Table of z values and probabilities for the standard normal distribution. z is the first column plus the top row. Each cell shows P( X z). For example P( X 1.04) = .8508. For z < 0 subtract the value from 1, e.g., P( X 1.04) = 1 .8508 = .1492. z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 1 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 2 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 3 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 1 0.09 0.5359 0.5753 0.6141 0.6517 0.6879 0.7224 0.7549 0.7852 0.8133 0.8389 0.8621 0.8830 0.9015 0.9177 0.9319 0.9441 0.9545 0.9633 0.9706 0.9767 0.9817 0.9857 0.9890 0.9916 0.9936 0.9952 0.9964 0.9974 0.9981 0.9986 0.9990 0.4 0.3 0.2 P(X>2.201)=.025 0.1 density for t df=11 0.0 P(X<2.201)=.975 4 2 0 2 4 t=2.201 Table of t values and right tail probabilities. Degrees of freedom are in the first column (df). Right tail probabilities are in the first row. For example for d. f . = 7 and = .05 the critical t value for a two-tail test is 2.365 and for d. f . = 10 and = .1 the critical t value for a one-tail test is 1.372. df .1 .05 .025 .01 .005 1 3.078 6.314 12.706 31.821 63.657 2 1.886 2.920 4.303 6.965 9.925 3 1.638 2.353 3.182 4.541 5.841 4 1.533 2.132 2.776 3.747 4.604 5 1.476 2.015 2.571 3.365 4.032 6 1.440 1.943 2.447 3.143 3.707 7 1.415 1.895 2.365 2.998 3.499 8 1.397 1.860 2.306 2.896 3.355 9 1.383 1.833 2.262 2.821 3.250 10 1.372 1.812 2.228 2.764 3.169 11 1.363 1.796 2.201 2.718 3.106 12 1.356 1.782 2.179 2.681 3.055 13 1.350 1.771 2.160 2.650 3.012 14 1.345 1.761 2.145 2.624 2.977 15 1.341 1.753 2.131 2.602 2.947 16 1.337 1.746 2.120 2.583 2.921 17 1.333 1.740 2.110 2.567 2.898 18 1.330 1.734 2.101 2.552 2.878 19 1.328 1.729 2.093 2.539 2.861 20 1.325 1.725 2.086 2.528 2.845 21 1.323 1.721 2.080 2.518 2.831 22 1.321 1.717 2.074 2.508 2.819 23 1.319 1.714 2.069 2.500 2.807 24 1.318 1.711 2.064 2.492 2.797 25 1.316 1.708 2.060 2.485 2.787 26 1.315 1.706 2.056 2.479 2.779 27 1.314 1.703 2.052 2.473 2.771 28 1.313 1.701 2.048 2.467 2.763 29 1.311 1.699 2.045 2.462 2.756 30 1.310 1.697 2.042 2.457 2.750 40 1.303 1.684 2.021 2.423 2.704 50 1.299 1.676 2.009 2.403 2.678 75 1.293 1.665 1.992 2.377 2.643 100 1.290 1.660 1.984 2.364 2.626 2 Table of F values for right tail probabilities of .05. Degrees of freedom for denominator are in the first column and degrees of freedom for the numerator are in the top row. denom. numerator df df 1 2 3 4 5 7 10 15 50 100 1 161.45 199.5 215.71 224.58 230.16 236.77 241.88 245.95 251.77 253.04 2 18.51 19 19.16 19.25 19.3 19.35 19.4 19.43 19.48 19.49 3 10.13 9.55 9.28 9.12 9.01 8.89 8.79 8.7 8.58 8.55 4 7.71 6.94 6.59 6.39 6.26 6.09 5.96 5.86 5.7 5.66 5 6.61 5.79 5.41 5.19 5.05 4.88 4.74 4.62 4.44 4.41 6 5.99 5.14 4.76 4.53 4.39 4.21 4.06 3.94 3.75 3.71 7 5.59 4.74 4.35 4.12 3.97 3.79 3.64 3.51 3.32 3.27 8 5.32 4.46 4.07 3.84 3.69 3.5 3.35 3.22 3.02 2.97 9 5.12 4.26 3.86 3.63 3.48 3.29 3.14 3.01 2.8 2.76 10 4.96 4.1 3.71 3.48 3.33 3.14 2.98 2.85 2.64 2.59 11 4.84 3.98 3.59 3.36 3.2 3.01 2.85 2.72 2.51 2.46 12 4.75 3.89 3.49 3.26 3.11 2.91 2.75 2.62 2.4 2.35 13 4.67 3.81 3.41 3.18 3.03 2.83 2.67 2.53 2.31 2.26 14 4.6 3.74 3.34 3.11 2.96 2.76 2.6 2.46 2.24 2.19 15 4.54 3.68 3.29 3.06 2.9 2.71 2.54 2.4 2.18 2.12 16 4.49 3.63 3.24 3.01 2.85 2.66 2.49 2.35 2.12 2.07 17 4.45 3.59 3.2 2.96 2.81 2.61 2.45 2.31 2.08 2.02 18 4.41 3.55 3.16 2.93 2.77 2.58 2.41 2.27 2.04 1.98 19 4.38 3.52 3.13 2.9 2.74 2.54 2.38 2.23 2 1.94 20 4.35 3.49 3.1 2.87 2.71 2.51 2.35 2.2 1.97 1.91 21 4.32 3.47 3.07 2.84 2.68 2.49 2.32 2.18 1.94 1.88 22 4.3 3.44 3.05 2.82 2.66 2.46 2.3 2.15 1.91 1.85 23 4.28 3.42 3.03 2.8 2.64 2.44 2.27 2.13 1.88 1.82 24 4.26 3.4 3.01 2.78 2.62 2.42 2.25 2.11 1.86 1.8 25 4.24 3.39 2.99 2.76 2.6 2.4 2.24 2.09 1.84 1.78 26 4.23 3.37 2.98 2.74 2.59 2.39 2.22 2.07 1.82 1.76 27 4.21 3.35 2.96 2.73 2.57 2.37 2.2 2.06 1.81 1.74 28 4.2 3.34 2.95 2.71 2.56 2.36 2.19 2.04 1.79 1.73 29 4.18 3.33 2.93 2.7 2.55 2.35 2.18 2.03 1.77 1.71 30 4.17 3.32 2.92 2.69 2.53 2.33 2.16 2.01 1.76 1.7 40 4.08 3.23 2.84 2.61 2.45 2.25 2.08 1.92 1.66 1.59 60 4 3.15 2.76 2.53 2.37 2.17 1.99 1.84 1.56 1.48 100 3.94 3.09 2.7 2.46 2.31 2.1 1.93 1.77 1.48 1.39 1000 3.85 3 2.61 2.38 2.22 2.02 1.84 1.68 1.36 1.26 3 xi /n and = wi xi /wi ( x i )2 ( xi x )2 and s2 = . N n1 Probability Rules: P( A) = 1 P( AC ) Counting Rule for Permutations \u0012 \u0013 N N! N Pn = n! = . n ( N n)! Chapter 4: Counting Rule for Combinations \u0012 \u0013 N N! N Cn = = . n n!( N n)! xy = xy /(x y ) and r xy = s xy /(s x sy ). Population and Sample Pearson Correlation xy 100 % \u0011 ( xi x )(yi y ) ( xi x )(yi y ) = and s xy = N n1 z-Score: zi = Population and Sample Covariance xi x s . Mean Coefficient of Variation \u0010 Standard deviation Population and sample standard deviation = 2 and s = s2 . 2 = xi /N and x g = [( x1 )( x2 ) . . . ( xn )]1/n . Interquartile Range: IQR = Q3 Q1 . Population and sample variance x = Weighted mean and geometric mean x = Chapter 1: no key formulas. Chapter 2: Relative Frequency=freq. of the class/n. Approx. Class Width: =(largest value-smallest value) /number of classes. Chapter 3: sample and population means ASU ECN 221 ASWCC P( A B) P( B) ( x )2 f ( x ). x e . x! f (x) = ( Nn ) r ( xr )( N n x ) and E( x ) = = nr . N Hypergeometric Probability Mass Function and Expected Value: P( X = x |) = f ( x ) = Expected Value for Binomial Distribution: E( x ) = = np. Variance for Binomial Distr.: Var ( x ) = 2 = np(1 p). Poisson Probability Mass Function: Binomial Probability Mass Function \u0012 \u0013 n x P( X = x ) = f ( x ) = p (1 p ) ( n x ) . x Number of Experimental Outcomes Providing Exactly x Successes in n Trials \u0012 \u0013 n n! . = x!(n x )! x Var ( x ) = 2 = Chapter 5: Discrete Uniform Probability Mass Function: f ( x ) = 1/n. Expected Value of a Discrete R. V.: E( x ) = = x f ( x ). Variance of a Discrete R. V.: P ( Ai ) P ( B | Ai ) P ( A1 ) P ( B | A1 ) + P ( A2 ) P ( B | A2 ) + + P ( A n ) P ( B | A n ) P ( Ai | B ) = Bayes' Theorem P ( A B ) = P ( B ) P ( A ). Multiplication Law for Independent Events P ( A B ) = P ( B ) P ( A | B ) = P ( A ) P ( B | A ). P( A| B) = P( A B) = P( A) + P( B) P( A B) Chapter 4 continued: Key Formula Sheet 1 ba 0 otherwise if a x b \u0010 ( x )2 \u0011 . exp 22 22 1 x . n= (z/2 )2 2 E2 Necessary Sample Size for Interval Estimate of s x z/2 and x t/2 n n Expected Value and Std Dev (Standard Error) of p r p (1 p ) E( p ) = p and p = n p Finite Pop. Correction Factor: ( N n ) / ( N 1). Chapter 8: Interval Estimate of Population Mean, known and unknown x = . n Standard Deviation of x (Standard Error) E( x ) = . Chapter 7: expected value of x f ( x ) = 1 e x/ and P( x x0 ) = 1 e x0 / . Exponential PDF and CDF for x 0 z= Converting to the Standard Normal rv: f (x) = Normal PDF The density function is f (x) = ( Chapter 6: Uniform PDF Chapter 5 continued: Variance for the Hypergeometric Distribution: \u0010 r \u0011\u0010 r \u0011\u0010 N n \u0011 1 . Var ( x ) = 2 = n N N N1 p (1 p ) n (z/2 )2 p (1 p ) E2 x 0 x 0 and t = / n s/ n df = 1 n1 1 \u0010 s1 n1 s21 n1 2 \u00112 \u0010 + + \u00112 1 n2 1 s22 n2 \u0010 s22 n2 \u00112 Degrees of Freedom for t, Two Independent Random Samples 1 Interval Estimate and Test Statistic for Difference in Two Means with Unknown Variances s s21 s2 x x D0 x1 x2 t/2 + 2 and t = 1r 2 n1 n2 s21 s22 n + n2 1 Interval Estimate and Test Statistic for Difference in Two Means with Known Variances s 12 2 x x D0 + 2 and z = 1r 2 x1 x2 z/2 n1 n2 12 22 n + n2 Chapter 10: Point Estimate and Standard Error for Difference in Two Population Means s 12 2 + 2 x1 x2 and x1 x2 = n1 n2 p0 (1 p0 ) n p p0 z= q Test Stat for Hypothesis About p z= Chapter 9: Test Statistic for Hypothesis Tests About , known and unknown n= Necessary Sample Size for Interval Estimate of p p z/2 Chapter 8 continued: Interval Estimate of p n d d sd / n nj j =1 (xij x )2 nj j =1 i =1 k j =1 (yi y)2 SSR = ( xi x )(yi y ) ( xi x )2 SST=SSR+SSE (yi yi )2 (yi y)2 SSE = y = b0 + b1 x b0 = y b1 x SST=SSTR+SSE SSE nT k ( xi x )2 sb1 = p ( xi s x )2 t= b1 sb1 spred = s s 1+ 1 ( x x )2 + n ( x1 x )2 Confidence Interval for E(y ): y t/2 sy For simple regression, MSR = SSR because there is only one independent variable. s MSR 1 ( x x )2 F= sy = s + MSE n ( x1 x )2 b1 = p SSR SSE r2 = r xy = (sign of b1 ) r2 s2 = MSE = SST n2 Standard Error of the Estimate, s = MSE. SST = b1 = E(y) = 0 + 1 x MSE = F = MSTR/MSE Chapter 11: not covered in this course Chapter 12: y = 0 + 1 x + e SST = k n j (x j x )2 (n j 1)s2j k SSTR SSTR = k1 SSE = MSTR = j j kj=1 i=1 xij xij 2 ( xij x j )2 x j = i=1 s j = i =1 x = nj nj 1 nT n ANOVA Related: t= Chapter 10 continued: Test Statistic (Matched Samples) SSR p t= MSE = R2 = SSE n p1 bi sbi 0! = 1 x0 = 1 F= ln1 = 0 lne = 1 x! = ( x )( x 1)( x 2) (2)(1) e x = exp( x ) Other Math Rule Reminders: MSR = SST = SSR + SSE SSR SST n 1 R2a = 1 (1 R2 ) n p1 y = b0 + b1 x1 + b2 x2 + + b p x p MSR MSE E ( y ) = 0 + 1 x1 + 2 x2 + + p x p y = 0 + 1 x1 + 2 x2 + + p x p + e Residual for Observation i: yi yi Chapter 13: y t/2 spred Chapter 12 continued: Prediction Interval for y : Use the output below to answer the appropriate questions in the exam. Be sure to check which table the question references. ANOVA Table 1 Treatments Error SS DF MS F p 31.162 46.91 4 16 7.791 2.932 2.657 0.0712 Here is regression output from a model where attendance is the independent variable and regular course points (excludes extra credit from polls) is the dependent variable. The data are from my Spring 2016 classes where I am using poll everywhere points as a proxy for attendance. Many students choose not to come to class but I rather think attending the lectures is helpful. Let's see what the data say. Regression Table 2 Regression Statistics R Square Adjusted R Square Standard Error Observations 0.2196685 0.2185966 99.0777428 730 ANOVA Regression Residual intercept attendance SS DF MS F p 2011741.4 7146338.563 1 728 2011741.4 9816.399 204.937 0 coefficients standard error t stat p-value 717.4464638 6.4083776 0.4687815 111.95446 14.31561 0 0 6 ECN221 Exam 3 VERSION A Summer 2016 (Chapters 1-10, 12), ASU-COX VERSION A Choose the best answer. Do not write letters in the margin or communicate with other students in any way. If you have a question note it on your exam and ask for clarification when your exam is returned. In the meantime choose the best answer. Neither the proctors nor Dr. Cox will answer questions during the exam. Please check each question and possible answers thoroughly as questions at the bottom of a page sometimes run onto the next page. Please verify that your test version and scantron version are the same. This exam has 25 questions. 7 1. I have checked that my ID is bubbled in correctly. If it is bubbled in incorrectly I will get this question wrong. I also understand that questions and their possible answers may run onto the next page and so I should always check the top of the next page for possible answers. I understand that if I have a question I should simply make a note on my exam and ask Dr. Cox afterwards. I should always choose the best answer. (a) False. (b) I didn't read the directions. (c) True. 2. Consider the null hypothesis H0 : 1 = 2 and a separate hypothesis H0 : 1 = 2 +.5. Suppose that you are going to conduct hypothesis tests for both of those hypotheses at the .05 level of significance. The critical value(s) for each test is (are) the same. (a) true (b) false 2.5 2.0 1.5 size 3.0 3.5 3. The graphs shown here are A B C drug 8 D (a) histograms (b) scatter plots (c) stem and leaf displays (d) box and whiskers plots 4. The probability that a z value is less than 0.5 is (a) 0.5 (b) 0.8413447 (c) 0.4640237 (d) 0.6616608 (e) 0.6914625 5. Consult Table 1. From the table you can conclude that the total sum of squares is ? (a) 46.91 (b) 78.072381 (c) 15.747619 (d) 39.0361905 6. Consult Table 1. From the table you can conclude that the total number of observations used in this analysis/experiment was? (a) 19 (b) 21 (c) 4 (d) 16 7. Consult Table 1. From the table what can you conclude concerning the null hypothesis? (a) fail to reject (b) rejcet the null (c) cannot be determined (d) depends on the number of observations. 8. Consult Table 1. What is the test statistic? 9 (a) 0.071234. (b) 7.7905952. (c) 2.6572058. (d) 1.3286029. 9. Consult Table 1. What is the critical value for a test at the .05 level? (a) 1.96 (b) 3.84 (c) 6.39 (d) 3.01 10. Suppose that the number of homework assigments a professor gives follows a Poisson distribution with a mean of 10. What is the probability of drawing a professor that gives exactly 12 assignments? (a) 0.012764 (b) 0.1143679 (c) 0.072765 (d) 0.0947803 11. Suppose that you collect data on apartment prices in Tempe. You look at 36 different apartments and find a mean of 738 and a standard deviation of 193.5. Construct a 95% confidence interval for the mean apartment price. The interval is (a) [ 672.5290193, 803.4709807 ] (b) [ 676.5290193, 799.4709807 ] (c) [ 666.5290193, 809.4709807 ] (d) [ 605.2761174, 723.1238826 ] 12. Suppose that you collect data on apartment prices in Tempe. You look at 36 different apartments and find a mean of 738 and a standard deviation of 193.5. Test the hypothesis, H0 : = 705 at the .05 level of significance. (a) the test statistic is 1.0232558 so we rejcet the null (b) the test statistic is 0.7409302 so we fail to reject (c) the test statistic is 1.0232558 so we fail to reject (d) the test statistic is 1.2355814 so we rejcet the null 10 13. Suppose you have a random variable that is exponentially distributed with a mean of 50. What is the probability of observing a random variable drawn from this distribution with a value of less than 50? (a) 0.6321206 (b) 0.6988058 (c) 0.550671 (d) 0.8646647 14. The greater the value of the greater the risk of committing a Type II error. (a) True. (b) False. 15. Consider the regression output in Table 2. What is the estimated increase in total points that a student earns (on average) when attending 1 additional lecture? (a) 5.1622272 (b) 99.0777428 (c) 6.7108954 (d) 204.937 (e) 9.5869935 16. Consider the regression output in Table 2. What is the estimated variance of the error term? (a) 9816.3991246 (b) 49.5388714 (c) 99.0777428 (d) 204.937 17. Consider the regression output in Table 2. What is the percentage of variation in total regular class points that can be explained by the variation in attendance? (a) 20.9668468% (b) 26.3602162% (c) 21.9668468% (d) 21.8596585% 11 18. Consider the regression output in Table 2. What is the predicted or estimated total regular class points for a student who came 15 times? (a) 838.2425814 (b) 885.2188493 (c) 818.1098951 (d) 804.6881043 (e) 797.9772088 19. Consider the regression output in Table 2. Suppose you want to test the hypothesis that attendance has no impact on grades, i.e. H0 : 1 = 0. What is the test statistic for this hypothesis? (a) 14.3156138 (b) 17.6082049 (c) 0.4687815 (d) 0 (e) 111.9544623 20. Consider the regression output in Table 2. Suppose you want to test the hypothesis that attendance has no impact on grades, i.e. H0 : 1 = 0. What is you conclusion for this hypothesis test? (Use = .05.) (a) This is inconclusive unless we know whether it is a right tail or a left tail test. (b) This cannot be determined without the appropriate df. (c) rejcet the null (d) fail to reject 21. Suppose that for all values of your eplanatory variable that are above the 75th percentile you notice that the residuals are positive. (a) This cannot happen because the residuals must average to 0. (b) The relationship between the X and Y variables is probably not linear. (c) There is an indicator that there may be an issue with heteroskedasticity. (d) This is an indicator that the error terms are probably correlated. (e) none of the above. 12 22. You are in a business meeting and a colleague presents regression results in the following form y = 12 + 2.3X. You know that the regression was performed with over 100,000 observations and your colleague reports that the p-value was .011 so that it is statistically significant. However, the executive to whom you are presenting asks what the standard error was. The value is not in your power point slides but you tell her you will give it to her in just a moment. While your colleague moves to the next slide you find that it is: (a) 1.0618639. (b) 0.4719395. (c) 1.2585053. (d) 0.9045507. (e) 1.4158185. 23. In our treatment of regression analysis we made which assumption(s) (a) Y is normally distributed. (b) X is normally distributed. (c) \u000f is normally distributed. (d) \u000f and X are normally distributed. (e) all of the above. 24. The following graph shows the total points and poll everywhere points together. What kind of graph is this? 13 900 800 850 points 950 students with at least 800 points 0 5 10 15 20 25 attendance (a) scatter plot (b) har chart (c) dot plot (d) stem and leaf display 25. The line that is superimpossed in the graph above is the regression line from Table 2. Looking at the graph below concerning total class points and poll everywhere points (a proxy for attendance), which of the following is most accurate? Note: the graph below uses a subset of the observations in the graph above. 14 940 900 920 points 960 students with at least 900 points 0 5 10 15 20 25 attendance (a) The residuals are normally distributed (b) The residuals are not normally distributed. (c) Students with 15-20 times attending tend to do better in the class than the regression line predicts. (d) Students that got over 900 points (A range) attended over half the time on average. 15 Key 1. c 2. a 3. d 4. e 5. b 6. b 7. a 8. c 9. d 10. d 11. a 12. c 13. a 14. b, refers to the probability of a Type I error not a Type II error. 15. c for all of the questions requiring reading the regression output you obviously need to know how to read the output in excel as shown in class. For this one you needed to work backwards from the test statistic and the standard error. 16. a, 17. c 18. c 19. a 20. c 21. b, this suggests that the relationship might be different at the 75th percentile and beyond and so it is not linear through out. 16 22. d, with over 100,000 observations the t distribution is essentially the same as the z distribution. If p = .011 the you can look on the z table and find that the test statistic was 2.54 and then you back this out to find a standard error of 2.3/2.54 is approximately 0.9055118. 23. c 24. a 25. d 17