wages = B1 + B2educ + ßzexper + ewhere wages denotes hourly wages. We estimate the regression in R and obtain the output## Coefficients:##Estimate Std. Error t value Pr(>|t|)## (Intercept)2.01.020.0455 *## educ0.50.510.3173## exper2.00.54 6.33e-05 ***## ---## Signif. codes:****' 0. 001 '**' 0.01**' 0.05 ' 0.1 ' ' 1Build a 90% confidence interval for B3 using a normal approximation. (Use that if Z ~ N(0, 1) and z1-a satisfies P(Z > z1-a) = a, then zo9 = 1.28, zo95 = 1.64,Z0.975 = 1.96, zo.99 = 2.33, and z0.995 = 2.58).Oa.[2 – 1.64 x (0.5), 2 + 1.64 x (0.5)]O b. [2 – 1.28 × (0.5), 2 + 1.28 x (0.5)]c.[2 – 1.28 x (0.5)², 2 + 1.28 × (0.5)²]O d. [2 – 1.64 × (0.5)², 2 + 1.64 × (0.5)²]O e. [2 – 1.96 × (0.5)², 2 + 1.96 × (0.5)²]O f. [2 – 1.96 × (0.5), 2 + 1.96 × (0.5)]

The coefficient of variable exper is β3. The estimated value of β3 is 2, i.e., β3 =2 The standard error of β3 is 0.5, i.e. S.E(β3) = 0.5 Construct 90% confidence interval for β3: Z0.9 = 1.28 Confidence interval: CI = [β3 - Z0.9 * S.E(β3) , β3 + Z0.9 * S.E(β3)] => CI = [2 - 1.28 * 0.5 , 2 + 1.28*0.5] Answer: Option (B)

Business

Economics

wages = ß1 + Bzeduc + Bzexper + e where wages denotes hourly wages. We estimate the regression in R and obtain the output ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) 0.0455 * 1 0.3173 ## (Intercept) 2.0 1.0 2 ## educ 0.5 0.5 ## exper 2.0 0.5 4 6.33e-05 *** ## --- ## Signif. codes: 0 '**** 0.001 '**' 0.01 '*' 0.05 '.' 0.1 '' 1 Build a 90% confidence interval for ß3z using a normal approximation. (Use that if Z ~ N(0, 1) and z1-a satisfies P(Z > z1-q) = a, then zo,9 = 1.28, zo.95 = 1.64, Z0,975 = 1.96, Z0,.99 = 2.33, and z0.995 = 2.58). O a. [2 – 1.64 × (0.5), 2 + 1.64 × (0.5)] O b. [2 – 1.28 × (0.5), 2 + 1.28 × (0.5)]

wages = ß1 + Bzeduc + Bzexper + e where wages denotes hourly wages. We estimate the regression in R and obtain the output ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) 0.0455 * 1 0.3173 ## (Intercept) 2.0 1.0 2 ## educ 0.5 0.5 ## exper 2.0 0.5 4 6.33e-05 * ## --- ## Signif. codes: 0 ' 0.001 '' 0.01 '*' 0.05 '.' 0.1 '' 1 Build a 90% confidence interval for ß3z using a normal approximation. (Use that if Z ~ N(0, 1) and z1-a satisfies P(Z > z1-q) = a, then zo,9 = 1.28, zo.95 = 1.64, Z0,975 = 1.96, Z0,.99 = 2.33, and z0.995 = 2.58). O a. [2 – 1.64 × (0.5), 2 + 1.64 × (0.5)] O b. [2 – 1.28 × (0.5), 2 + 1.28 × (0.5)]

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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