.09)x The fire simulation consumption when treadmill consumption 40 is obtained by substituting x =40 in the above equation. y=bo+b1x =-11.37+(1.09)40 =-11.36+43.6 =32.24 The predicted fire simulation is 32.24 when treadmill consumption is 40. From the SPSS output, the value of R2 is, 0.59. This means that the percentage of variability in response variable that can be explained by the regression model involving the given predictor variable. Thus, the oxygen consumption during treadmill test can predict around 59% of the variability in the oxygen consumption during fire-suppression simulation. This suggests that the straight line is moderately effective in summarizing the relationship between two variables.The original data has the following information, ∑i=110xi=388.8,∑i=110yi=310.3 ∑i=110xiyi=12306.58, ∑i=110xi2=15338.54 and ∑i=110yi2=10072.41 By deleting first observation (x1,y1)=(51.3,49.3), the new data has the following, ∑i=19xi(new)= ∑i=110xi-x1 =388.8-51.3 =337.5 ∑i=19yi(new)=∑i=110yi-y1 =310.3-49.3 =261 ∑i=19xi2=∑i=110xi2-x12 =15338.54-(51.3)2 =15338.54-2631.69 =12706.85 ∑i=19yi2=∑i=110yi2-y12 =10072.41-(49.3)2 =10072.41-2430.49 =7641.92 ∑i=19xiyi=∑i=110xiyi-(xy11)=12306.58-51.3*49.3=12306.58-2529.09=9777.49 The least square regression equation is obtained below. The slope is computed below. b1=n∑i=19xiyi-∑i=19xi×∑i=19yin∑i=19xi2-∑i=19xi2 =9(9777.49)-337.5×2619(12706.85)-337.52=87997.41-88087.5114361.65-113906.25=-90.09455.4 =-0.1978=-0.20(rounded to two decimals) The intercept is computed below. bo=∑i=19yi×∑i=19xi2-∑i=19xiyi×∑i=19xin∑i=19xi2-∑i=19xi2 =(261)(12706.85)-9777.49×337.59(12706.85)-337.52=3316487.85-3299902.875114361.65-113906.25=16584.975455.4 =36.41848=36.42(rounded to two decimals) The regression equation is displayed below. y=bo+b1x =-0.20+(36.42)x The correlation coefficient is, r=n∑i=19xiyi-∑i=19xi×∑i=19yin∑i=19xi2-∑i=19xi2n∑i=19yi2-∑i=19yi2 =9(9777.49)-337.5×261(9(12706.85)-(337.5)2)(9(7641.92)-(261)2) =87997.41-88087.5(114361.65-113906.25)(68777.28-68121) =-90.9(455.4)×656.28 =-90.9546.690=-0.1648 The Coefficient of determination is, R2=(-0.1648)2 =0.0272 The value of R^2 is part(b) is,0.59 and the value of R^2 in part(d) is, 0.0272. By comparing two models, the new model is not the approximate least square regression model because R^2 is very less in the new model. The old model is the best model for least square regression.