Human capital theory suggests that those who have invested in higher levels of education will be able to command higher wages. A labour economist collected data on annual wages (Y, in $'000) and years of study (X) from a random sample of 12 employees to test this proposition. Assuming a linear relationship between Y and X, the labour economist used a least-squares method and found that the Y intercept = -23.50 and the slope = 9.73. Also, the sum of squares total (SST) and the sum of squares regression (SSR) were equal to 11132.92 and 8726.91, respectively. What is the lower bound of a 95% prediction interval of annual wages (in $'000) for an employee with 10 years of study. Assumer the h Statistic is 0.10. a. greater than $20 but less than $36 b. less than $20 c. greater than $20 but less than $30 d. greater than $36
Question
Human capital theory suggests that those who have invested in higher levels of education will be able to command higher wages. A labour economist collected data on annual wages (Y, in '000) for an employee with 10 years of study. Assumer the h Statistic is 0.10. a. greater than 36 b. less than 20 but less than 36
Solution
To calculate the lower bound of a 95% prediction interval, we first need to calculate the predicted wage for an employee with 10 years of study using the regression equation Y = a + bX, where a is the Y-intercept, b is the slope, and X is the years of study.
Substituting the given values, we get:
Y = -23.50 + 9.73 * 10 Y = 71.8
Next, we need to calculate the standard error of the estimate (SEE), which is the square root of the mean square error (MSE). The MSE is calculated as (SST - SSR) / (n - 2), where n is the number of observations.
MSE = (11132.92 - 8726.91) / (12 - 2) MSE = 240.6
SEE = sqrt(240.6) = 15.5
The lower bound of the 95% prediction interval is then calculated as Y - t * SEE * sqrt(1 + h), where t is the t-value for a 95% confidence interval with n - 2 degrees of freedom, and h is the leverage. Assuming a t-value of 2.228 (approximate value for 10 degrees of freedom), and h = 0.10, we get:
Lower bound = 71.8 - 2.228 * 15.5 * sqrt(1 + 0.10) Lower bound = 71.8 - 34.5 Lower bound = 37.3
So, the lower bound of the 95% prediction interval for an employee with 10 years of study is greater than 36.
Similar Questions
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