Course Hero has millions of student submitted documents similar to the one

below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

512 Stat 2
Solutions to Homework #8
Dr. Simonsen
Due October 26, 2005 by 4:30pm The first three problems use the Filling Machines dataset from Problem 16.11 of KNNL described on page 725, and continue the analysis begun on Homework #7. 1. Use the Tukey multiple comparison method to determine which pairs of machines differ significantly. Summarize the results. The Tukey comparison method shows that machines 3 and 4 are significantly different from machines 1, 2, 5, and 6 (on a pairwise basis).
The GLM Procedure Tukey's Studentized Range (HSD) Test for wtdev NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher Type II error rate than REGWQ. Alpha 0.05 Error Degrees of Freedom 114 Error Mean Square 0.03097 Critical Value of Studentized Range 4.09949 Minimum Significant Difference 0.1613 Means with the same letter are not significantly different. Tukey Grouping Mean N machine A A A B B B B B B B 0.46000 0.36550 0.19050 0.15150 0.12500 0.07350 20 20 20 20 20 20 3 4 2 6 5 1
2. Suppose you want to compare the average of the first two machines with the average of the last four. Use the ESTIMATE and CONTRAST statements in PROC GLM to test the appropriate hypothesis. Report the estimated value of this contrast with its standard error; state the null and alternative hypotheses, the test statistic with degrees of freedom, the Pvalue and your conclusion. The contrast is L = 2 4 ^ ^ The estimated value of this contrast is L = -0.1435 with standard error s L = 0.03408 . We test
( 1 + 2 ) - ( 3 + 4 + 5 + 6 ) .
{}
the null hypothesis ( 1 + 2 ) - ( 3 + 4 + 5 + 6 ) = 0 vs H : H0: a 2 4 [These could also be correctly written as H0: L = 0 vs. Ha: L 0 or
( 1 + 2 ) - ( 3 + 4 + 5 + 6 ) 0 .
2 4
H0:
( 1 + 2 ) = ( 3 + 4 + 5 + 6 )
2 4
vs. Ha:
( 1 + 2 ) ( 3 + 4 + 5 + 6 ) ].
2 4
The test statistic is either F = 17.73 with 1,114 df or t = 4.21 with 114 df, and the P-value is P < 0.0001. We reject H0 and conclude that the mean for the first two machines is not the same as the mean for the last four machines.
The GLM Procedure Dependent Variable: wtdev Contrast prob2 Parameter prob2 DF 1 Contrast SS 0.54912667 Mean Square F Value Pr > F 0.54912667 17.73 <.0001 Standard Error t Value Pr > |t| 0.03407905 -4.21 <.0001
Estimate -0.14350000
3. Check assumptions using the residuals. Turn in the plots/output you used to check the assumptions and state your conclusions.
r esi d 0. 4
0. 3
0. 2
0. 1
0. 0
- 0. 1
- 0. 2
- 0. 3
- 0. 4 1 2 3 m achi ne 4 5 6
The assumptions of normality and constant variance appear to be satisfied, as shown by the residual plot and qqplot. [Grader: the residual plot may be plotted vs. machine or vs. predicted value. Only one is necessary, plus the qqplot.]
The remaining problems use the Helicopter Service dataset from Problem 18.15 on page 804 of KNNL. Helicopter service. An operations analyst in a sheriff's department studied how frequently their emergency helicopter was used during a recent 20-day period, by time of day (shift 1: 2 am 8 am; shift 2: 8 am 2 pm; shift 3: 2 pm 8 pm; shift 4: 8 pm 2 am). The data follow (in time order). Since the data are counts, the analyst was concerned about the normality and equal variances assumptions of ANOVA model (16.2). 4. Here we will examine the equal variances assumption. (a) Fit a one-way ANOVA model and give the (four) predicted values.. The fitted values are the means for the four levels as shown below.
The GLM Procedure Level of shift 1 2 3 4 -------------use------------Mean Std Dev 3.90000000 1.15000000 2.00000000 3.40000000 1.97084006 1.08942283 1.45095250 1.78885438
N 20 20 20 20
(b) Prepare suitable residual plots to study whether or not the error variances are equal for the four shifts. What do you conclude?
r esi d 4
r esi d 4
3
3
2
2
1
1
0
0
-1
-1
-2
-2
-3
-3
-4 1 2 pr edi ct 3 4
-4 1 2 shi f t 3 4
The residual plot shows that the variances are not constant. The plot of residuals vs. predictors shows that the variances tend increase with the predicted value. [Grader: only one of these plots is required] (c) Use the modified Levene's test whether the error variances are equal. Report the Pvalue and a conclusion. Are your results consistent with your conclusion from (b)? The P-value of the modified Levene test is 0.1344, so we do not reject the null hypothesis of equal variances; however note that the P-value is fairly small This failure to reject does not support the conclusion of unequal variances reached in part b.
The GLM Procedure Levene's Test for Homogeneity of use Variance ANOVA of Absolute Deviations from Group Means Sum of Squares 4.8394 64.0395 Mean Square 1.6131 0.8426
Source shift Error
DF 3 76
F Value 1.91
Pr > F 0.1344
(d) Section 18.5 of KNNL gives a simple guide to selecting a suitable transformation for constant variance problems. Prepare a table similar to the one on page 791 where si2 si si , , are tabulated for each level of the factor. Choose an appropriate Yi Yi Yi2 transformation (square root, log, reciprocal) of Y depending on which of these is the most stable (i.e. approximately constant across levels). shift Yi i level 1 2 3 4 The ratio 3.90 1.15 2.00 3.40 si 1.971 1.089 1.451 1.789
si2 Yi i 0.996 1.032 1.053 0.941
si Yi i
si Yi i2
0.130 0.824 0.363 0.154
0.505 0.947 0.725 0.526
si2 appears the most consistent across levels. This indicates that a square root Yi i transformation may be appropriate.
m e a n u s e s q 15.2100 1.3225 4.0000 11.5600 l o g m e a n u s e 1.36098 0.13976 0.69315 1.22378 l o g s t d u s e 0.67846 0.08565 0.37222 0.58158 r o o t r a t i o 0.99595 1.03204 1.05263 0.94118 l o g r a t i o 0.50534 0.94732 0.72548 0.52613 r e c i p r a t i o 0.12958 0.82376 0.36274 0.15475
O b s 1 2 3 4
s h i f t 1 2 3 4
_ T Y P E _ 0 0 0 0
_ F R E Q _ 20 20 20 20
m e a n u s e 3.90 1.15 2.00 3.40
v a r u s e 3.88421 1.18684 2.10526 3.20000
s t d u s e 1.97084 1.08942 1.45095 1.78885
5. A rather simple approximation of the Box-Cox procedure is the following: (i) Compute the mean and standard deviation for each treatment factor level. (ii) Take the log of both the mean and standard deviation. (iii) Fit the regression model log ( i ) = 0 + 1 log ( i ) + using the observed means and standard deviations as the data for i and i respectively (there are 4 "observations" in this dataset).
^ (iv) Set = 1 - b1 where b1 is the estimate for 1 obtained in (iii).
Use the Helicopter service data to perform this approximation. What value of appears reasonable according to this method?
^ The regression model gives the estimate b1 = 0.47, and so = 1 - 0.47 = 0.53 0.5 . This also suggests a square root transformation.
The REG Procedure Model: MODEL1 Dependent Variable: logstduse Parameter Estimates Parameter Standard Estimate Error 0.02766 0.02261 0.47029 0.02305
Variable Intercept logmeanuse
DF 1 1
t Value 1.22 20.40
Pr > |t| 0.3458 0.0024
6. Define a new response variable by adding 1 to the original response. (This will avoid 0's which mess up the log and reciprocal transformations.) Then use SAS's Box-Cox procedure to determine an appropriate transformation. Proc transreg can be used to perform ANOVA if we tell it shift is a class variable, as in the following: proc transreg data=helicopter; model boxcox(usesplus1) class(shift);
The = TRANSREG Procedure Transformation Information for BoxCox(usep1) Lambda -3.00 -2.75 -2.50 -2.25 -2.00 -1.75 -1.50 -1.25 -1.00 -0.75 -0.50 -0.25 0.00 0.25 0.50 + 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 R-Square 0.14 0.15 0.15 0.16 0.17 0.18 0.20 0.21 0.23 0.24 0.26 0.28 0.30 0.31 0.32 0.32 0.33 0.33 0.32 0.32 0.31 0.30 0.29 0.28 0.26 Log Like -182.106 -165.922 -150.178 -134.942 -120.291 -106.320 -93.144 -80.897 -69.735 -59.830 -51.364 -44.517 -39.442 -36.250 * -34.991 < -35.645 * -38.129 -42.313 -48.035 -55.124 -63.413 -72.748 -82.989 -94.017 -105.728
< - Best Lambda * - Confidence Interval + - Convenient
The Box-Cox procedure suggests that = 0.5, i.e. a square root transformation, would be appropriate. 7. NKNW 18.16 (Omit the coefficient of correlation in part b). In part (c), Brown-Forsythe is the same as Modified Levene).): Refer to Helicopter service Problem 18.15. The analyst decided to apply the square root transformation Y = Y and examine its effectiveness. (a) Obtain the transformed response data, fit ANOVA model (16.2), and obtain the residuals.
The GLM Procedure Dependent Variable: rootuse Source Model Error Corrected Total R-Square 0.288945 Sum of DF Squares Mean Square F Value 3 13.60854916 4.53618305 10.29 76 33.48888826 0.44064327 79 47.09743742 Coeff Var Root MSE rootuse Mean 46.66182 0.663810 1.422597 The GLM Procedure Level of shift 1 2 3 4 -----------rootuse----------Mean Std Dev 1.87136414 0.84265033 1.22925287 1.74712038 0.64725840 0.68051097 0.71740569 0.60486655 Pr > F <.0001
N 20 20 20 20
(b) Prepare suitable plots of the residuals to study the equality of the error variances of the transformed response variable for the four shifts. Also obtain a normal probability plot and the coefficient of correlation between the ordered residuals and their expected values under normality. What are your findings? Does the transformation appear to have been effective?
r esi d 1 r esi d 1
0
0
-1
-1
-2 0. 8 0. 9 1. 0 1. 1 1. 2 1. 3 1. 4 1. 5 1. 6 1. 7 1. 8 1. 9
-2 1 2 shi f t 3 4
pr edi ct
There has been an improvement in the variances of the transformed variable, but there are still signs that the variance increases with the predicted value. [Grader: only one of the two above plots is necessary, either vs. shift or vs. predictor]
re d si 1
0
-1
-2 -3 -2 -1 Rn fo a k r 0 1 2 3
V ri a l e re d a b si
The QQ-plot shows some skewness away from normality. The transformation stabilized the variance somewhat, however, it also appears to have destroyed the normality of the residuals. (c) Test by means of the modified Levene test whether or not the treatment error variances for the transformed response variable are equal; use = 0.10. State the alternatives, decision rule, and conclusion. Are your findings in part (b) consistent with your conclusion here?
The GLM Procedure Levene's Test for Homogeneity of rootuse Variance ANOVA of Absolute Deviations from Group Means Sum of Squares 0.3652 11.3771 Mean Square 0.1217 0.1497
Source shift Error
DF 3 76
F Value 0.81
Pr > F 0.4905
The Levene test is now very far from significant. The P-value is 0.49, which is a big increase over the 0.13 for the untransformed data. This gives some evidence that the errors are more homogeneous after the transformation. (However, the Levene test is sensitive to the normality assumption, and since we know normality is violated, we cannot really trust the results.) 8. Use the Tukey multiple comparison method for differences in means on both the untransformed and square-root transformed Helicopter service data (from Problem 7) to determine which shifts differ significantly. Summarize and compare the results. Using the transformed data we find that the following pairs are significantly different: 1 with 2 and 3 2 with 1 and 4
Tukey's Studentized Range (HSD) Test for rootuse Alpha 0.05 Error Degrees of Freedom 76 Error Mean Square 0.440643 Critical Value of Studentized Range 3.71485
Minimum Significant Difference Tukey Grouping Mean N A 1.8714 20 A B A 1.7471 20 B B C 1.2293 20 C C 0.8427 20
0.5514 shift 1 4 3 2
Using the untransformed data we find that shifts 1 and 4 differ significantly from shifts 2 and 3.
Tukey's Studentized Range (HSD) Test for use Alpha 0.05 Error Degrees of Freedom 76 Error Mean Square 2.594079 Critical Value of Studentized Range 3.71485 Minimum Significant Difference 1.3379 Tukey Grouping Mean N shift A A A B B B 3.9000 3.4000 2.0000 1.1500 20 20 20 20 1 4 3 2
It is interesting that in the transformed data 3 and 4 are not significantly different but in the untransformed data they are.
SAS Code
*Homework 8 * Problems 1-3; title1 'detergent'; data detergent; infile 'H:\Stat512\Datasets\Ch16pr14.dat'; input wtdev machine j; proc glm data = detergent; class machine; model wtdev = machine; output out=det2 r=resid; means machine / tukey; estimate 'prob2' machine 0.5 0.5 -0.25 -0.25 -0.25 -0.25; contrast 'prob2' machine 0.5 0.5 -0.25 -0.25 -0.25 -0.25; symbol1 v=circle; proc gplot data=det2; plot wtdev*machine; plot resid*machine/vref=0; proc univariate data=det2 noprint; var resid; qqplot /normal(mu=est sigma=est L=1); run; *Problems 4 - 5; * Read in the data; title1 'helicopter'; data helicopter; infile 'h:\stat512\datasets\ch18pr15.dat'; input use shift; proc print data=helicopter; * Run ANOVA, store the residuals and predictors, do the modified Levene test; proc glm data=helicopter; class shift; model use=shift; output out=diags p=predict r = resid; means shift / hovtest=levene(type=abs); * Plot the residuals vs. factor or vs. predictor; symbol1 v = circle; proc gplot data=diags; plot resid*shift/vref=0; plot resid*predict/vref=0; proc univariate data=diags normal; var resid; qqplot / normal (mu=est sigma=est); * Store the means, variances, and standard deviations for each shift; proc means data=helicopter; by shift; output out=heli2 mean = meanuse var = varuse std = stduse; * Store variations on the basic variables; data heli2; set heli2; meanusesq = meanuse*meanuse; logmeanuse = log(meanuse); logstduse = log(stduse); rootratio = varuse/meanuse; logratio = stduse/meanuse; recipratio = stduse/meanusesq; proc print data=heli2;
* Run regression on log(sigma) vs. log(mu) to get b1 = 1 - lambda; proc reg data=heli2; model logstduse = logmeanuse; * Problem 6; data helicopter; set helicopter; usep1 = use + 1; proc print data=helicopter; run; proc transreg data=helicopter; model boxcox(usep1) =class(shift); run; *Problem 7 (18.16); * Use the square root transformation on the data; title1 'helitrans'; data helitrans; set helicopter; rootuse = sqrt(use); * Run ANOVA, store the residuals and predictors, do the modified Levene test; proc glm data=helitrans; class shift; model rootuse = shift; output out=heli3 r = resid p = predict; means shift / hovtest = levene(type = abs); * Plot residuals vs. factor level or vs. predictor; symbol1 i = none; proc gplot data=heli3; plot resid*(shift predict)/vref=0; * Check normality with a qqplot; proc univariate data=heli3 normal; var resid; qqplot / normal (mu = est sigma = est L = 1); run; * Use Tukey to find which means differ significantly; proc glm data=helitrans; class shift; model rootuse = shift; means shift / tukey; run; proc glm data=helicopter; class shift; model use = shift; means shift / tukey; run; quit;

**Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.**

Below is a small sample set of documents:

Purdue - MA - 373

Chapter 2, Section 21.Yu invests 1000 at a nominal interest rate of 6% compounded monthly. At time T , Yu has 2000. Calculate T in years. Nic invests 700 in an account earning an annual effective interest rate of i . After 8 years, Nic's account has a v

Purdue North Central - STAT - stat512

Stat 512 2Solutions to Homework #9Dr. SimonsenDue November 2, 2005 by 4:30pm 1. KNNL 19.4: In a two-factor study, the treatment means ij are as follows: Factor B Factor A B1BB2BB3BA1 A234 4023 2936 42a. Obtain the factor A level means.^ 1i =

Purdue - MA - 373

Chapter 2, Section 35.Kunqi takes a loan 4000 with an annual effective interest rate of 4%. The loan is repaid with a payment of P at the end of one year and 2000 at the end of two years. Calculate P. Zach repays a loan of 100,000 with a payment of 60,0

Purdue North Central - STAT - stat512

Stat 512 2Solutions to Homework #10Dr. SimonsenDue Wednesday, November 9, by 4:30 pm Problems 1 3 continue the analysis begun on Homework #9. 1. Refer to the Hay Fever Relief dataset from problem 19.14. The table below gives the exact quantities of ing

Purdue - MA - 373

Chapter 2, Section 412.Kyle has agreed to pay Aaron 1000 now. In two years, Aaron will pay Andrew a payment of 600. At the end of four years Aaron will pay Kyle a payment of 700. Also, at the end of four years, Andrew will make a payment of 800 to Kyle

Purdue North Central - STAT - stat512

Stat 512 2Solutions to Homework #11Dr. SimonsenDue Wednesday, November 16, 2005, by 4:30pm For these problems use the Electronics Assembly data described in problem 24.12 on page 1025 of KNNL (this is 23.12 in NKNW). 1. Run the full three-way analysis

Purdue - MA - 373

Chapter 2, Section 515.Kelli loans 2000 to Adam today. Adam repays the loan by making two payments: One payment of 1500 at the end of two years and a second payment of 1200 at the end of four years. Calculate the interest rate on this loan from Adam's s

Purdue North Central - STAT - stat512

Stat 512 2Solutions to Homework #12Dr. SimonsenDue Friday, December 2, 2005 by 4:30pm1. KNNL 25.3: In each of the following cases, indicate whether ANOVA model I or model II is more appropriate and state your reasons: (a) In a study of absenteeism at

Purdue - MA - 373

Chapter 2, Section 6Use the following information about an investment fund for problems 17 19: Time 0 1/4 1/2 3/4 117. 18.Fund Value Before Contributions 10,000 8,000 8,000 12,000 15,000Contributions 0 -3,000 3,000 3,000 0Calculate the exact annual d

Purdue North Central - STAT - stat512

PHYS 272: Matter and Interactions II - Electric And Magnetic Interactionshttp:/www.physics.purdue.edu/academic_programs/courses/phys272/PHYSICS 272 Electric & Magnetic InteractionsSection 1 MW 8:30-9:20AM Prof. Yong P. Chen yongchen@purdue.edu Room:74,

Purdue - MA - 373

Chapter 2, Section 722.Sheila invests 1000 in a fund. One year later her investment is worth 800. At that time, Sheila invests an additional 500 into the fund. Two years after her initial investment of 1000, her fund is worth 2000. Calculate Sheila's ti

Purdue North Central - STAT - stat512

PHYS 272: Matter and Interactions II - Electric And Magnetic Interactionshttp:/www.physics.purdue.edu/academic_programs/courses/phys272/PHYSICS 272 Electric & Magnetic InteractionsLecture 2 Superposition Principle & Electric Dipoles [EMI 14.4-14.8]Fal

Purdue - MA - 373

Chapter 3 Section 21.(S12HW) Calculate the present value of an annuity that pays 1000 at the end of each year for 10 years using an annual effective interest rate of 8%. (S12HW) Calculate the present value of an annuity that pays 100 at the end of each

Purdue North Central - STAT - stat512

PHYS 272: Matter and Interactions II - Electric And Magnetic Interactionshttp:/www.physics.purdue.edu/academic_programs/courses/phys272/PHYSICS 272 Electric & Magnetic InteractionsLecture 3 Charged Objects; Polarization of Atoms; Induced Dipoles [EMI 1

Purdue - MA - 373

Chapter 3 Section 314.(S12HW) Shuda is the beneficiary of a trust fund which will pay her 1000 at the beginning of each month for the next 5 years. Calculate the present value of these payments assuming an annual effective interest rate of 7.2%. (S09Q2)

Purdue North Central - STAT - stat512

PHYS 272: Matter and Interactions II - Electric And Magnetic Interactionshttp:/www.physics.purdue.edu/academic_programs/courses/phys272/PHYSICS 272 Electric & Magnetic InteractionsLecture 4 Conductors and Insulators [EMI 15.4-15.7]Fall 2010 Prof. Yong

Purdue - MA - 373

Chapter 3 Section 421.(S12HW) A perpetuity immediate pays 10,000 per month.The interest rate earned by the perpetuity is 8% compounded monthly. Calculate the present value of this perpetuity. (S12HW) Mohamad is the beneficiary of a trust fund. Mohamad a

Purdue - MA - 373

Chapter 3, Section 528.(S11HW) A monthly annuity immediate pays 100 per month for 12 months. Calculate the accumulated value 12 months after the last payment using a nominal rate of 4% compounded monthly. (S11HW) A monthly annuity due pays 100 per month

Purdue - MA - 373

Chapter 3, Section 634.(S12HW) Yue bought a house with a 200,000 mortgage for 15 years being repaid with payments at the end of each month at an interest rate of 6% compounded monthly. What is the outstanding balance at the end of 10 years immediately a

Purdue - MA - 373

Chapter 3, Section 744.(S11HW) An annuity pays 100 at the end of each month for 4 years and then 200 a month at the end of each month for the next four years. Calculate the present value at i(12) = 0.09. (S11HW) An annuity pays 100 at the end of each mo

Purdue - MA - 373

Chapter 3, number 843.(S08T1) Kurt is the beneficiary of a trust. Under the trust, he will receive payments at the end of each year for the next 20 years. The payment will be 2000 at the end of one year. Each subsequent payment will increase by 8%. In o

Purdue - MA - 373

Math 373 Spring 2011 Quiz 1January 27, 20111. Jacob invests 1000 in an account earning simple interest of 4%. Cassie invests X in an account earning 2% compounded annually. In year Y, Cassie and Jacob earn the same annual effective interest rate. At the

Purdue - MA - 373

Math 373 Spring 2011 Quiz 2February 8, 20111. Danny, Sandy, and Julia enter into a financial agreement. Today, Danny will pay Sandy 1000. He will also pay Julia 2000 today. At the end of 3 years, Julia will pay Danny 2500. Julia will also pay Sandy 2500

Purdue - MA - 373

Math 373 Spring 2011 Quiz 3March 1, 20111. Elaine is repaying a loan of 55,000 with annual payments of 6000 plus a smaller final payment. The annual effective interest rate on the loan is 7%. Calculate Elaine's smaller final payment.2. Bryce invests X

Purdue - MA - 373

Math 373 Quiz 4March 29, 20111. A 10 year callable bond has a par value of 1500. The bond matures for par and has a coupon rate of 6% convertible semi-annually. The bond may be called on the following dates and amounts: End of Year 5 6 7 8 9 Amount 1568

Purdue - MA - 373

Math 373 Quiz 5April 13, 20111. Bryce wants to sell 500 shares of Guo International stock. At the exact same moment, Austin wants to buy 500 shares of Guo International. Bryce's broker charges a commission of 1% of the proceeds from the sale of the stoc

Purdue - MA - 373

Math 373 Quiz 7April 21, 20111. The stock of Hunter Corporation currently sells for 150. Edyta enters into a one year floor on Hunter stock. You are given: a. The annual effective risk free interest is 6%. b. The premium for a one year Put on Hunter wit

Purdue - MA - 373

Quiz 1 Math 373September 2, 2010 1. Ryan invests 1200 at a simple interest rate of 5%. Lena invests X at a compound interest rate of 10%. At the end of 10 years, Ryan and Lena each have the same amount. Calculate X.2. Candace grandfather has promised to

Purdue - MA - 373

Quiz 2 Math 373September 14, 2010 1. The Chin Chip Company is looking at building a new factory. Chin's actuary, Taylor, has estimated the following cash flows from the new factory:Time of Cash Flow 0 1 2 3 4 5Cash Flow (in millions) -48 -4 +20 +30 +15

Purdue - MA - 373

Quiz 3 Math 373September 23, 20101. Adam has won the lottery! He will receive 20 annual payments of 100,000 beginning today. Adam takes each payment and invests it at an annual effective interest rate of 8%. How much will Adam have at the end of 20 year

Purdue - MA - 373

Quiz 4 Math 373October 13, 20101. A loan of 100,000 is being repaid with level annual payments of 13,000 for 10 years. Calculate the principle in the 5th payment.2. A nine year continuous annuity pays at a rate of 70t at time t. The discount function i

Purdue - MA - 373

Quiz 5 Math 373November 2, 20101. Li Life Insurance Company sells preferred stock with a quarterly dividend of 6.50. The next dividend is payable in 3 months. Calculate the price of the preferred stock using an annual effective yield of 12%.2. The comm

Purdue - MA - 373

Quiz 6 Math 373November 18, 20101. Aiman sells 200 shares of TAK LTD at the same time that Liyana purchases 200 shares. The Bid Price for the shares of TAK was 9.75. Aiman paid a commission of 60. Liyana paid a commission of 2%. The total transaction co

Purdue - MA - 373

Quiz 7 Math 373December 1, 20101. The current price of Pace Industries is S0. You enter into a synthetic forward by buying a call and selling a put. The put and call both have a strike price of 100 and an expiration date in one year. The net premium pai

Purdue - MA - 373

1. Adam invests in a fund that earns 5% over the next year. During that year, the rate of inflation is 8%. Calculate Adam's real interest rate on his investment.2. Linda borrows 25,000 to buy a car. The loan will be repaid with 60 monthly payments of 634

Purdue - MA - 373

Math 373 Test 2October 21, 20101. A 20 year zero coupon bond has a price of 100,000 at an annual effective interest rate of 3.6%. Calculate the maturity value of the bond.2. Candace is repaying a loan with level monthly payments of 1000. The interest r

Purdue - MA - 373

Math 373 Test 3November 9, 20101. A callable bond matures in 10 years for 10,000. The bond has semi-annual coupons of 400. The bond is callable based on the following schedule:Call Date 6 Years 7 Years 8 Years 9 YearsCall Premium 700 500 300 150Calcu

Purdue - MA - 373

Purdue - MA - 373

Purdue - MA - 373

Purdue - MA - 373

Purdue - MA - 373

QUIZ 2Math 373Spring 20101. You are given that i(4) = 8.0%. Provide all answers to 6 decimal places. a. Calculate the quarterly effective interest rate.b. Calculate the annual effective interest rate.c. Calculate the monthly effective interest rate.

Purdue - MA - 373

Purdue - MA - 373

Purdue - MA - 373

City College of San Francisco - CS - CS270

A. HISTORY OF THE SOFTWARE=Python was created in the early 1990s by Guido van Rossum at StichtingMathematisch Centrum (CWI, see http:/www.cwi.nl) in the Netherlandsas a successor of a language called ABC. Guido remains Python'sprincipal author, altho

Purdue - MA - 373

Purdue - MA - 373

Purdue - MA - 373

QUIZ 5Math 373Spring 20101. Chupp Corporation pays a quarterly dividend of 10 later today. Future dividends are expected to increase at a rate of 2% per quarter. Calculate the price of Chupp Stock (prior to the payment of the dividend today) in order t

Purdue - MA - 373

Purdue - MA - 373

QUIZ 6Math 373Spring 20101. Circle any of the following that are NOT a derivative: An automobile insurance policy IBM stock The right to sell corn in six months for a price of $7 per bushel A United States Treasury Bond2. Jordan short sells Hardwick I

Purdue - MA - 373

Purdue - MA - 373

Purdue - MA - 373

InstructionsDO NOT OPEN THE TEST UNTIL TOLD TO DO SO. Put your answer (A, B, C, D, or E) on the Answer Sheet. Do NOT remove the answer sheet from the packet. If you want full credit for those correct or Partial Credit for those that are wrong You must sh

Purdue - MA - 373

Purdue - MA - 373

Test 3 Math 373 Spring 20101. Tian purchased a 10 year callable bond with a par value of 10,000. The bond matures for par and pays semi-annual coupons at a rate of 8% per year. The bond is also callable at the end of years 6 through 9. The bond is callab

Purdue - MA - 373

Purdue - MA - 373

1. Wang Corporation invests 10 million today to build a new factory. Thereafter, Wang expects to receive the following profits from the factory: End of Year 1 2 3 4 5 6 Profits - 1 million 2 million 4 million 4 million 4 million 2 millionThe factory will

Purdue - MA - 373

Math 373 Test 2March 10, 2011InstructionsDO NOT OPEN THE TEST UNTIL TOLD TO DO SO.If you want full credit for those correct or Partial Credit for those that are wrong You must show your work. You may only use the approved calculators. Use of other cal

Purdue - MA - 373

Math 373 Test 3April 5, 2011InstructionsDO NOT OPEN THE TEST UNTIL TOLD TO DO SO.If you want full credit for those correct or Partial Credit for those that are wrong You must show your work. You may only use the approved calculators. Use of other calc

Purdue - MA - 373