5. You want to investigate how unemployment rates and city crime rates are relat

Question

5. You want to investigate how unemployment rates and city crime rates are related and so set up the following model additionally controlling population and income: log(crime) = 0 + 1 log(pop) + 2log(inc) + 3unemp + u where crime is total number crimes, pop is population, inc is per capita income, and unemp is unemployment rate for each city. You have a R output but by mistake some parts are teared-off call: 1m(formulalog crime1og_poplog inc unemp, data-new data) Residuals: -0.5276-0.1805-0.0357 0.1551 0.6866 Coefficients: Min 1Q Median 30 Max Estimate 1.6658 0.9460 -0.3527 -0.0150 std. value 1.26 21.82 0.1512-2.33 Pr(>|t!) 0.210

Explanation / Answer

since we have got the results of the multi-linear regression and so we can make the final equation as

log(crime)=1.6658+ 0.9460 * log(pop) -0.3527 * log(inc) - 0.0150 * unemp

Where B1 = 0.9460

and B3=-0.0150

so from the B1 we can say that a unit channge in log(pop) value will increase the value of log(crime) by 0.9460 by keeping rest of the parameters constant.And so we can say that increase in log(pop) will result in increase in log(crime) and so we can say that by increasing pop the crime also increases.

At the same time by B3 we can says that

a unit channge in unemp value will decrease the value of log(crime) by 0.015 by keeping rest of the parameters constant.And so we can say that increase in unemp will result in decrease in log(crime) and so we can say that by increasing unemp the crime will decrease.

(ii)here the value of Rsquare is 0.813 and so we can say that 81.3% of the variation in log(crime) can be explained by log(pop) ,log(inc) and unemp.

(iii) At overall significance of 5% we can say that the variation in log(crime) is due to the variation in log(pop) and log(inc) but the variation in unemp is not related to the variation is log(crime) which is very much evident from the negative relationships they share as discussed in problem(i). Since you can notice the same by the p values of log(pop),log(inc) and unemp where only for unemp the p value is more than 0.05 (for 95% significance p value should be less than 0.05) and so we have derived that except unemp rest of the parameters are significat.

(iv) As we know that the formula of std error=estimate/t-value and so for log(pop) std error = estimate(logpop)/tvalue(logpop) =0.9460/21.82=0.043355

I hope this has helped your understanding. To to time crunch here I've explained first 4 questions. Kindly upvote the ans if it has really helped you understanding the proble. Good Luck!!

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