{r}GasTurbine=read.table(file="C:/Users/Unico/Downloads/GasTurbine19095.txt",header=T)
attach(GasTurbine)
My answer for this question is: t.test(HeatRate) The mean of the gas turbine data I have is 11328.25kJ/kWh, higher than that of a standard gas turbine. My t-test returned a p-value of far less than 0.05 and therefore shows the null hypothesis should be rejected.
The mark for this question is:
My answer for this question is: par(mfrow=c(2,4))
plot(HeatRate~ExhTemp)
plot(HeatRate~Airflow)
plot(HeatRate~CPRatio)
plot(HeatRate~Engine)
plot(HeatRate~InletTemp)
plot(HeatRate~Power)
plot(HeatRate~RPM)
plot(HeatRate~Shafts)
CPRatio, RPM and InletTemp.
The mark for this question is:
My answer for this question is: When predicting HeatRate by means of AirFlow we find that for each unit increase in AirFlow, HeatRate will decrease by 5.453 units, giving these two variables an inverse relationship with the formula “HeatRate=12291.077-5.453(AirFlow).” This formula also indicates a HeatRate of 12291.077 when AirFlow is equal to 0.
The mark for this question is:
My answer for this question is: I have chosen to use InletTemp as a sole predictor of HeatRate. The simple linear regression formula is “HeatRate=23230.324-10.345(InletTemp).”This formula indicates that for an InletTemp value of zero, HeatRate would be 23230.324, decreasing by 10.345 for each unit increase in InletTemp.This model has a greater R squared value than that of the previous model which would indicate that InletTemp is a greater sole predictor of HeatRate.
The mark for this question is:
My answer for this question is:
The mark for this question is:
My answer for this question is: This multiple regression model can be considered more accurate as it has a very small p-value and an F-statistic value greater than that needed to prove significance at 9 and 22 DF.
The mark for this question is:
My answer for this question is:
The mark for this question is:
My answer for this question is: The R squared value for this multiple regression model is relatively high, at 0.9251, and would indicate that these variables are fairly viable in predicting HeatRate.
The mark for this question is:
My answer for this question is:
The mark for this question is:
My answer for this question is: Points 4 and 6 have relatively high leverage values, considerably more than most other points. Point 13 is somewhat untoward but not anything too major.
The mark for this question is:
My answer for this question is:
The mark for this question is:
My answer for this question is:
The mark for this question is:
My answer for this question is:
The mark for this question is:
My answer for this question is:
The mark for this question is:
Your total mark is:
HeEx.lm=lm(HeatRate~ExhTemp)
summary(HeEx.lm)
E[HeatRate]=14932.286-6.872(ExhTemp)
HeAi.lm=lm(HeatRate~Airflow)
summary(HeAi.lm)
E[HeatRate]=12291.077-5.453(Airflow)
HeCP.lm=lm(HeatRate~CPRatio)
summary(HeCP.lm)
E[HeatRate]=-258.27(CPRatio)+15132.14
HeEn.lm=lm(HeatRate~Engine)
summary(HeEn.lm)
HeIn.lm=lm(HeatRate~InletTemp)
summary(HeIn.lm)
E[HeatRate]=23230.324-10.345(InletTemp)
HePo.lm=lm(HeatRate~Power)
summary(HePo.lm)
E[HeatRate]=1.216e+04-1.318e-02(Power)
HeRP.lm=lm(HeatRate~RPM)
summary(HeRP.lm)
E[HeatRate]=9.356e+03+1.978e-01(RPM)
HeSh.lm=(HeatRate~Shafts)
summary(HeSh.lm)
HeatRate.lm=lm(HeatRate~.,data=GasTurbine)
par(mfrow=c(2,2))
plot(HeatRate.lm)
summary(HeatRate.lm)
t.test(HeatRate)