Nuclear and Particle Physics- using liquid drop nuclear mass formulaNuclear and Particle PhysicsA typical binding energy for an alpha particle in a heavy nucleus is about 28.3 MeV. Using the liquid drop nuclear mass formula, determine for what nuclear mass number alpha decay is energetically allowed. Ignore the spin pairing term. Assume that for heavy nuclei, Z≈A/2.25.Hint: Parametrize the change in binding energy of the heavy nuclei as (screenshot of equation was uploaded)ETF5952 Questions for Final Exam INSTRUCTIONS TO STUDENTS 1. Answer ALL questions and type all your answers. Any answers by image or handwriting will receive zero mark. 2. When certain decimal places for answers are stated, your answer should be rounded. 3. When word limits for answers are stated, your answers should satisfy the limits. If your answer exceeds the stated limit, you will receive zero mark for the question. 1 Question 1 (25 points=5+5+5+10) We consider a data set on cars. The data set includes the following variables: • mpg: miles per gallon • cylinders: Number of cylinders between 4 and 8 • displacement: Engine displacement (cu. inches) • horsepower: Engine horsepower • weight: Vehicle weight (lbs.) • acceleration: Time to accelerate from 0 to 60 mph (sec.) • year: Model year (modulo 100) • origin: Origin of car (1. American, 2. European, 3. Japanese) • name: Vehicle name 1. We estimate a regression as below. Explain the effect of weight on mpg (no more than 20 words, 4 decimal places). glm(formula = mpg ~ weight + cylinders + displacement + horsepower + acceleration, data = DATA) Deviance Residuals: Min 1Q Median 3Q Max -11.5816 -2.8618 -0.3404 2.2438 16.3416 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 4.626e+01 2.669e+00 17.331 <2e-16 *** weight -5.187e-03 8.167e-04 -6.351 6e-10 *** cylinders -3.979e-01 4.105e-01 -0.969 0.3330 displacement -8.313e-05 9.072e-03 -0.009 0.9927 horsepower -4.526e-02 1.666e-02 -2.716 0.0069 ** acceleration -2.910e-02 1.258e-01 -0.231 0.8171 — Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 2. We estimate a regression as below. Explain the effect of origin on mpg (no more than 30 words, 4 decimal places). glm(formula = mpg ~ weight + cylinders + displacement + horsepower + acceleration + origin, data = DATA) Deviance Residuals: Min 1Q Median 3Q Max -12.6303 -2.8009 -0.2871 2.0945 14.8931 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 44.7687911 2.6398457 16.959 < 2e-16 *** weight -0.0048119 0.0008089 -5.948 6.1e-09 *** cylinders -0.5661876 0.4042069 -1.401 0.162100 displacement 0.0114270 0.0095737 1.194 0.233376 2 horsepower -0.0613339 0.0168679 -3.636 0.000314 *** acceleration -0.0319841 0.1232529 -0.259 0.795389 origin2 1.1255451 0.7015566 1.604 0.109458 origin3 2.9325397 0.6955675 4.216 3.1e-05 *** — Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 3. We estimate a regression as below. Explain the effect of weight on mpg (no more than 30 words, 4 decimal places). glm(formula = mpg ~ weight * origin + cylinders + displacement + horsepower + acceleration + origin, data = DATA) Deviance Residuals: Min 1Q Median 3Q Max -13.3865 -2.6755 -0.4621 2.0962 14.5518 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 43.9946277 2.7178880 16.187 < 2e-16 *** weight -0.0038512 0.0009205 -4.184 3.56e-05 *** origin2 4.1783349 2.9695460 1.407 0.160222 origin3 11.9740394 3.6294558 3.299 0.001061 ** cylinders -0.6249505 0.4029298 -1.551 0.121726 displacement 0.0041280 0.0100292 0.412 0.680867 horsepower -0.0580462 0.0168325 -3.448 0.000627 *** acceleration -0.0774625 0.1243493 -0.623 0.533694 weight:origin2 -0.0012627 0.0011885 -1.062 0.288710 weight:origin3 -0.0040237 0.0015922 -2.527 0.011900 * — Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 4. We estimate a regression as below. Explain the effect of weight on mpg (no more than 30 words, 4 decimal places). glm(formula = log(mpg) ~ log(weight) * origin + cylinders + displacement + horsepower + acceleration + origin, data = DATA) Deviance Residuals: Min 1Q Median 3Q Max -0.51675 -0.10216 -0.00386 0.10077 0.51237 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 9.0859329 0.7847542 11.578 < 2e-16 *** log(weight) -0.6930137 0.1096725 -6.319 7.34e-10 *** origin2 -1.7792391 0.9726568 -1.829 0.0681 . origin3 -1.2022192 1.1471960 -1.048 0.2953 cylinders -0.0268622 0.0149654 -1.795 0.0735 . displacement 0.0004781 0.0003746 1.276 0.2026 horsepower -0.0033636 0.0006076 -5.536 5.76e-08 *** acceleration -0.0051126 0.0045646 -1.120 0.2634 log(weight):origin2 0.2302492 0.1248918 1.844 0.0660 . log(weight):origin3 0.1633573 0.1485229 1.100 0.2721 — Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 3 Question 2 (25 points=5+5+5+10) 1. We obtain a correlation matrix as below. Answer the variables that have the highest positive and negative correlation with lprice (no more than 20 words). cor(DATA) crime nox rooms dist radial proptax stratio lowstat lprice crime 1.0000000 0.4211523 -0.2188157 -0.3799093 0.6254423 0.5828192 0.2886909 0.4470330 -0.5274947 nox 0.4211523 1.0000000 -0.3028280 -0.7702225 0.6103279 0.6669806 0.1868634 0.5856131 -0.5087672 rooms -0.2188157 -0.3028280 1.0000000 0.2054095 -0.2097727 -0.2921202 -0.3540075 -0.6096048 0.6329095 dist -0.3799093 -0.7702225 0.2054095 1.0000000 -0.4950646 -0.5343788 -0.2292694 -0.4956025 0.3420084 radial 0.6254423 0.6103279 -0.2097727 -0.4950646 1.0000000 0.9102282 0.4642446 0.4760376 -0.4809716 proptax 0.5828192 0.6669806 -0.2921202 -0.5343788 0.9102282 1.0000000 0.4542378 0.5276241 -0.5596710 stratio 0.2886909 0.1868634 -0.3540075 -0.2292694 0.4642446 0.4542378 1.0000000 0.3654023 -0.4976345 lowstat 0.4470330 0.5856131 -0.6096048 -0.4956025 0.4760376 0.5276241 0.3654023 1.0000000 -0.7914387 lprice -0.5274947 -0.5087672 0.6329095 0.3420084 -0.4809716 -0.5596710 -0.4976345 -0.7914387 1.0000000 2. We run K-mean clustering (K=3) and obtain the result as below. Explain the result related to lprice (no more than 20 words). > group5$centers crime nox rooms dist radial proptax stratio lowstat lprice 1 55.6069989 6.747500 5.758750 1.585000 24.000000 66.60000 20.20000 22.54875 9.022747 2 0.3914715 5.122358 6.387453 4.442195 4.455285 31.19268 17.81436 10.48515 10.070670 3 9.5978914 6.698140 6.020853 2.083721 23.224806 66.77442 20.19612 18.43054 9.627254 3. From the K-mean clustering result, explain the relation between lprice and crime (no more than 20 words, 2 decimal places). 4. We apply the principal component analysis for the data and obtain the result, as below. According the estimation result of the first principal component, explain the relation among crime, nox and lprice through the factor (no more than 40 words). > pc = prcomp(DATA, scale=TRUE) > round(pc$rotation, 2) PC1 PC2 PC3 PC4 PC5 PC6 PC7 PC8 PC9 crime 0.31 0.15 -0.26 -0.67 0.47 -0.31 0.04 0.16 -0.12 nox 0.35 0.27 0.41 0.13 -0.10 0.06 0.56 0.53 -0.04 rooms -0.25 0.60 -0.21 -0.01 0.29 0.67 0.03 -0.03 -0.03 dist -0.31 -0.33 -0.43 -0.34 -0.43 0.24 0.23 0.43 -0.09 radial 0.37 0.30 -0.33 0.00 -0.37 -0.02 -0.23 0.07 0.68 proptax 0.39 0.25 -0.23 0.04 -0.45 0.01 -0.04 -0.31 -0.66 stratio 0.25 -0.26 -0.55 0.61 0.37 0.00 0.08 0.22 -0.08 lowstat 0.36 -0.28 0.27 -0.11 0.06 0.47 -0.60 0.31 -0.13 lprice -0.37 0.37 -0.01 0.19 -0.08 -0.41 -0.45 0.51 -0.23 4 Question 3 (25 points=5+5+5+10) • Income: Income in $10,000s • Limit: Credit limit • Rating: Credit rating taking 1 for good customers and 0 for bad customers • Cards: Number of credit cards • Age: Age in years • Education: Number of years of education • Gender: A factor with levels Male and Female • Student: A factor with levels No and Yes indicating whether the individual was a student • Married: A factor with levels No and Yes indicating whether the individual was married • Ethnicity: A factor with levels African American, Asian, and Caucasian indicating the individuals ethnicity • Balance: Average credit card balance in $. 1. We obtain an estimation result as below. Explain the effect of Education on Rating. glm(formula = Rating ~ ., family = “binomial”, data = DATA) Deviance Residuals: Min 1Q Median 3Q Max -2.50873 0.00000 0.00000 0.00001 2.07528 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -116.63752 74.03912 -1.575 0.1152 Income -0.54971 0.90624 -0.607 0.5441 Limit 0.03335 0.03126 1.067 0.2862 Cards 1.89375 2.33613 0.811 0.4176 Age -0.05553 0.11428 -0.486 0.6270 Education -0.24079 0.25734 -0.936 0.3494 GenderFemale 1.21706 1.49002 0.817 0.4140 StudentYes 27.48922 45.32207 0.607 0.5442 MarriedYes 4.72963 2.19800 2.152 0.0314 * EthnicityAsian -0.76642 1.73813 -0.441 0.6593 EthnicityCaucasian 1.24550 1.73212 0.719 0.4721 Balance -0.04779 0.09097 -0.525 0.5993 — Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 554.358 on 399 degrees of freedom Residual deviance: 22.972 on 388 degrees of freedom AIC: 46.972 Number of Fisher Scoring iterations: 13 2. Using the above estimation result, explain the effect of Student on Rating. 3. We obtain the estimation result as below. Explain the effect of Limit on Rating. 5 glm(formula = Rating ~ . * Student, family = “binomial”, data = DATA) Deviance Residuals: Min 1Q Median 3Q Max -2.349 0.000 0.000 0.000 2.071 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.276e+02 7.934e+01 -1.609 0.1077 Income -6.945e-01 9.247e-01 -0.751 0.4526 Limit 3.817e-02 3.269e-02 1.168 0.2430 Cards 2.283e+00 2.480e+00 0.921 0.3572 Age -7.293e-02 1.180e-01 -0.618 0.5367 Education -2.659e-01 2.477e-01 -1.073 0.2832 GenderFemale 1.320e+00 1.462e+00 0.903 0.3667 StudentYes -3.495e+02 1.941e+05 -0.002 0.9986 MarriedYes 5.325e+00 2.623e+00 2.030 0.0423 * EthnicityAsian -5.208e-01 1.874e+00 -0.278 0.7810 EthnicityCaucasian 2.008e+00 1.957e+00 1.026 0.3049 Balance -6.242e-02 9.258e-02 -0.674 0.5002 Income:StudentYes -1.875e+01 8.781e+03 -0.002 0.9983 Limit:StudentYes 6.206e-01 2.854e+02 0.002 0.9983 Cards:StudentYes 4.215e+01 2.031e+04 0.002 0.9983 Age:StudentYes -2.199e+00 8.374e+02 -0.003 0.9979 Education:StudentYes -2.578e+00 7.695e+03 0.000 0.9997 GenderFemale:StudentYes 1.975e+01 1.062e+04 0.002 0.9985 StudentYes:MarriedYes -3.319e+00 9.854e+03 0.000 0.9997 StudentYes:EthnicityAsian -2.865e+01 2.266e+04 -0.001 0.9990 StudentYes:EthnicityCaucasian -2.950e+01 9.090e+03 -0.003 0.9974 StudentYes:Balance -1.822e+00 8.712e+02 -0.002 0.9983 — Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 4. We obtain an estimation result as below. Explain the effect of Limit on Rating and the effect of Gender on Rating. > library(gamlr) > x = sparse.model.matrix(Rating ~ .*Student, data=DATA)[,-1] > y = DATA$Rating > sclasso = gamlr(x, y, family = “binomial”, nfold=10) > coef(sclasso) 22 x 1 sparse Matrix of class “dgCMatrix” seg100 intercept -24.479492961 Income . Limit 0.004813959 Cards 0.119308039 Age . Education -0.024437478 GenderFemale 0.458552143 StudentYes . MarriedYes 1.253561287 EthnicityAsian -0.117659315 EthnicityCaucasian . Balance 0.002295561 Income:StudentYes . 6 Limit:StudentYes . Cards:StudentYes . Age:StudentYes . Education:StudentYes . GenderFemale:StudentYes . StudentYes:MarriedYes . StudentYes:EthnicityAsian . StudentYes:EthnicityCaucasian . StudentYes:Balance . 7 Question 4 (25 points=5+5+5+10) We consider a data set on credit card default. The variables in the data set are • default: A factor with levels No and Yes indicating whether the customer defaulted on their debt • student: A factor with levels No and Yes indicating whether the customer is a student • balance: The average balance that the customer has remaining on their credit card after making their monthly payment • income: Income of customer 1. We estimate a classification tree as below. Explain “Yes”, 0.76 and 1% at the far right node (no more than 30 words). fit1 = rpart(default~balance+income+student, data = DATA, method=”class”) 2. From the estimation result, explain the probability that balance is strictly less than 1800 in the sample (no more than 30 words). 3. Explain characters of individuals in the second node from the left. 4. We compare actual and predicted outcomes as below. Obtain false positive and negative rates. > fit1 = rpart(default~balance+income+student, data = DATA, method=”class”) > pred.tree = predict(fit1, type=”class”) > table(pred.tree, DATA$default) pred.tree No Yes No 9611 203 Yes 56 130 8
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