Suppose that the LFPR in 1968 was 0.58 (or 58 percent). On the basis of the regression results given above, what is the mean LFPR in 1972? Establish a 95% confidence interval for the mean prediction. Consider the following regression output:Ŷt= 0.2033 +0.656Xtse (0.0976) (0.1961)r² = 0.397 RSS=0.0544 ESS = 0.0358Where:Y = labor force participation rate (LFPR) of women in 1972 andX = LFPR of women in 1968. The regression results were obtained from a sample of 19 cities in theUnited States.

Answered: Image /qna-images/answer/accc8aed-f698-4fbc-a9ea-784171939ce0.jpg

Consider the following regression output: Ît = 0.2033 +0.656Xt se (0.0976) (0.1961) r² = 0.397 RSS=0.0544 ESS = 0.0358 Where: Y = labor force participation rate (LFPR) of women in 1972 and X = LFPR of women in 1968. The regression results were obtained from a sample of 19 cities in the United States.

Consider the following regression output: Ît = 0.2033 +0.656Xt se (0.0976) (0.1961) r² = 0.397 RSS=0.0544 ESS = 0.0358 Where: Y = labor force participation rate (LFPR) of women in 1972 and X = LFPR of women in 1968. The regression results were obtained from a sample of 19 cities in the United States.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.1: Measures Of Center

Problem 9PPS

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