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.

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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.

College Algebra

1st Edition

ISBN:9781938168383

Author:Jay Abramson

Publisher:Jay Abramson

Chapter4: Linear Functions

Section: Chapter Questions

Problem 30PT: For the following exercises, use Table 4 which shows the percent of unemployed persons 25 years or...

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