ISS PREVIOUS YEAR 2016 PAPER-2 SOLUTION SET-A Q.NO. 9
- Shivani Rana
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ISS PREVIOUS YEAR 2016 PAPER-2 SOLUTION SET-A
ISS PREVIOUS YEAR 2016 PAPER-2 SOLUTION SET-A Q.NO. 9
Question:/iss-previous-year-2016-paper-2-solution-set-a-q-no-9
For a simple linear regression modely=β0+β1x+εy = \beta_0 + \beta_1 x + \varepsilony=β0+β1x+ε,consider the following statements where β^0\hat{\beta}_0β^0 and β^1\hat{\beta}_1β^1 are estimates of β0\beta_0β0 and β1\beta_1β1:
Cov(y,β^1)≠0\text{Cov}(y, \hat{\beta}_1) \neq 0Cov(y,β^1)=0
Cov(β^0,β^1)=−xˉ Var(β^1)\text{Cov}(\hat{\beta}_0, \hat{\beta}_1) = -\bar{x} \, \text{Var}(\hat{\beta}_1)Cov(β^0,β^1)=−xˉVar(β^1)
E(y)=β0+β1xE(y) = \beta_0 + \beta_1 xE(y)=β0+β1x
β^0=yˉ−β^1xˉ\hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x}β^0=yˉ−β^1xˉ
Which of the above are correct?
Options:(a) 1, 2 and 3(b) 1, 3 and 4(c) 2, 3 and 4(d) 1, 2 and 4
Solution:
We check each statement:
Statement 1:Cov(y,β^1)≠0\text{Cov}(y, \hat{\beta}_1) \neq 0Cov(y,β^1)=0Since β^1\hat{\beta}_1β^1 is calculated using sample values of yyy, both are dependent.Hence, covariance is not zero.✔ Correct
Statement 2:Cov(β^0,β^1)=−xˉ Var(β^1)\text{Cov}(\hat{\beta}_0, \hat{\beta}_1) = -\bar{x} \, \text{Var}(\hat{\beta}_1)Cov(β^0,β^1)=−xˉVar(β^1)This is a standard result in simple linear regression.✔ Correct
Statement 3:E(y)=β0+β1xE(y) = \beta_0 + \beta_1 xE(y)=β0+β1xSince E(ε)=0E(\varepsilon) = 0E(ε)=0, we get:E(y)=β0+β1xE(y) = \beta_0 + \beta_1 xE(y)=β0+β1x✔ Correct
Statement 4:β^0=yˉ−β^1xˉ\hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x}β^0=yˉ−β^1xˉThis is the standard formula for estimating the intercept.✔ Correct
Conclusion:
All statements (1, 2, 3, and 4) are correct.
Final Answer:
Since all four are correct but not given in options, the closest correct answer is:
(a) 1, 2 and 3
Note:
The question likely contains a misprint because option including all four correct statements is missing.
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ISS PREVIOUS YEAR 2016 PAPER-2 SOLUTION SET-A
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