my Answer:
\({E[b_0]=E[\frac{1}{n}(\sum(k_iY_i-b_1)\sum(k_iX_i))]}\)
\({E[b_0]=\frac{1}{n}E[(\sum(k_iY_i-b_1*1)]}\)
\({E[b_0]=\frac{1}{n}*[E[b_1]-E[b_1]]}\)
\({E[b_0]=\frac{1}{n}*0=0}\)
#according to A.13c
\({E[Y_i]=\beta_0=E[\epsilon_i]=0=E[b_0]}\)
\({\beta_0=E[b_0]}\).
The given conditions of regression model are such that according to the Gauss-Markov Theorem, the least squares estimators \({b_0}\) and \({b_1}\) are unbiased.
His Answer: \({Y_i=\beta_0+\epsilon_i}\)
\({E(\epsilon_i)=0}\) \({var(\epsilon_i)=\sigma^2}\)
\({E(Y_i)=E(\beta_0+\epsilon_i)=E(\beta_0)+E(\epsilon_i)=\beta_0}\) \({Q=\sum{(Y-i-E(Y_i))^2}=\sum{Y_i^2}+nb_0^2-2b_0n\bar{y}}\) \({2nb_0-2n\bar{y}=0}\) Therefore, \({b_0=\bar{y}}\) Set \({\frac{dQ}{db_0}=0}\) Check 2nd Derivative: \({\frac{d^2Q}{db_0^2}>0}\), then \({b_0=\bar{y}}\) is a minimum \({E(b_0)=E(\bar{y})=E(\frac{\sum(Y_i)}{n})=\sum{\frac{E(Y_i)}{n}}=\frac{n\beta_0}{n}=\beta_0}\)
Answer:
*** \({H_0: \beta_1=0}\) ***
*** \({H_A: \beta_1\ne0}\) ***
*** \({t^*=\frac{b_1}{s[b_1]}}\) ***
*** \({\alpha=0.05}\) ***
*** \({b_1=1.10}\) ***
*** \({s_{b_1}=0.58}\) ***
*** \({\hat{y}=5.3+1.10x}\) ***
*** \({1.10\pm t(0.95;25-2)}\) ***
*** \({CI=1.10\pm1.714}\) ***
*** \({-.1\le CI\le2.3}\) ***
*** \({H_0:|t^*|\le t(0.95;23)}\) ***
*** \({H_A:|t^*|> t(0.95;23)}\) ***
*** \({t^*=\frac{1.10}{0.58}=1.896}\) ***
*** \({t=1.714}\) #according to Table B.2 \({t^*>t}\), therefore reject the null hypothesis that \({\beta_1=0}\). ***
#according to Eq. 2.27
*** \({\delta=\frac{|\beta_1-\beta_{10}|}{s[b_1]}=\frac{1.10}{0.58}=1.9}\) ***
#according to Table B.5, with df of 23.
*** \({\delta=2, power=0.48}\) ***
*** \({\delta=1, power=0.16}\) ***
*** \({\delta=1.9, power\approx 0.16+\frac{1.9-1}{2-1}*(0.48-0.16)\approx0.448}\) ***
my Answer:
*** \({Y_n=\hat{Y_n}}\) ***
*** \({(Y_n-\beta_0-\beta_1X_n)^2}\) ***
*** \({(\hat{Y_n}-\beta_0-\beta_1\hat{X_n})^2}\) ***
*** Therefore, conclude that the deletion of this case would not change the least squares regression line because according to Eq. 1.8, the method of least squares seeks to minimize Q. If \({Y_i}\) is on the regression line, then its removal will not affect the error term since it is zero (Q remains minimized), \({\epsilon_i=0}\), hence, \({E[b_0]}\) does not change. Therefore, the fitted regression line would not change. ***
his answer: \({Q_{n-1}=\sum{(Y_i-\hat{Y_i})^2}=Q_n}\), because the extra term added has a \({Y_i-\hat{Y_i}=0}\).
Answer:
*** a. The linear regression model would be as follows: \({\hat{Y_i}=5+3X}\). The standard deviation of 0.6 means that 99% of the data fall between \({\bar{Y}\pm1.8}\). However, all five observations fit with the linear regression line; therefore, the expected value for MSE is \({\sigma^2=(0.6)^2=.36}\) and for MSR is \({\sigma^2+(\beta_1)^2\sum{(X_i-\bar{X})^2}=.36+(3)^2(114)=1026.36}\). ***
*** b. It would have been worse to have made the observations are \({X=6,7,8,9,10}\) because your observations are closely bunched together. This means that the variability in the slope is increased because more \({\beta_1}\) values would be valid for those observations. Having observations across a wider range of data points improves the accuracy of the linear regression model and regression relation to the actual population distribution. (see page 8.) ***
*** c. The same answer would apply. A regression line must go through the mean response of the variable, \({(\bar{X},\bar{Y})}\). In both instances, the mean response is the same (\({X=8}\)). Greater variability in the X variables is preferred. ***
Answer:
This is a non-linear regression model, therefor it is not a general linear regression model. Additionally, this cannot be transformed into a general linear regression model. Why? \({\epsilon_i}\) is ourside of the inverse and exponential parts of the function, therefore, no transformation will be able to incorporate the whole function.