1. (Twelve parts; 40 marks in total) Though global climate change is partly a na

Question

1. (Twelve parts; 40 marks in total) Though global climate change is partly a natural phenomenon, it appears to be largely exacerbated and accelerated by various human activities. In particular, carbon dioxide emissions due to industries, automobiles, and deforestation may be largely responsible for the rise in global surface temperature, which in turn results in numerous other environmental consequences. Below is partial SPSS output analyzing the relationship between global atmospheric CO2 (in ppm), as a predictor variable and the minimum summer coverage of Arctic sea ice (in millions of square kilometers), as a response variable. The analysis is based on data recorded over a 20-year period from 1998 to 2017, inclusive (from NOAA, an organization which has been collecting global climate data since 1880). Using the computer output provided, answer parts (a) to (k) based on the simple linear regression model, (Arctic sea ice | CO,)-B + ,CO, . Assume that all the required assumptions for this model are satisfied Descriptive Statistics Mean Std. Deviation Arctic sea ice CO2 8637 12.3129 5.4200 20 384.6500 20 Stat 252-Homework # 3 Questions-Winter 2018 Model Summary ModelR R SquareAdjusted R Std. Error of the Estimate Square a. Predictors: (Constant), CO2 b. Dependent Variable: Arctic sea ice ANOVA Model Sum of Squaresdf Mean Sig Square Regression 8.2610 0.0000898 1 Residual Total a. Dependent Variable: Arctic sea ice b. Predictors: (Constant), CO2 14.1720

Explanation / Answer

(d)since the p-value of the F test is less than alpha=0.05, so model is significant.

here F=MS(regression)/MS(error)=8.261/0.3283=25.16, with df= (n1=1,n2=18 )

residual SS=Total SS- regression SS

total df=N-1=20-1=19,

there there are p=2 regression coefficient is estimate regression df=p-1=2-1=1

error df=total df- regression df=19-1=18

following ANOVA is completed

(e) R2=regression SS/ total SS=8.261/14.172= 0.5829

for simple linear regression , the required correlation coefficeint=r=sqrt(R2)=sqrt(0.5829)=0.7635

since the regression coefficient of Co2=-0.0536 is negative sign , so sign of r would also be negative

so r=-0.7635

(f) here we use t-test for slope of the regression and t=B/SE(B)=-0.0536/0.107=-5, which is significant at 5% level of significance( critical t(0.05,18)=2.1)

p-value=0.0000928 ( using ms-excel =tdist(5,18,2))

model sum of square(SS) df mean square(MS) F sig regression 8.261 1 8.261 25.16 0.0000898 residual 5.911 18 0.3283 total 14.172 19

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.