Research Topics: Math Samples

Math/Statistic Assignment Questions

Math/Statistic Assignment Questions

Quantitative Methods for Business 1) George & Jim were brothers who were each left a will inheritance of ,000. G...

Maths/Statistic Assignment

Quantitative Methods for Business   1)      George & Jim were brothers who were each left a will inheritance of ,000.  George invested his money in an account for 5 years at an annual interest rate of 6.0% compounded quarterly.  Jim invested his money for 5 years at an annual simple interest rate of 6.75%.  At the end of that time they agreed to poor their accumulated principals into one account for a further 5 years with an annual interest rate of 5.0% compounded monthly.   a)            How much accumulated principal did George and Jim each have after 5 years?  Hence, determine how much money they had to pool.   Answers:   To determine the accumulated principal did George and Jim each have after 5 years, we will use the simple interest formula and compound interest formula.   i.e. for George we have, Where F is the final amount or the sum due at the end of the transaction; P is the original principal; r is the interest rate in one period of time; and t is the time in which the number of period during which the principal is used expressed in years. Then we have On the other hand, we have to use the compound interest formula for Jim’s investment for 5 years at an annual simple interest rate of 6.75% to determine his final amount. Then we have, Where F is the final amount or the sum due at the end of the transaction; P, is the original interest rate per period; and n, the number of conversion periods within the entire length of the transaction. The compound interest rate j is usually expressed as the rate per year and the interest is understood to be converted once a year if the frequency of conversion is not specified.  The frequency of conversion m is the number of times interest compounded in year. The interest rate per period is expressed as: From this we have,                         Thus the amount of money they have to pool is ,000 + , 949.97 = ,949.97.   b)            What was the accumulated principal of their money at the end of the 10-year period? Answer:   Using the formula  we have,   c)            In exercise (b), when George and Jim received their inheritance, the executor of the will suggested that they pool their resources immediately for 10 years and invest in an account that paid an annual interest rate of 6.75% compounded monthly.  In view of your answer (b), should they have taken this advice? Explain.   Answer:               With respect to the answer in (b), let’s try to compare the final amount to be accumulated in years using 6.75% of interest compounded monthly of the principal ,000. Then we have,                         From these values George and Jim should take the advice of the executor since the amount of money to be accumulated at the end of the 10th year is much bigger than their preferred investment. Actually, the suggested investment preference of the executor is ,175.7 bigger compared to preference of George and Jim at the end of 10th year.   2)      A cricket team decide to purchase a house where they can hold their functions near their oval.  They take out a mortgage of 0,000 from the Ruptcy Bank, which charges on annual interest rate of 5%, compounded semi-annually over 10 years.   a)            What payments will the team have to make every 6 months on this loan?   Answers:                         Let       I = ?                 P= 0,000                t=10                 j = 5%              m = 2         i = 0.025              n = 20   Then we have,     6,633.97/20 = ,831.6985   The team have to pay ,831.6985 every 6 months on this loan.   b)            What is the total amount of interest they will pay?   Answers:           3)      In Exercise 2, the rival Fleece Finance offers the cricket team a loan for which it will charge on annual interest rate of 4.8% compounded monthly over 10 years.   a)            What payments will the team have to make every month on this loan?   Answers:                         Let       I = ?                 P= 0,000                t=10                 j = 4.8%                       m = 12         i = 0.004              n = 120   Then we have,     3,743.34/120 = ,614.53   The team have to pay ,614.53 every month on this loan.   b)            What is the total amount of interest they will pay?   Answers:             4)      In view of your answers to Exercise 2 and 3, from which lending institution should the cricketers take their loan?   From the previous answers, Fleece Finance should be considered by the cricket team for this load since the interest is much lower compared to Ruptcy Bank.   5)      Scott, Anne and Jane form a partnership in which Scott invests ,000.  Anne invests ,000 and Jane invests ,000.   a) Scott and Jane are each to receive a salary of ,000 out of the profit, while the remaining profit is to be divided between the three in the ratio of their investments.  Find out how much each partner will receive in total if the profit is 0,000.   Answers:               From the given amount of investment (i.e. 0,000), Scott actually invested 35% of their total investment, 18% for Anne and 47% for Jane.   Thus, their shares in the profit of 0,000 are:   Scott:  120,000 x 0.35 = 42,000 Anne:  120,000 x 0.18 = 21,600 Jane:  120,000 x 0.47 = 56,400   Since Scott and Jane also receive a salary of ,000 each out of the profit, then they will have:   Scott:  120,000 x 0.35 = 42,000 + 25,000 = ,000 Anne:  120,000 x 0.18 = ,600 Jane:  120,000 x 0.47 = 56,400 + 25,000 = ,400           <p class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph; text-indent: -.25in; text-autospace: none; line-height: 200%; margin-left: .25in; margin-right: 0in; margin-top: 6.0pt; margin-bottom: 6.0pt"> 6)      Next week you have to travel around parts of the country on business and have decided to hire a car from a car-hire company. You have contacted two companies which offer different services. The first company will rent you a car for 5 per day. The second company will charge per day, but with an additional charge of 50 cents per kilometre travelled. You know you will require the car for four days but are unsure of the number of kilometres that you will cover.   (a) Determine mathematically what distance you need to cover to make the second company’s charge cheaper.   Answer:             Let us consider that the car will be used in 4 days.  Consider the following breakdown per company.   Company A (5/day) = $ 500 in 4 days Company B (/day+.50/per km.) = 0 in 4 days.               With these values, there is 0 discrepancy between Company A and B regardless of the per km charges of company B.  Meaning to say, I only need to travel (i.e. 180/0.5) 360km to make the total charges of Company B become 0. Thus, I only need to cover 359km. to make the second company’s charge cheaper.                     (b) Confirm this using Excel graphs.   Answer:     (c) Both companies now realise that they will need to add 10% GST to the charge for the daily car hire, but not to the charge for the kilometres covered. How will this affect your decision? Justify your answer with both the calculations and Excel graphs.   Answer:               Let us consider that the car will be used in 4 days.  Consider the following breakdown per company in accordance to 10% additional GST. Company A (5/day +10%GST) = $ 550 in 4 days Company B (/day+10%GST +.50/per km.) = 2 in 4 days.               With these values, there are now 8 discrepancy between Company A and B regardless of the per km charges of company B.  Meaning to say, I only need to travel (i.e. 198/0.5) 396 km to make the total charges of Company B become 0. Thus, I only need to cover 395 km. to make the second company’s charge cheaper.   7) You are the manager of a toy company. Your company manufactures a variety of soft toys and you are trying to improve the efficiency of your operations. You wish to adhere to an annual operating budget and to develop operating plans using quantitative methods wherever possible. The overall budget for the year for production of two different soft toys is 0,000. The soft toys are koalas and kangaroos. The budget is intended to pay for labour and materials. Processing requirements for the two toys, on a per unit basis, are shown in the table.   Product Grey cloth White cloth Stuffing Labour Koala 1.5 metres 0.5 metres 350gms 1.75 hours Kangaroo 1.0 metres 1.0 metres 400gms 1.50 hours   The production costs for the toys are:   Material Cost Labour/hour .50 Grey material/metre .50 White material/metre .50 Stuffing/100gms .35   The company has a contractual agreement to produce a minimum of 3,000 toys per year. There must be at least 200 of each toy produced. The company can only get a maximum of 5,000 metres of each material. You wish to minimize the number of labour hours required to meet your contractual obligation.   Use Excel Solver to obtain a solution to this problem. Your submission should include: ·         your mathematical set-up of the objective function and constraints ·         your Excel spreadsheet ·         the sensitivity report ·         the answer report   Use your Excel output to answer the following questions: <li class="MsoNormal" style="margin-top: 3.0pt; margin-bottom: 12.0pt"> What is the amount of labour hours required to meet the contractual agreement? </li> <li class="MsoNormal" style="margin-top: 3.0pt; margin-bottom: 12.0pt"> What are the optimal quantities of each toy? </li> <li class="MsoNormal" style="margin-top: 3.0pt; margin-bottom: 12.0pt"> If you only wish to employ staff for a maximum of 5 hours per day. How many staff should you employ if there are 48 working weeks in the year? How many hours per week would each staff member work? </li> <li class="MsoNormal" style="margin-top: 3.0pt; margin-bottom: 3.0pt"> As the manager you realize that labour per unit is actually the same for koalas as it is for kangaroos (1.50 hours). Does this change the optimum number of toys of each kind to be produced? Does the optimum number of labour hours change? </li>   Credit:ivythesis.typepad.com

Maths/Statistic Assignment

Quantitative Methods for Business 1) George & Jim were brothers who were each left a will inheritance of ,000. G...

SPSS

1. Using the T test of means, the difference of the data can be measured. The table 1 is to measure the difference of the weights using the program 1. The analysis of the data is given below: One-Sample Test   Test Value = 0               t df Sig. (2-tailed) Mean Difference 95% Confidence Interval of the Difference               Lower Upper   WBP 11.422 14 .000 86.0667 69.9057 102.2276   WAP 27.142 14 .000 67.8000 62.4423 73.1577     The program 2 is given below: One-Sample Test   Test Value = 0               t df Sig. (2-tailed) Mean Difference 95% Confidence Interval of the Difference               Lower Upper   WBP 28.332 14 .000 80.6000 74.4983 86.7017   WAP 26.115 14 .000 69.0000 63.3332 74.6668     The program 3 is given below: One-Sample Test   Test Value = 0               t df Sig. (2-tailed) Mean Difference 95% Confidence Interval of the Difference               Lower Upper   WBP 35.610 14 .000 77.8000 73.1141 82.4859   WAP 28.365 14 .000 63.8000 58.9759 68.6241     Based on the data above the most effective way of weight loss is the program 1 for having the most number of the weight loss.   Question 1.B Correlation One of the statistical techniques that can be shown on the relationship of the two variables is related is through the use of correlation. This means that the correlation analysis is the method used in measuring the strength and its relationship between the two variables with the aid of the single number which is so called the correlation coefficient. As an example, there is a relationship between the height and the weight of the people thought it is not perfect. This means that the people of the same height will vary on their weight. This can also signifies that the correlation can also tell the variability of the weight of the people as it relates in their height. There are also different types of correlation techniques as the Survey System’s Optional Statistics Module that includes most of the common type as the product-moment and the Pearson. The correlation is also appropriate for any data and it can works for the data wherein the numbers are meaningful as it contains in the quantities of some sort. Most of the statistician tells that it is not applicable for the rating scale but some still using it though it is required to have so much care. The correlation coefficient is the main result of the correlation analysis or simply “ r “ which ranges from + 0.01 to + 0.99 this means that as the r is closer to +1 of -1, the variables are also closely related. This means that if the r is close to 0, then, the variables have no correlation. In getting positive value of r, it only means that the one variable gets larger the other also gets larger. In the case of negative value of r, it is sometimes called the inverse correlation. Squaring the value r yield to the percentage of the variation of one variable as it relates to the other variable. This means that an r value of 0.7 has a 49% of the related variance. The second part is the technique of Pearson correlation which is usually use in the linear relationship wherein as the variable gets larger the other also gets larger or it can be smaller but in the direct proportion (Creative Research System, 2002). The formula for finding the value of r is:                         Source: Research Methods Knowledge Base   In doing the said technique, it is important to determine that the correlation only affects only one subject and not to the others. This means that violations occur if one chose the half of the groups’ subject and other half of the other because there will affect also the half subjects and to the other half. There is also an invalid calculations if the value of X and Y are intertwined as the value of the midterm to the overall scores wherein the midterm is one of the criteria in the overall scores. Controlling the values of X signifies that it is better to calculate the linear regression and not the correlation.     Simple Linear Regression The regression projects the distribution of the variable which is called response and with the aide of the one or more predictors. In the incident of studying one predictor to its relationship to the response variable, then it is called the simple regression analysis. The components of the model of simple linear regression is forecasted to be the relationship between the variable of X and Y or some of the transformation of X. This means that air of he (X,Y) is being observed for unit of n and can yield to a sample of the pair (x1, y1), (x2,y2)…..(xn,yn). This can be graph in the scatter plot and can also provide the clue for the possible relationship of X and Y. This also means that the scatter plot will be a huge clue for eh possible relationship of the Y and X. In order to understand the analysis and the concept of regression, it is advisable to determine the simple linear regression model for the investigation of the response relationship variable of Y and X.  (Astro Temple, n.d.). The standard regression equation is: y = mx +b  where: y = predicted value Fitting the regression equation into the set of data has a reason of describing the data and to predict the response of the carrier. The regression line had been calculated because it can be seen and use for to the prediction. This line can also gives the good fit for the set of data when the points are close to it. This means that value which are obtained from the line must be close to it that to other line. In assessing for the fit of the line, the vertical distances of the points and to the other lines are the only one that matters. The line of regression is also called to be the least square regression equation. This also means that the sum of the squares in its residual must be least. There is also importance of fitting large number of lines trough the aide of trial and error in finding the best fit. This can also be minimized trough the line for which   or it can also be computed manually (Dallal, 2000).  The regression can also be use for prediction wherein the confidence interval can be a best way in assessing the quality of the prediction. In the prediction using the regression, the confidence interval in the single forecast value of the Y is also corresponds to the chosen value of X and to the straight pint of line. Estimating the value of the m and b needs to use the criterion of least-squared-error that will want to find in estimating the value of m and Y which can also minimize the difference between the predicted and the observed values for all of the value of X. The error can be found through searching for procedure in the evaluated and propose different values of m and b. The other way is trough the approach of using the calculus in finding the equations for estimating the m and b (IBM, 2003). To make it short, the simple linear regression is the modeling technique to study further due to fact that it is the main way for the understanding to the more advanced forms of modeling statistics. This can also be a versatile modeling technique which can be use in modeling the curvilinear data through the transformation of the raw of data which are commonly having the logarithmic or the transformation power. The said transformation can also make the data linear in order for using the simple linear regression in modeling the data. The results of the model which is linear can be expressed through the linear formula that relates to the transformed values (Ibid). In the given problems being solved, the regression and the correlation had been responsible in determining the relationship of the programs and the weights of the clients. This can also be used in forecasting the value of the weights as it relates to the given programs.   Question 2. Chi-Square Technique a. The chi-square technique is use for the investigation of the difference of the categorical variable distribution. Normally, the variable can yield the data into numerical variable or to categories in the numerical form. On the other hand, the chi square statistics had been done in comparing for the counts and for the tallies of the categorical response between the independent group of two or more (Eck, 2001). The formula for the chi-square statistics is: The technique of chi-square can be use in determining for the worth of the effort of the researcher in interpreting the contingency table. This is therefore required in the interpretation of the results.   b. For the more precise use of chi-square, the example below shows the incidence of the three types of Malaria in the three tropical regions. The table of values are given below:    Asia Africa South America Totals  Malaria A 31 14 45 90  Malaria B 2 5 53 60  Malaria C 53 45 2 100  Totals  86 64 100 250                                     Source: Mathbeans Project by Eck, 2001 The following table had now been set up, Observed Expected |O -E|  (O — E)2  (O — E)2/ E  31  30.96  0.04  0.0016  0.0000516  14  23.04  9.04 81.72 3.546  45  36.00  9.00 81 2.25  2  20.64  18.64 347.45 16.83  5  15.36  10.36 107.33 6.99  53  24.00  29.00 841 35.04  53  34.40  18.60 345.96 10.06  45  25.60  19.40 376.36 14.7  2  40.00  38.00  1444.00 36.1                                     Source: Mathbeans Project by Eck, 2001 The computed value of the chi square is 125.516 with the degree of freedom of (c-1)(r-1) = 4. In this particular problem, the chi-square distribution table is given below: Df 0.5 0.1 0.05 0.02 0.01 0.001 1 0.455 2.706 3.841 5.412 6.635 10.827 2 1.386 4.605 5.991 7.824 9.21 13.815 3 2.366 6.251 7.815 9.837 11.345 16.268 4 3.357 7.779 9.488 11.668 13.277 18.465 5 4.351 9.236 11.07 13.388 15.086 20.517                         Source: Mathbeans Project by Eck, 2001     In this particular study, the decision is that the null hypothesis needs to reject because there is no relationship between the malaria and to the location. In the sample problem, it had been expected that the there is a relationship of the place and the decease which resulted of having no relationship base on the chi square method.   c. The technique of using the chi-square is that it can be seen as to general. This can also be applied for any of the distribution either trough continuous or for discrete wherein the function of cumulative distribution is can be computed. This means that the data plot can support the chi-square for all of the distribution whereas it can support the function of CDF. Nevertheless, the chi square can be sensitive on the performance of binning the data as well as it requires enough sample size in order to get 5 as the minimum expected frequency. In using the chi square technique, misleading or incorrect analysis can be done if there is a violation on the assumptions of goodness of fit. Some of the examples of the potential assumption violations are the lack of independence, the structural zero, the outliers, special problems which has the continuous variables, and the expected cell frequencies for the chi-square test expected are small.      Bibliography Chi Square 2001, Math Beans Project, Eck, viewed 21 April, 2008, http://math.hws.edu/javamath/ryan/ChiSquare.html. Correlation 2002, Creative Research Systems, viewed 21 April, 2008, http://www.surveysystem.com/correlation.htm. Correlation 2006, Research Methods Knowledge Base, viewed 21 April, 2008, http://www.socialresearchmethods.net/kb/statcorr.php. Simple Linear Regression 2003, IBM, viewed 21 April, 2008, http://www.ibm.com/developerworks/web/library/wa-linphp2/. Simple Linear Regression 2000, TUFTS, viewed 21 April, 2008, http://www.tufts.edu/~gdallal/slr.htm. Simple Linear Regression n.d., Astro Temple, viewed 21 April, 2008, http://astro.temple.edu/~jagbir/regression1.pdf. Credit:ivythesis.typepad.com

SPSS

1. Using the T test of means, the difference of the data can be measured. The table 1 is to measure the difference of the weights using t...