Question

Simple Linear Regression For Questions 1 to 10, consider the following simple linear regression model: Yi = Bo + Bix; tui for

6. (4 points) Suppose n = 82. How much is σˆ 2 , the estimated variance of the error term ui?

A. 0.00625

B. 0.0125

C. 0.025

D. 0.05

7. (4 points) Suppose βˆ 1 = 0.75, SE(βˆ 1) = 0.01 and n = 52. Then the 90% confidence interval for βˆ 1 would be

A. [0.4224, 1.1994]

B. [0.6112, 0.9112]

C. [0.7336, 0.7664]

D. [0.7229, 0.7661]

8. (4 points) Suppose one tested and rejected the null hypothesis H0 : β1 = 0 against H1 : β1 > 0 at the significance level α = 0.05. Then which of the following is TRUE?

A. She/He will reject the null hypothesis H0 : β1 = 0 against H1 : β1 ̸= 0 at the significance level α = 0.05.

B. She/He will reject the null hypothesis H0 : β1 = 0 against H1 : β1 ̸= 0 at the significance level α = 0.1.

C. She/He will not reject the null hypothesis H0 : β1 = 0 against H1 : β1 ̸= 0 at the significance level α = 0.05.

D. She/He will not reject the null hypothesis H0 : β1 = 0 against H1 : β1 ̸= 0 at the significance level α = 0.1.

9. (4 points) Suppose you know that βˆ 0 = 2.5, x¯ = 2.5 and y¯ = 2. Then

A. βˆ 1 = 1.8

B. βˆ 1 = −1.8

C. βˆ 1 = 0.2

D. βˆ 1 = −0.2

10. (4 points) For the 100(1 − α)% confidence interval of βˆ 0, which of the following is TRUE?

A. The larger is α, the larger is the length of the interval.

B. The smaller is α, the larger is the length of the interval.

C. α does not affect the length of the interval.

D. The length of the interval depends on the value of βˆ 0.

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Answer #1

6. The correct option would be

  • C. 0.025

The residual standard error is \sigma^2 = RSE = \frac{\sum u^2}{n-k} , for k be the number of parameters to be estimated, ie \sigma^2 = \frac{\sum u^2}{n-2} or \sigma^2 = \frac{2}{82 - 2} or \sigma^2 = \frac{2}{80} = \underline{0.025} .

7. The correct option would be

  • C. (0.7336, 0.7664)

For 10% significance, we have t_{0.95,50} = 1.6759 , since the df=n-k=52-2=50. The confidence interval would be \beta_1 \pm t_{0.95,50}*SE(\beta_1) or 0.75 \pm 1.6759*0.01 or 0.75 \pm 0.016759 or (0.733241, 0.766759) , which is closest to the option C. In option C, since df>30, the t-statistic is substituted by the normal z statistic of 1.64 (which is not recommended but can be done nonetheless). In that case, the CI would be the same as in option C.

8. The correct option would be

  • B. She/He will reject the null hypothesis H_0 : \beta_0 = 0 against H_1 : \beta_0 \ne 0 at the significance level \alpha = 0.1

The reason being that, the test for \beta > 0 would be one tailed, and the test would be conducted as t > t_{0.95,df} . Now, for two tailed test at 5%, we would require |t| > t_{0.975,df} . Since t_{0.975,df} > t_{0.95,df} , the condition |t| > t_{0.975,df} is not necessarily true. Hence, option A is incorrect. However, at 10%, the condition required would be |t| > t_{0.95,df} , and since we already know t > t_{0.95,df} , which implying that |t| > t_{0.95,df} , the null would be rejected at 10%. Hence, option B is correct.

9. The correct option would be

  • D. \beta_1 = - 0.2

We have \beta_0 = \bar y - \beta_1*\bar x , and for the given values, we have 2.5 = 2 - \beta_1*2.5 or - \beta_1*2.5 = 0.5 or \beta_1 = - 0.2 .

10. The correct option would be

  • B. The smaller is \alpha , the larger is the length of the interval.

The smaller the alpha, the smaller is the region to reject the null, and larger is the region to accept the null, and hence, larger is the CI that may include 0.

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