Mean Square Error from the Trend in Temperature over Time

Statistics
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Question1
a. The trend in temperature over time
b. Sample size
Residuals df=n-p
345=n-2
N=347
The sample size is 347
c. F=MSE/MSR
F-statistic (Variance Ration)= Mean Square Error(MSE)/Mean Square due to residuals
(MSR)
Step 1: Find the Mean Square Error (MSE=SSE/DF)
Step 2: Find Mean Square Error (MSR)=SSR/DF)
Step 3: Calculate F=MSE/MSR
MSE=56.604
MSR=2.024
F=56.604/2.024=27.966
A=27.966
Mean Square Error(MSE)=Sum Square Error/DF
Sum Square Error= Mean Square Error(MSE)*DF
Step 1: Find MSE
Step 2: Find DF
Step 3: Calculate B
SSE=MSE*DF=2.024*345=698.28
B=698.28
d. Temperature change
Temperature change=−2.082+0.004∗years
After every century (t=10), the temperature changes by;
Temperature change=−2.082+0.004∗10=−2.042
e. Yes, the output is statistically significant (p<0.05).
f. Mean Square Error(MSE)=Sum of Square (SSE) /Degree of Freedom(DF)
Step 1: Find MSE (Mean Square Error(MSE)=Sum Square Error/DF=Sxx/DF
Step 2: Calculate the upper and lower bounds
MSE=3481798/345=10092.168

Upper bound=mean + MSE
1832+10092.168=11924.168
Lower bound=Mean-MSE
1832-10092.168=-8260.168
g. Constant variance of the errors
h. Prediction intervals have larger standard errors associated with them than confidence
intervals.