Question4 Consider a simple regression in which the dependent variable MIM- mean

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Question4 Consider a simple regression in which the dependent variable MIM- mean income of males who are 18 years of age or older, in thousands of dollars. The explanatory variable PMHS- percent of males 18 or older who are not high school graduates. The data consists of 51 observations on the 50 states plus the District of Columbia. Thus MIM and PMHS are "state averages". The estimated regression, along with standard errors (se) and t-statistics (t) are shown below: MTM = 2.733-0180PMHS (se) (So) (t) (1.257) (-5.754) R2 0.873 (Sb) a. Calculate So b. Interpret the R value. Based on this value, explain briefly whether the model fits the data You want to confirm that there is indeed a negative link between the two variables. Now conduct a relevant hypothesis test using the p-value method and state your conclusion at the 5% significance level. c. State the hypothesis H0: H1: Test statistic P-value for the test statistic 6

Explanation / Answer

Part (a)

To test = 0 = 0

Tb = (b - 0)/SE(b)

= (- 0.180 – 0)/sb

= - 5.574 (given)

Hence, sb = - 0.180/(- 5.574)

= 0.0323 ANSWER 1

To test = 0 = 0

t = (a - 0)/SE(a)

= (2.733 – 0)/sa

  = 1.257

=> sa = 2.733/1.257

= 2.174 ANSWER 2

Part (b)

A value of 0.873 for R2 => 87.3% of the variation in dependent variable (MIM) is explained by the variation in predictor (independent) variable, PMHS.

This, in turn implies that mean income of males 28 years of age or older is impacted by their education, namely whether they are high school graduates or not. ANSWER

Part (c)

Test statistic, tb for testing H0: = 0 = 0, Vs HA: < 0 is given to be - 5.754.

Under the hypothesis, H0: = 0 = 0, the test statistic has t-distribution with (n – 2) degrees of freedom. In the given question, n, the sample size = 51 and hence

p-value = P(t49 < - 5.754) = 2.79E-07 which is very small, smaller than the specified level of significance, 5%.

=> H0 is rejected => HA is accepted => < 0 => the link is negative. ANSWER

Part (d)

Test statistic, tb for testing H0: = 0 = - 0.2, Vs HA: - 0.2 is

t = (b - - 0.2)/SE(b).

= (- 0.180 + 0.2)/0.0323 [from Answer 1 of Part (1)]

= - 0.02/0.0323

= - 0.619.

Under the hypothesis, H0: the test statistic has t-distribution with (n – 2) degrees of freedom. In the given question, n, the sample size = 51 and hence critical value = upper 5% point of t49 = 2.01 [using Excel Function of t-distribution]   

We find that |tcal| = 0.619 < tcrit = 2.01

=> H0 is accepted

=> Conclusion:

There is not enough evidence to suggest that the slope is not equal to – 0.2 ANSWER

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