Can you please answer the problem below: Thanks 2. Planets. (Pretend that Pluto is still a planet.)Distance(millionPlanet miles)MercuryVenusEarthMarsJupiterSaturnUranusNeptunePluto366793142484887176527913654Use the linear model -sun.Diameter(miles) Revolution(days) Position12330307520792642178883874896317623077414288822536568743321076030684601889046745678(a) Create a scatterplot of distance from the sun versus position.Is the relationship linear?If not, transform the distance using both the square root of distance and the log of distance.Create two new scatterplots using the transformed distances.Which transformation is better?Choose the better transformation. Then create a linear model. Graph the residual plot. Explain themeaning of the residual plot.9(b) Create a scatterplot of period of revolution versus distance from the sun.Is the relationship linear?If not, find a power of distance so that the relationship is more linear.Then create a linear model.Create a residual plot. Explain the meaning of the residual plot.predict the period of an imaginary planet that is 1,000 million miles from the

Given the data asPlanetDistance (million miles)Diameter (miles)Revolution (days)PositionMercury363030881Venus6775202252Earth9379263653Mars14242176874Jupiter4848883843325Saturn88774896107606Uranus176531762306847Neptune279130774601888Pluto36541428904679a. The scatterplot distance from the sun versus position:The steps for Scatter Diagram in EXCEL :Step 1 : Upload the data into EXCELStep 2 : Select Insert in Home barStep 3 : Select Charts - Scatter tab in Insert barStep 4 : Click OKFrom the scatterplot it is visible that the relationship distance versus position is not linear.Scatter plot for Square root of Distance Vs Position :Square root of DistancePosition618.185429.6437311.9164422529.7825642.0119752.8299860.44839Scatter plot for log of Distance Vs Position:log of DistancePosition1.556311.826121.968532.152342.684852.947963.246773.445883.56289The better transformation is log of distance transformation since the scatterplot becomes linear.Let the random variables x and y denotes the log of distance and position respectively.The estimated linear regression model iswhere and Distance361.55631-1.0427-44.17081.0872671.82612-0.7729-32.31880.5974931.96853-0.6305-21.26100.39761422.15234-0.4467-10.44670.19964842.684850.0858000.00748872.947960.348910.34890.121717653.246770.647721.29550.419627913.445880.846832.54030.717036543.562890.963843.85510.928823.39124516.23714.4763The estimated linear regression equation is Data used for the residual plot :1.556311.2178-0.21781.826122.1964-0.19641.968532.71300.28702.152343.37970.62032.684855.3115-0.31152.947966.2658-0.26583.246777.3497-0.34973.445888.0716-0.07163.562898.49610.5039The steps for Residual plot in EXCEL :Step 1: Take the difference of actual and predicted valuesStep 2: Select the x as x-axis values and the obtained difference as y-axis valuesStep 2 : Select Insert in Home barStep 3 : Select Charts - Scatter tab in Insert barStep 4 : Click OKThe linear model is good fit for the given data since the the residuals are randomly scattered around the residual.b. The scatterplot of period of revolution versus distance from the sun:Yes, the relationship is linear.Scatter plot for Revolution Vs square of DistanceThe above scatter plot is more linear than the first one.The data used :RevolutionDistance288129622544893658649687201644332234256107607867693068431152256018877896819046713351716The steps for Regression output in EXCEL :Step 1 : Upload the data into EXCELStep 2 : Select Data in Home barStep 3 : Select Data analysis tab in Data barStep 4 : Select Regression in Data analysis tool barStep 5 : Select input Y range as dependent variable and input X range as independent variableStep 6 : Tick Residual Plots Step 7 : Click OKThe estimated regression equation isGiven, The predicted period of an imaginary planet that is 1,000 million miles from the sun is 2394 days.The residual plot :The linear model is not a right fit for the given data since the residuals are clustered in a curved pattern.a. The scatterplot distance from the sun versus position:From the scatter plot it is visible that the relation distance versus position is not linear.The better transformation is log of distance transformation since the scatterplot becomes linear.The estimated linear regression model isThe residual plot isThe linear model is good fit for the given data since the the residuals are randomly scattered around the residual.b. Yes, the relationship is linear.The square distance transformation scatter plot is more linear than the first one.The estimated regression equation isThe predicted period of an imaginary planet that is 1,000 million miles from the sun is 2394 days.The residual plot :The linear model is not a right fit for the given data since the residuals are clustered in a curved pattern.