Chapter 5 Answers - Chapter 5 Answers 1(a The 95 confidence interval for 1 β is 5 82 1 96 2 21 Î ª that is 1 10 152 1 4884 β μ(b)Calculate

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Unformatted text preview: Chapter 5 Answers 1 (a) The 95% confidence interval for 1 β is { 5 82 1 96 2 21},- . Î . ª . that is 1 10 152 1 4884. β- . μ- . (b)Calculate the t-statistic: 1 1 5 82 2 6335 SE( ) 2 21 act t β β-- . = = = - . . . The p-value for the test 1 H β = vs. 1 1 H β ≠ is-value 2 ( | |) 2 ( 2 6335) 2 0 0042 0 0084 act p t = Φ - = Φ - . = . = . . The p-value is less than 0.01, so we can reject the null hypothesis at the 5% significance level, and also at the 1% significance level. (c) The t-statistic is 1 1 ( 5.6) 0 22 0.10 SE( ) 2 21 act t β β- - . = = = . The p-value for the test 1 : 5.6 H β = - vs. 1 1 : 5.6 H β ≠ - is-value 2 ( | |) 2 ( 0.10) 0.92 act p t = Φ - = Φ - = The p-value is larger than 0.10, so we cannot reject the null hypothesis at the 10%, 5% or 1% significance level. Because 1 5.6 β = - is not rejected at the 5% level, this value is contained in the 95% confidence interval. (d)The 99% confidence interval for β is {520.4 2.58 20.4}, that is, 467.7 573.0. β ≤ ≤ 2. (a) The estimated gender gap equals $2.12/hour. (b)The hypothesis testing for the gender gap is 1 H β = vs. 1 1 0. H β ≠ With a t-statistic 1 1 2.12 5.89 ( ) 0 36 act t SE β β- = = = , . the p-value for the test is-value 2 ( | |) 2 ( 5.89) 2 0 0000 0 000 act p t = Φ - = Φ - = . = . (to four decimal places) The p-value is less than 0.01, so we can reject the null hypothesis that there is no gender gap at a 1% significance level. (c) The 95% confidence interval for the gender gap 1 β is {2 12 1 96 0 36}, . ) . ¼ ....
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This note was uploaded on 05/03/2010 for the course ECON 303 taught by Professor Grant during the Spring '10 term at Lewis and Clark Community College.

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Chapter 5 Answers - Chapter 5 Answers 1(a The 95 confidence interval for 1 β is 5 82 1 96 2 21 Î ª that is 1 10 152 1 4884 β μ(b)Calculate

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