Hypothesis Testing and Confidence Intervals, Exams of Statistics

Download Hypothesis Testing and Confidence Intervals and more Exams Statistics in PDF only on Docsity! Questions and Answers on Hypothesis Testing and Confidence Intervals L. Magee Fall, 2008 ———————————————————– 1. Using 25 observations and 5 regressors, including the constant term, a researcher estimates a linear regression model by OLS and finds b2 = 4.21 , st.err. of b2 = 1.80 b3 = −0.116 , st.err. of b3 = 0.0388 Let t20df be a random variable having a t distribution with n −K = 25 − 5 = 20 d.f.. Use the following probabilities: Prob(t20df > 1.75) = .05 Prob(t20df > 2.086) = .025 For each of the pairs of hypotheses given below, use a t-test to accept or reject H0 at the 5% significance level. Show your reasoning. (a) H0 : β2 = 0 ; Ha : β2 < 0 (b) H0 : β2 = 0 ; Ha : β2 6= 0 (c) H0 : β2 = 0 ; Ha : β2 > 0 (d) H0 : β3 = 0 ; Ha : β3 < 0 (e) H0 : β3 = 0 ; Ha : β3 6= 0 (f) H0 : β3 = 0 ; Ha : β3 > 0 (g) H0 : β2 = 1 ; Ha : β2 6= 1 2. Suppose that you estimate the model y = β1ι+ β2x2 + β3x3 + β4x4 + ε (1) using 13 observations (n = 13), and compute the sum of squared residuals to be SSE1 = 126.1. Then the following smaller models are estimated using the same 13 observations. y = β1ι+ β2x2 + β3x3 + ε, with SSE2 = 166.1 (2) y = β1ι+ β2x2 + β4x4 + ε, with SSE3 = 204.4 (3) y = β1ι+ β2x2 + ε, with SSE4 = 221.4 (4) 1 (a) Compute F statistics for each of the following (i) H0 : β3 = 0 ; Ha : β3 6= 0 (ii) H0 : β4 = 0 ; Ha : β4 6= 0 (iii) H0 : β4 = 0 ; Ha : β4 6= 0 as in part (ii), but this time impose the restriction β3 = 0 in both H0 and Ha. (iv) H0 : β3 = 0 and β4 = 0 ; Ha : β3 6= 0 and/or β4 6= 0 Parts (b) and (c) require the probabilities: Prob(t9df > 2.262) = .025 Prob(t10df > 2.228) = .025 Prob(t9df > 1.383) = .10 Prob(t9df > 1.833) = .05 (b) What are the absolute values of the t statistics for testing the hypotheses in parts (i), (ii) and (iii)? What would the t-test accept/reject conclusions be at the 5% level? (c) Suppose the OLS estimate of β4 in equation (1) above is b4 = 6.13. (i) Use some of the above results to obtain a 90% confidence interval for β4. (Use the t-statistic from part (b) to get the standard error of b4.) (ii) Use some of the above results to test H0 : β4 = 4 against Ha : β4 > 4 at the 10% significance level. 3. The model y = β1ι+ β2x2 + β3x3 + β4x4 + ε (5) is estimated by OLS using 15 observations (n = 15), and compute the sum of squared residuals to be SSE1 = 55. Then the following smaller models are estimated using the same 15 observations. y = β1ι+ β2x2 + β3x3 + ε, with SSE2 = 75 (6) y = β1ι+ β2x2 + β4x4 + ε, with SSE3 = 72 (7) y = β1ι+ β2x2 + ε, with SSE4 = 79 (8) (a) Using the above information, compute F statistics for each of the following hypotheses (i) H0 : β4 = 0 ; Ha : β4 6= 0 (ii) H0 : β4 = 0 ; Ha : β4 6= 0, impose the restriction β3 = 0 in both H0 and Ha (iii) H0 : β3 = 0 and β4 = 0 ; Ha : β3 6= 0 and/or β4 6= 0 (b) Suppose you wish to test the restriction β2 + β3 + β4 = 1. 2 sums of squared residuals are yi = β1 + β2wi + εi, SSE1 = 400 (13) yi = β1 + β2wi + β3gi + εi, SSE2 = 360 (14) yi = β1 + β2wi + β3gi + β4(wigi) + εi, SSE3 = 350 (15) yi = β1 + β2wi + β3hi + εi, SSE4 = 340 (16) In equations (2) and (3), gi = 1 if wi > 10 and gi = 0 if wi ≤ 10. In equation (4), hi = 0 if wi < 10 and hi = wi − 10 if wi ≥ 10. The β values can differ across models. (a) Compute F statistics for testing the following pairs of hypotheses. (i) H0: The same linear relationship between yi and wi applies when wi ≤ 10 as when wi > 10 Ha: The linear relationship between yi and wi jumps to a different level at wi = 10, but the slope of the linear relationship is the same, regardless of whether wi ≤ 10 or wi > 10 (ii) H0: The same linear relationship between yi and wi applies when wi ≤ 10 as when wi > 10 Ha: The linear relationship between yi and wi jumps to a different level at wi = 10, and/or the slope of the linear relationship depends on whether wi ≤ 10 or wi > 10 (iii) H0: The same linear relationship between yi and wi applies when wi ≤ 10 as when wi > 10 Ha: The slope of the linear relationship between yi and wi depends on whether wi ≤ 10 or wi > 10, but these two lines join at wi = 10 (b) Sketch four diagrams, one for each model, each with wi on the horizontal axis and E(yi|wi) on the vertical axis. On each one, draw an example of the type of relation between yi and wi that is allowed by that model. Answers 1. (a) t = (4.21 - 0)/1.80 = 2.34 Reject if t < −1.725. Since t > −1.725, then accept H0 at the 5% significance level (b) t = 2.34 Reject if |t| > 2.086. Since t = 2.34, then |t| > 2.086, so reject H0 at the 5% level (c) t = 2.34 Reject if t > 1.725. Since t > 1.725, then reject H0 at the 5% level 5 (d) t = (−0.116− 0)/.0388 = −2.99 Reject if t < −1.725. Since t < −1.725, then reject H0 at the 5% level (e) t = -2.99 Reject if |t| > 2.086. Since t = −2.99, then |t| > 2.086, so reject H0 at 5% level (f) t = -2.99 Reject if t > 1.725. Therefore accept H0 at the 5% level (g) t = (4.21− 1)/1.80 = 1.78 Reject if |t| > 2.086. Since t = 1.78, then |t| < 2.086, so accept H0 at the 5% level 2. (a) (i) F = (SSE3−SSE1)/1 SSE1/(13−4) = (204.4−126.1)/1 126.1/9 = 5.59 (ii) F = (SSE2−SSE1)/1 SSE1/(13−4) = (166.1−126.1)/1 126.1/9 = 2.86 (iii) Because of the restriction that β3 = 0, the unrestricted model is now (11) and the restricted model is (12), so F = (SSE4−SSE3)/1 SSE3/(13−3) = (221.4−204.4)/1 204.4/10 = 0.83 (iv) F = (SSE4−SSE1)/2 SSE1/(13−4) = (221.4−126.1)/2 126.1/9 = 3.40 (b) When testing one restriction, the t statistic and the F statistic are related by t2 = F . So from the F statistics in (a), we know: in (i) |t| = 2.36 ; in (ii) |t| = 1.69 ; and in (iii) |t| = 0.91. Reject H0 when |t| > 2.262 at the 5% level for a two-tailed test with 9 d.f. Reject H0 when |t| > 2.228 at the 5% level for a two-tailed test with 10 d.f. So the test in (i) rejects H0, and the tests in (ii) and (iii) accept H0. (c) (i) From (b)(ii), |t| = 1.69 for testing H0 : β4 = 0. Since b4 = 6.13 and t = (b4 − 0)/[st.err(b4)] then 1.69 = 6.13/[st.err(b4)] so that st.err(b4) = 3.63. The 90% c.i. is 6.13± t.05 × 3.63 = 6.13± 1.833× 3.63 = 6.13± 6.65 = (−0.52, 12.78) (ii) t = (b4 − 4)/st.err(b4) = (6.13− 4)/3.63 = 0.59. Reject H0 if t > t.10 ⇒ Reject if t > 1.383. Accept H0 at the 10% level, since t = 0.59 < 1.383. 3. (a) Use F = (SSER−SSEU )/J SSEU/(n−K) (i) Compare (5) and (6). F = (75−55)/1 55/(15−4) = 4 (ii) Compare (7) and (8). F = (79−72)/1 72/(15−3) = 7/6 (iii) Compare (5) and (8). F = (79−55)/2 55/(15−4) = 2.4 (b) (i) R = [ 0 1 1 1 ] ; q = 1 (ii) Substitute out one of the β’s from the original model, then re-arrange the variables. There should be one less regressor since there is one restriction imposed. For example, 6 replacing β4 with 1− β2 − β3 gives y = β1ι+ β2x2 + β3x3 + (1− β2 − β3)x4 + ε y − x4 = β1ι+ β2(x2 − x4) + β3(x3 − x4) + ε 4. (a) linear. R = [ 0 0 0 1 0 ] ; q = 3 (b) nonlinear (c) linear. R = [ 0 1 −1 0 0 ] ; q = 0 (d) nonlinear (e) linear. R = [ 0 1 −1 0 0 0 0 1 −1 0 ] ; q = [ 0 0 ] 5. (a) Let the coefficient be β and its estimate be β̂. “Statistically significant at the 5% significance level” means “The null hypothesis H0 : β = 0 is rejected at the 5% significance level.” Usually this means that | β̂ st.err.(β̂) | > (critical value), or P -value < .05, where P -value = Prob(|t| > | β̂ st.err.(β̂) |) and t represents a random variable with a distribution that one hopes is close to the one that the test statistic would have when H0 is true. (This is the “null distribution”.) (b) Suppose that if β really was equal to β̂ in the model, the economic implications would be much the same as if β = 0. In that case, β̂ is not significant in an economic sense. For example, suppose β̂ is estimating an income elasticity and β̂ = .01. Such a β̂ still could be statistically significant, because its standard error might also be very small. In other words, | β̂ st.err.(β̂) | might be large enough for statistical significance, even when β̂ is small enough to be economically insignificant. 6. (a) Compare (11) with (9). F = (59−49)/1 49/(20−3) = 3.47 (b) Compare (12) with (9). F = (59−46)/2 46/(20−4) = 2.26 (c) Compare (12) with (10). F = (54−46)/2 46/(20−4) = 1.39 7. (a) t = 3.1/1.7 = 1.82 (i) Reject if t < −1.66. Therefore, do not reject H0. (ii) Reject if |t| > 1.99. Therefore, do not reject H0. (iii) Reject if t > 1.66. Therefore, reject H0. (iv) Now t = (3.1− 1)/1.7 = 1.24. Reject if |t| > 1.99. Therefore, do not reject H0. 7