Meteorologists use sophisticated models to predict the occurrence, duration, and trajectory of weather events. They build their models based on observations that they have made in the past. By understanding how previous weather events evolved, meteorologists can apply that knowledge to future weather events.Parametric equations can be used to graph the path of an object in space. For example, they can be used to describe the path of a storm moving through an area. In this portfolio, you will use historical storm data to trace the path of a hurricane. From this data, you will use parametric equations to model the path of the storm.

Question 1 (1 point) SavedWhich of the following equations are linear?Question 1 options:y+7=3xy=6x2+83y=6x+5y2y-x=8x2Question 2 (1 point) Which of the following equations are linear?Question 2 options:4y=8y2=6x3+83y=6x+5y2y-x=8x2Question 3 (1 point) You decided to join a fantasy Baseball league and you think the best way to pick your players is to look at their Batting Averages.You want to use data from the previous season to help predict Batting Averages to know which players to pick for the upcoming season. You want to use Runs Score, Doubles, Triples, Home Runs and Strike Outs to determine if there is a significant linear relationship for Batting Averages.You collect data to, to help estimate Batting Average, to see which players you should choose. You collect data on 45 players to help make your decision.x1 = Runs Score/Times at Batx2 = Doubles/Times at Batx3 = Triples/Times at Batx4 = Home Runs/Times at Batx5= Strike Outs/Times at BatIs there a significant linear relationship between these 5 variables and the Batting Average?If so, what is/are the significant predictor(s) for determining the Batting Average?See Attached Excel for Data.Baseball data.xlsxQuestion 3 options:No,Triples/Times at Bat, p-value = 0.291004037 > .05, No, Triples, are not a significant predictor for Batting Average.Home Runs/Times at Bat, p-value = 0.114060301 > .05, No, Home Runs are not a significant predictor for Batting Average.Yes,Triples/Times at Bat, p-value = 0.291004037 > .05, No, Triples, are not a significant predictor for Batting Average.Home Runs/Times at Bat, p-value = 0.114060301 > .05, No, Home Runs are not a significant predictor for Batting Average.No,Runs Score/Times at Bat, p-value = 0.000219186 < .05, Yes, Runs Score is a significant predictor for Batting Average.Doubles/Times at Bat, p-value = 0.00300543 < .05, Yes, Doubles are a significant predictor for Batting Average.Strike Outs/Times at Bat, p-value = 0.00000258892 < .05, Yes, Strikes Outs are a significant predictor for Batting Average.Yes,Runs Score/Times at Bat, p-value = 0.000219186 < .05, Yes, Runs Score is a significant predictor for Batting Average.Doubles/Times at Bat, p-value = 0.00300543 < .05, Yes, Doubles are a significant predictor for Batting Average.Strike Outs/Times at Bat, p-value = 0.00000258892 < .05, Yes, Strikes Outs are a significant predictor for Batting Average.Question 4 (1 point) You are thinking about opening up a Starbucks in your area but what to know if it is a good investment. How much money do Starbucks actually make in a year? You collect data to, to help estimate Annual Net Sales, in thousands, of dollars to know how much money you will be making.You collect data on 27 stores to help make your decision.x1 = Rent in Thousand per monthx2 = Amount spent on Inventory in Thousand per monthx3 = Amount spent on Advertising in Thousand per monthx4 = Sales in Thousand per monthx5= How many Competitors stores are in the AreaApproximately what percentage of the variation in Annual Net Sales is accounted for by these 5 variables in this model?See Attached Excel for Data.Starbuck Sales data.xlsxQuestion 4 options:27.62% of variation in the Annual Net Sales is accounted for by Rent, Inventory, Advertising, Sales per Month and # of Competitor store in this model. 98.24% of variation in the Annual Net Sales is accounted for by Rent, Inventory, Advertising, Sales per Month and # of Competitor store in this model. 99.11% of variation in the Annual Net Sales is accounted for by Rent, Inventory, Advertising, Sales per Month and # of Competitor store in this model. 99.25% of variation in the Annual Net Sales is accounted for by Rent, Inventory, Advertising, Sales per Month and # of Competitor store in this model. Question 5 (1 point) With Obesity on the rise, a Doctor wants to see if there is a linear relationship between the Age and Weight and estimating a person's Systolic Blood Pressure. Is there a significant linear relationship between Age and Weight and a person's Systolic Blood Pressure?If so, what is/are the significant predictor(s) for Systolic Blood Pressure?See Attached Excel for Data.BP dataQuestion 5 options:No,Age, p-value = 0.9388 > .05, No, Age is not a significant predictor for Systolic BPWeight, p-value = 0 .3092 > .05, No, Weight is not a significant predictor for Systolic BPNo,Age, p-value = 0.001303023 < .05, No, Age is not a significant predictor for Systolic BPWeight, p-value = 0.023799395 < .05, No, Weight is not a significant predictor for Systolic BPYes,Age, p-value = 0.9388 > .05, Yes, Age is a significant predictor for Systolic BPWeight, p-value = 0 .3092 > .05, Yes Weight is a significant predictor for Systolic BPYes, Age, p-value = 0.001303023 < .05, Yes, Age is a significant predictor for Systolic BPWeight, p-value = 0.023799395 < .05, Yes Weight is a significant predictor for Systolic BPQuestion 6 (1 point) You move out into the country and you notice every Spring there are more and more Deer Fawns that appear. You decide to try and predict how many Fawns there will be for the up coming Spring.You collect data to, to help estimate Fawn Count for the upcoming Spring season. You collect data on over the past 10 years.x1 = Adult Deer Countx2 = Annual Rain in Inchesx3 = Winter SeverityWhere Winter Severity Index:1 = Warm2 = Mild3 = Cold4 = Freeze5 = SevereEstimate Fawn Count when Adult Deer Count = 10, Annual Rain = 13.5 and Winter Severity = 4See Attached Excel for Data.Deer data.xlsxQuestion 6 options:53.853.062.95Question 7 (1 point) In the context of regression analysis, what is the definition of an influential point?Question 7 options:Observed data points that are far from the least squares lineObserved data points that are far from the other observed data points in the horizontal directionObserved data points that are close to the least squares lineObserved data points that are close to the other observed data points in the horizontal directionQuestion 8 (1 point) The least squares regression line for a data set is yˆ=2.3−0.1x and the standard deviation of the residuals is 0.13.Does a case with the values x = 4.1, y = 2.34 qualify as an outlier?Question 8 options:YesNoCannot be determined with the given informationQuestion 9 (1 point) The following data represent the weight of a child riding a bike and the rolling distance achieved after going down a hill without pedaling. Weight (lbs.)Rolling Distance (m.)59268443974856201035987448848924653286632713910049Can it be concluded at a 0.05 level of significance that there is a linear correlation between the two variables?Question 9 options:yesnoCannot be determinedQuestion 10 (1 point) A negative linear relationship implies that larger values of one variable will result in smaller values in the second variable.Question 10 options:TrueFalseQuestion 11 (1 point) You determine there is a strong linear relationship between two variables using a test for linear regression. Can you immediately claim that one variable is causing the second variable to act in a certain way?Question 11 options:No, the correlation would need to be a perfect linear relationship to be sure.No, you should examine the situation to identify lurking variables that may be influencing both variables.No, you must first decide if the relationship is positive or negative.Yes, a strong linear relationship implies causation between the two variables.Question 12 (1 point) Which of the following describes how the scatter plot appears? Select all that apply.Question 12 options:positivenegativeneither positive or negativeQuestion 13 (1 point) The following data represent the weight of a child riding a bike and the rolling distance achieved after going down a hill without pedaling. Weight (lbs.)Rolling Distance (m.)59268443974856201035987448848924653286632713910049 Regression StatisticsMultiple R0.956806R Square0.915477Adjusted R Square0.907025Standard Error3.483483Observations12ANOVAdfSSMSFSignificance FRegression11314.321314.32108.31131.1E-06Residual10121.346612.13466Total111435.667CoefficientsStandard Errort StatP-valueLower 95%Upper 95%Intercept-8.564724.7892-1.788340.104007-19.23572.106284Weight (lbs.)0.6116910.05877510.407271.1E-060.4807310.742651Find the standard error of estimate. Round answer to 4 decimal places.___Question 13 options:Question 14 (1 point) Body frame size is determined by a person's wrist circumference in relation to height. A researcher measures the wrist circumference and height of a random sample of individuals.Model SummarybModelRR SquareAdjusted R SquareStd. Error of the Estimate1.734a.539.5254.01409a. Predictors: (Constant), Wrist Circumferenceb. Dependent Variable: HeightANOVAaModelSum of SquaresdfMean SquareFSig.1Regression621.7931621.79338.590.000bResidual531.7263316.113Total1153.51934a. Dependent Variable: Heightb. Predictors: (Constant), Wrist CircumferenceModelUnstandardized CoefficientsStandardized CoefficientstSig.BStd. ErrorBeta1(Constant)38.1775.0897.502.000Wrist Circumference4.436.714.7346.212.000What is the value of the test statistic to see if the correlation is statistically significant?Question 14 options:0.5397.5024.4365.0896.2120.734Question 15 (1 point) What are the hypotheses for testing to see if a correlation is statistically significant?Question 15 options:H0: ρ = 0 ; H1:ρ =1H0: ρ = ±1 ; H1:ρ ≠ ±1H0: r = 0 ; H1: r ≠ 0 H0: ρ = 0 ; H1:ρ ≠ 0H0: r = ±1 ; H1: r ≠ ± 1Question 16 (1 point) A teacher believes that the third homework assignment is a key predictor in how well students will do on the midterm. Let x represent the third homework score and y the midterm exam score. A random sample of last terms students were selected and their grades are shown below. Assume scores are normally distributed. Calculate the correlation coefficient using technology (you can copy and paste the data into Excel). Round answer to 4 decimal places. Make sure you put the 0 in front of the decimal.HW3Midterm13.159.81121.987.5398.853.72824.395.2835.439.17413.266.09220.989.72918.578.9852086.215.473.2742593.259.752.2576.443.98420.279.76221.884.25823.192.9112287.8211.454.03414.971.86918.476.70415.170.4311565.1516.877.208Answer:______Question 16 options:Question 17 (1 point) Choose the correlation coefficient that is represented in the scatterplot.Question 17 options:0.830.15-0.82Question 18 (1 point) The correlation coefficient, r, is a number between:Question 18 options:0 and ∞-10 and 10-∞ and ∞0 and 100 and 1- 1 and 1Question 19 (1 point) A teacher believes that the third homework assignment is a key predictor in how well students will do on the midterm. Let x represent the third homework score and y the midterm exam score. A random sample of last terms students were selected and their grades are shown below. Assume scores are normally distributed.HW3Midterm13.159.81121.987.5398.853.72824.395.2835.439.17413.266.09220.989.72918.578.9852086.215.473.2742593.259.752.2576.443.98420.279.76221.884.25823.192.9112287.8211.454.03414.971.86918.476.70415.170.4311565.1516.877.208Find the y-intercept and slope for the regression equation using technology (you can copy and paste the data into Excel). Round answer to 3 decimal places.ŷ=___+___xQuestion 19 options:Blank # 1Blank # 2Question 20 (1 point) Bone mineral density and cola consumption has been recorded for a sample of patients. Let x represent the number of colas consumed per week and y the bone mineral density in grams per cubic centimeter. Assume the data is normally distributed. A regression equation for the following data is ŷ=0.8893-0.0031x. Which is the best interpretation of the slope coefficient?xy10.88320.873430.889840.885250.881660.86370.863480.864890.8552100.8546110.862Question 20 options:For every additional average weekly soda consumption, a person's bone density increases by 0.0031 grams per cubic centimeter.For every additional average weekly soda consumption, a person's bone density decreases by 0.0031 grams per cubic centimeter.For an increase of 0.8893 in the average weekly soda consumption, a person's bone density decreases by 0.0031 grams per cubic centimeter. For every additional average weekly soda consumption, a person's bone density decreases by 0.8893 grams per cubic centimeter.