Females, it is said, make 70 cents to the dollar in the United States. To invest

Question

Females, it is said, make 70 cents to the dollar in the United States. To investigate this Phenomenon, you collect data on weekly earnings from 1, 744 individuals, 850 females and 894 males. Next, you calculate their average weekly earnings and find that the females in your sample earned $346.98, while the males made $517.70. (a) You recall from your textbook that additional years of experience are supposed to result in higher earnings. You reason that this is because experience is related to "on the job training." One frequently used measure for (potential) experience is "Age Education-6." Explain the underlying rationale. Assuming, heroically, that education is constant across the 1, 744 individuals, you consider regressing earnings on age and a binary variable for gender. You estimate two specifications initially: Earn = 323.70 + 5.15 times Age - 169.78 times Female, R^2 = 0.13, SER = 274.75 (21.18) (0.55) (13.06) Ln(Earn) = 5.44 + 0.015 times Age - 0.421 times Female, R^2= 0.17, SER = 0.75 (0.08) (0.002) (0.036) where Earn are weekly earnings in dollars, Age is measured in years, and Female is a binary variable, which takes on the value of one if the individual is a female and is zero otherwise. Interpret each regression carefully. At age 30, how much less do females earn on average on a dollar basis and a percent basis? (b) Your peer points out to you that age-earning profiles typically take on an inverted U shape. To test this idea, you add the square of age to your log-linear regression. Ln(Earn) = 3.04 + 0.147 times Age - 0.421 times Female - 0.0016 Age^2, (0.18) (0.009) (0.033) (0.0001) R^2 = 0.28, SER = 0.68 Interpret the results again. Are there strong reasons to assume that this specification is superior to the previous one?

Explanation / Answer

Given that earn is dependent variable and age and gender are independent variables.

Assume male = 0 and female= 1

Given that the regression equation is,

Earn = 320.70 + 5.15*Age - 169.78*female

Here coefficient of age = 5.15

Coefficient of female = -169.78

Here we see that there is positive relationship between earn and age.

There is negative relationship between earn and female.

Now we have given that standard error of estimates.

Standard error of intercept = 21.18

standard error of age = 0.55

standard error of female = 13.06

Now we can test individual slope from the given information.

Here we have to test the hypothesis that,

H0 : B = 0 Vs H1 : B not= 0

where B is population slope for independent variable.

Assume alpha = level of significance = 0.05

The test statistic follows t-distribution.

Here test statistic is,

t = b / SEb

where b is the sample slope of independent variable.

SEb is the standard eroro of the estimate.

Now we have to find P-value for taking decision.

P-value we can find by using EXCEL.

syntax :

=TDIST(x,deg_freedom)

where x is absolute value of test statistic

deg_freedom = n - 2

n = 850

deg_freedom = 850-2 = 848

Test statistic for age :

t = 5.15/0.55 = 9.36

P-value = 6.74879E-20 = 0.000

P-value < alpha

Reject H0 at 0.05 significance level.

COnclusion : The population slope for age is differ 0.

test statistic for female :

t = -169.78/13.06 = -13

P-value = 2.29E-35 = 0.000

P-value < alpha

Reject H0 at 0.05 significance level.

COnclusion : The population slope for female is differ 0.

R2 = 0.13

It expresses the proportion of variation in earn which is explained by variation in age and female.

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Second model is the log model

Here coefficient of age = 0.015

Coefficient of female = -0.421

Here we see that there is positive relationship between log of earn and age.

There is negative relationship between log of earn and female.

Now we have given that standard error of estimates.

Standard error of intercept = 0.08

standard error of age = 0.002

standard error of female = 0.036

Now we can test individual slope from the given information.

Here we have to test the hypothesis that,

H0 : B = 0 Vs H1 : B not= 0

where B is population slope for independent variable.

Assume alpha = level of significance = 0.05

The test statistic follows t-distribution.

Here test statistic is,

t = b / SEb

where b is the sample slope of independent variable.

SEb is the standard eroro of the estimate.

Now we have to find P-value for taking decision.

P-value we can find by using EXCEL.

syntax :

=TDIST(x,deg_freedom)

where x is absolute value of test statistic

deg_freedom = n - 2

n = 850

deg_freedom = 850-2 = 848

Test statistic for age :

t = 0.015/0.002 = 7.5

P-value = 6.74879E-20 = 0.000

P-value < alpha

Reject H0 at 0.05 significance level.

COnclusion : The population slope for earn is differ 0.

test statistic for female :

t = -0.421/ 0.036= -11.69

P-value = 2.29E-35 = 0.000

P-value < alpha

Reject H0 at 0.05 significance level.

COnclusion : The population slope for female is differ 0.

R2 = 0.17

It expresses the proportion of variation in log of earn which is explained by variation in age and female.

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