# Vn19 Assignment Submission

**Unformatted text preview: **10/2/2015 Graded homework E r d 1 Graded homework [feedback page] This attempt will impact your course performance This is a feedback page. You have NOT yet finished your assignment. Please read the following paragraph carefully. When you are ready, you must complete this assignment by clicking finish. You will then see your final score. Once you have clicked finish you will not be able to return to this feedback page, so please ensure that you print or save it to your computer if you want to refer to it later. If you do not click finish your score will not be displayed on your e-workbook home page. However, your results will be provided to your instructor. 1 of 10 ID: MST.CI.CIM.06.0010a As a stock analyst, your boss, Jerry, has asked you to compile some information on stock of Southern Infrastructure Corporation including a 95% confidence interval for the mean daily return that he needs to include in a report to senior management. He says that he is also not sure exactly what a 95% confidence interval means and would like you to add an explanation. You have been following the share price of Southern Infrastructure Corporation and have recorded the daily return (as a percentage) for the last 120 days. The data is presented here: Download the data 1.609 2.078 -0.143 0.236 2.1 1.806 0.598 0.378 -0.454 2.213 2.22 2.17 0.646 1.115 1.239 1.024 1.726 0.636 0.845 0.853 1.367 1.708 1.268 2.029 1.872 1.504 1.222 0.694 1.725 0.719 | Daily returns (% -0.133 0.554 0.703 1.029 0.523 0.513 1.157 1.308 0.105 1.011 0.932 2.214 1.914 1.116 0.15 0.591 -0.134 1.125 0.667 0.403 0.753 1.436 0.808 1.373 0.407 0.599 1.875 0.707 2.041 2.556 2.699 1.143 1.644 0.328 1.992 1.509 0.979 0.124 1.61 0.969 0.585 2.154 1.146 1.379 -0.285 0.155 0.584 0.784 0.818 1.091 1.43 -0.466 1.303 1.843 0.214 1.549 1.802 0.672 0.853 0.553 0.659 1.45 0.237 2.57 2.333 1.393 0.238 1.86 0.788 3.009 1.669 1.659 0.289 2.165 1.762 0.658 24 1.036 1.738 0.046 1.37 -0.228 0.121 1.177 2.689 0.306 0.434 1.266 1.541 Historically, the standard deviation in daily return for this stock is 0.8%. Complete the report to your boss. Give your numeric answers to 3 decimal places. Sent: October 2, 2015 11:10 AM To: Jerry Kendall Subject: Southern Infrastructure Corp. stock info hﬂp:/Mww.perdisco.comlel ms/qsam/htm l/qsam .ast 1/14 10/2/2015 Graded homework Dear Jerry, Here are the results gathered from the collected data: Assuming a population standard deviation in daily return of 0.8%, the 95% confidence interval for the mean daily return is: a) 0.973 S u S 1.259 b) This means that approximately 95% of sample means will be within the interval given above using a process that gives correct results in 95% of cases, the population mean daily return is within the interval given above the population mean daily return is definitely within the interval given above on approximately 95% of days in a given period the stock makes a return within the interval given above Feedback a) You are correct. b) You are correct. Discussion a) Using a statistical software package, the following results can be obtained: Sample Statistics sample mean 1.1159 population standard deviation 95% confidence interval (0.9727..., 1.2590...) Alternatively, the 95% confidence interval can be found in the following way: show variables _ o 0.8 :l: * — = 1.11 :I: 1. x z xvn 585 96 xvlzo 1.11585 :l: 1.96 x 0.07302967... 1.11585 :l: 0.14313816... 0.97271184... 1.25898816... 0.973 s p 5 1.259 Rounded as last step IA "C IA b) The idea behind confidence intervals is as follows: - the Central Limit Theorem says that, when sampling from a population with mean u and standard deviation 0, for large enough sample sizes the sample mean is http:/Mww.perdisco.comlel ms/qsam/htm l/qsam .ast 2/14 10/2/2015 Graded homework approximately normally distributed with mean u and standard deviation o/x/n - since 95% of the normal distribution is contained within 1.96 standard deviations of the mean, it follows that 95% of sample means are within 1.96xo/w/n of u - therefore, for 95% of samples, u is within 1.96xo/x/n of the sample mean So for the confidence interval found in part a) you can say that using a process that gives correct results in 95% of cases, the population mean daily return is within the interval given above. 2 of 10 ID: MST.HT.TM.02.0040b Each year, all final year students take a mathematics exam. It is hypothesised that the population mean score for this test is 90. It is known that the population standard deviation of test scores is 10. A random sample of 22 students take the exam. The mean score for this group is 84. You may find this standard normal table useful throughout the following questions. a) Calculate the 95% confidence interval for the population mean test score. Give your answers to 2 decimal places. 79.82 S u S 88.18 b) At a significance level a = 0.05, the null hypothesis that the population mean test score is 90 is rejected V . Feedback a) You are correct. b) You are correct. Calculation a) The significance level a = 0.05 confidence interval can be found in the following way: show variables _ o 10 :l: * — 84 :l: 1.96x X Z X_\/n 84 :l: 1.96 X 2.13200716... 84 :l: 4.17873404... 79.82126596... 88.17873404... 79.82 s u 5 88.18 Rounded as last step IA 'C IA b) The (two-sided) hypothesis that the population mean test score is 90 can be tested (at a significance level a = 0.05) by inspecting whether the hypothesized value falls within the (95%) confidence interval. If the hypothesized value lies in the interval then the null hypothesis is not rejected. If it lies outside of the interval then the null hypothesis is rejected. In this case, the hypothesized value of 90 is outside the confidence interval. Therefore, the null hﬂp:/Mww.perdisco.comlel ms/qsam/htm llqsam .ast 3/14 10/2/2015 Graded homework hypothesis is rejected. This means that, based upon the sample data, there is enough evidence to conclude that the mean test score is not equal to 90. That is, the hypothesized value of the mean score is not supported by the sample data. 3 of 10 ID: MST.CI.CIM.03.0010a The maternity ward of the Royal Prince of Persia Hospital is doing a study on the average gestation times of the babies born to its patients. A sample of 55 past patients in the ward have been randomly selected and the gestation times of their babies recorded. The sample mean gestation time was calculated as 268 days. The population standard deviation of gestation times of all babies is known to be 19 days. It is assumed that this standard deviation will also apply specifically to the patients of the maternity ward of the Royal Prince of Persia Hospital. Calculate the upper and lower bounds of the 95% confidence interval for the mean gestation time of babies born in the maternity ward of the Royal Prince of Persia Hospital. You may find this standard normal table useful. Give your answers in days to 2 decimal places. a) Upper bound = 273.02 days b) Lower bound = 262.98 days Feedback a) You are correct. b) You are correct. Calculation A 95% confidence interval is constructed such that if all possible samples of a certain size are collected and a confidence interval for each of those samples is calculated, approximately 95% of those confidence intervals would contain the population mean. Thus approximately 5% of the confidence intervals would not contain the population mean. This 5% can be divided into two equal parts, one for the upper 2.5% of samples whose sample means generate confidence intervals that do not contain the population mean and one for the lower 2.5%. The standardized values corresponding to an upper and lower tail of 2.5% can be obtained using the standard normal table. According to this table, the standardized value corresponding to the upper 5% (that is, the standardized value corresponding to a value of 0.975 in the table) is 1.96. By symmetry, the standardized value corresponding to the lower 5% is -1.96. a) The upper bound can be calculated using the following formula: show variables ;+*x—O Z \/ n U 19 268 + 1.96 x \/55 273.02144058... 273.02 days Rounded as last step b) The lower bound can be calculated using the following formula: show variables hﬂp:/Mww.perdisco.comlel ms/qsam/htm l/qsam .ast 4/14 10/2/2015 Graded homework — 0' L: +*x— X Z vn 19 268 + (-1.96) x #55 262.97855942... 262.98 days Rounded as last step 4 of 10 ID: MST.CI.CIM.02.0010a A sample of 21 Menso members have had their Intelligence Quotients (IQs) measured and recorded. The data were used to calculate a 98% confidence interval for the mean IQ of all Menso members. The confidence interval was calculated as 142 :I: 15. According to this confidence interval, it is most reasonable to conclude that: menso INTERNATIONAL you are 98% confident the interval between 127 and 157 contains the mean IQ of all Menso members there is a 98% chance that the mean IQ of all Menso members is between 127 and 157 98% of all Menso members have an IQ between 127 and 157 the mean IQ of all Menso members is between 127 and 157 Feedback You are correct. Discussion A 98% confidence interval for the mean is created using sample data to estimate bounds for the value of the population mean. Once the interval is constructed it is possible to make statements about where the population mean probably lies. It is not possible to make definite statements because a sample is only an approximate representation of the population. The 98% conﬁdence interval for the mean IQ of all Menso members was calculated as 142 :I: 15. This means that if every possible sample of size 21 was selected from the population of Menso members and a confidence interval was constructed for each of those samples, approximately 98% of those conﬁdence intervals would contain the population mean. In other words, you are 98% confident the interval calculated contains the mean IQ of all Menso members. 5 of 10 ID: MST.HT.TM.05.001O Select whether a one-tailed or two-tailed hypothesis test is most appropriate in the following situations: hﬂp:/Mww.perdisco.comlel ms/qsam/htm llqsam .ast 5/14 10/2/2015 Graded homework a) Fantra wants to test whether the mean amount of wastage during I its manufacturing process is less than 200 gallons per day. b) Holdem Motors wants to test whether the mean time to assemble a car is less than 24 hours. c) The Roads and Traffic Authority wants to test whether the mean number of speeding fines on a particular road is greater than 10 4 per day. Feedback Legend 9‘ You are correct. This option should have been selected. I This is not correct. Discussion A hypothesis test is constructed to test the validity of a proposed (or hypothesized) value of a population parameter. This proposed value is chosen based on the situation at hand. A hypothesis test involves two hypotheses; the null hypothesis and the alternative hypothesis. The null hypothesis represents the status quo. That is, what you assume to be true. The alternative hypothesis represents what is trying to be shown. One-tailed hypothesis tests A one-tailed hypothesis test is used when the statistician is specifically interested in whether the population mean (u) is greater than a hypothesized value (or alternatively, whether u is less than a hypothesized value). Two-tailed hypothesis tests A two-tailed hypothesis test is used when the statistician is interested in whether u is simply not equal to a hypothesized value (and the statistician is not concerned whether p is specifically greater than the hypothesized value or whether u is specifically less than the hypothesized value). 6 of 10 ID: MST.HT.TM.02.0060a Batteries R Us is a manufacturer of batteries and are testing a new production technique for their most popular battery type. It is known that the mean lifetime of these batteries made using the previous production technique was 68 hours. The standard deviation in battery lifetime is 5.8 hours and it is assumed that this has not changed. They would like to know whether mean battery lifetime has increased with the new production technique. The following sample data of battery lifetimes has been collected from a random sample of 40 batteries made using the new manufacturing technique. hﬂp:/Mww.perdisco.comlel ms/qsam/htm llqsam .ast 6/14 10/2/2015 Graded homework Download the data Battery lifetimes (hours) 676695 72.9 74.2 67.9 68.2 68.1 71.6 72.1 - 69.8 79.7 65.8 70.8 m 64.7 75.9 4.5 3. 66.8 76.4 79.4 69.8 70.5 68.3 65.8 71.8 70.362.4 68.9 62.9 63.3 1.4 72.7 66.4 61.4 Conduct a hypothesis test to test whether the new manufacturing process has increased mean battery lifetime. You may find this standard normal table useful. a) From the following options, select the correct null and alternate hypotheses for this test: A: H0: p = 68, Ha: p < 68 B: H0: 11 = 68, Ha: p > 68 C: H0: p > 68, Ha: p = 68 D: H0:p=68,Ha:p=#68 The correct null and alternate hypotheses for this test are: B V b) Calculate the test statistic. Give your answer to 2 decimal places. 2 = 1.43 c) Calculate the p-value for the test. Give your answer to 4 decimal places. P-value = 0.0764 d) Therefore, at a significance level of 0.05 the null hypothesis is not rejected V . e) At a significance level of 0.05, from the result of the test you can conclude that there is: significant evidence V to conclude that the population mean battery lifetime V has increased V . Feedback a) You are correct. b) You are correct. c) You are correct. d) You are correct. e) You are partly correct. From the result of the test you can conclude that there is not enough evidence to conclude that the http://www .perdisco.comlel m s/qsam lhtm llqsam .ast 7/14 10/2/2015 Graded homework population mean battery lifetime has increased. Discussion a) In a hypothesis test for the population mean (or any parameter), the null hypothesis typically represents the 'status quo' belief; it states that things are the way they have always been. In this question, the status quo belief is that the mean lifetime of these batteries made using the previous production technique was 68 hours. However, it is suspected that batteries made with the new production technique may have a higher mean lifetime. Therefore, the null hypothesis states that the mean lifetime is equal to 68 hours while the alternative hypothesis states that the mean battery lifetime is actually higher than 68 hours. That is, the correct null and alternate hypotheses for this test are those given in option B: Ho: p = 68 Ha:p > 68 b) Using a statistical software package, the following results can be obtained: Sample Statistics sample mean 69.3075 population standard deviation 5.8 test statistic P-value 0.07697011... Calculating the t Statistic c) The P-value can be found using software or calculated as the probability that Z > 1.43 where Z is a standard normal random variable. In this case since the test is one-sided and the alternate hypothesis is u > 68 you only consider Z > the test statistic. Using either software or a standard normal table you find that P(Z > 1.43) = 1 - P(Z < 1.43) z 1 - 0.9236 = 0.0764 d) The null hypothesis is rejected if the P-value is less than the level of significance of the test. This is because the level of significance defines the standard by which statisticians decide that there is enough evidence to suggest that the null hypothesis is false. In this test the level of significance is 0.05. Therefore, the null hypopthesis is not rejected at significance level of 0.05 since the P-value is greater than the level of signiﬁcance used in the test. e) This means that the sample does not provide enough evidence (as determined by the level of significance of the test) to conclude the alternate hypothesis. That is, from the result of the test you can say that there is not enough evidence to conclude that the population mean battery lifetime has increased. 7 of 10 ID: MST.HT.TM.02.0010 hﬂp:/Mww.perdisco.comlel ms/qsam/htm llqsam .ast 8/14 10/2/2015 Graded homework Painadol is a well trusted, popular pain-killer used by millions throughout the country. Recently an epidemic of headaches has hit the nation and the manufacturers of Painadol have created a new, stronger painkiller called Painadene. The manufacturers are interested in testing whether the speed of pain relief is different with Painadene. It is known that the mean time taken by Painadol tablets to relieve pain is 33 minutes and the standard deviation is 5 minutes. The manufacturers would like to construct a hypothesis test for the mean time (u) taken for Painadene tablets to relieve pain assuming the same population standard deviation (0) as the Painadol tablets. A random sample of 33 people with headaches tried the new Painadene tablets and the mean time taken to relieve the pain was calculated as 32 minutes. The hypotheses that will be used by the manufacturers are H0: p = 33 and Ha: p 1: 33. You may find this standard normal table useful throughout the following questions. a) Calculate the test statistic (2) that corresponds to the sample and hypotheses. Give your answer as a decimal to 3 decimal places. = l—1.149 b) Using the test statistic for Painadol's hypothesis test and level a = 0.1, Painadol should accept V the null hypothesis. a) You are correct. b) This answer is not possible. According to the process of statistical hypothesis testing, it is only possible to reject the null hypothesis (if there is enough evidence to suggest that the null is false) or not reject the null hypothesis (if there is not enough evidence to suggest that the null is false). The option of accepting the null hypothesis is not available because it is not possible to prove that the null hypothesis is true. Using the test statistic for Painadol's hypothesis test and level a = 0.1, Painadol should not reject the null hypothesis. Calculation a) The test statistic can be calculated using the following formula: show variables Y - No L \/n 32 - 33 i w/ 33 -1.14891253... -1.149 Rounded as last step hﬂp:/Mww.perdisco.comlel ms/qsam/htm llqsam .ast 9/14 10/2/2015 Graded homework b) The hypothesis test carried out by the manufacturers of Painadol is a two-tailed test because the alternative hypothesis is simply concerned with whether u is not equal to 33 minutes, as opposed to being specifically concerned with whether u is greater than 33 minutes (or alternatively, whether u is less than 33 minutes). For a two-tailed test, the null hypothesis is rejected if the P-value of the test statistic is less than the level of significance of the test. This is because the level of significance defines the standard by which statisticians decide that there is enough evidence to suggest that the null hypothesis is false. The P- value for this test statistic is equal to 2 x Probability(z > 1.149) = 0.2506. According to the test statistic for Painadol's hypothesis test, the null hypothesis should not be rejected at level a = 0.1 since 0.2506 > 0.1. This means that, based upon the results of the test, there is not enough evidence to conclude that the mean time taken to relieve pain is not equal to 33 minutes. That is, the hypothesized claim is not contradicted by the results of the test. 8 of 10 ID: MST.HT.TM.02.0020 The Mean Corporation would like to invest in the booming health food J— industry. It is considering the creation of a health-drink franchise called Goose Juice. The investment department of the Mean Corporation wants to investigate the feasibility of this venture by examining the profits of similar franchises. It believes that the venture will be feasible if an average annual profit of more than $73,000 can be expected from each Goose Juice that is opened. It is known that the annual profits earned by health-drink franchises has a population standard deviation of $7,800. Mean Cur oration The Mean Corporation's statisticians would like to construct a hypothesis test for the mean annual profit (u) earned by health-drink franchises. A random sample of 30 franchises were chosen and their annual profit for the previous financial year was recorded. The mean annual profit for the sample was calculated as $74,850. The hypotheses that will be used by the statisticians are Ho: u = 73,000 and Ha: p > 73,000. You may find this standard normal table useful throughout the following questions. a) Calculate the test statistic (2) that corresponds to the sample and hypotheses. Give your answer to 3 decimal places. 2 = 1.299 b) Using the test statistic for the Mean Corporation's hypothesis test and level a = 0.05, the Mean Corporation should not reject V the null hypothesis. c) If the sample size is increased to 70 (but the sample mean remains unchanged), the Mean Corporation should reject V the null hypothesis. Feedback a) You are correct. b) You are correct. c) You are correct. hﬂp:/Mww.perdisco.comlel ms/qsam/htm llqsam .ast 10/14 10/2/2015 Graded homework Calculation a) The test statistic can be calculated using the following formula: show variables X ' lJO = i x/n 74,850 - 73,000 = 7,800 x/ 30 1.29908555. .. 1.299 Rounded as last step Z b) The hypothesis test carried out by the Mean Corporation is a one-tailed test because the alternative hypothesis is specifically concerned with whether 0 is greater than $73,000. For this one-tailed test, the null hypothesis is rejected if the P-value of the test statistic is less than the level of significance of the test. This is because the level of significance defines the standard by which statisticians decide that there is enough evidence to suggest that the null hypothesis is false. The P- value for this test statistic is equal to Probability(z > 1.299) = 0.097. According to the test statistic for the Mean Corporation's hypothesis test, the null hypothesis should not be rejected at level a = 0.05 since 0.097 > 0.05. This means that there is not enough evidence to conclude that the mean annual profit is not equal to $73000. That is, the hypothesized claim that the mean annual profit is $73000 is not contradicted by the results of the test. c) If the sample size is 70 then (if all other variables remain constant) the new test statistic can be calculated using the following formula: show variables X-Ho 0' vnnew 74,850 - 73,000 7,800 \/ 70 1.984385%. .. 1.984 Rounded as last step Znew Resulting in a new P-value of 0.0236. Therefore, the null hypothesis should be rejected at level CI = 0.05 since 0.0236 5 0.05. This means that there is enough evidence to conclude that the mean annual profit is not equal to $73000. In fact, since this is a one-tailed test, it means that there is enough evidence to conclude that the mean annual proﬁt is greater than $73000. 9 of 10 ID: MST.HT.TM.01.0030 A climate researcher knows that it has previously been the case that the mean temperature in his city has been 77 °F. He suspects that this may no longer be the case and that the mean temperature may now have changed, hﬂp:/Mww.perdisco.comlel ms/qsam/htm llqsam .ast 1 1/14 10/2/2015 Graded homework and he suspects that is has risen in recent times. He collects sample data with the aim of conducting a hypothesis test to examine his suspicions. He will use a level of significance of 0.01 for the test. Select the correct null and alternate hypotheses for this test: Ho:u=77°F H0:u¢77°F Ha:u¢77°F Hazu=77°F H0:u=77°F H0:u>77°F Ha:u>77°F Ha:u=77°F Feedback You are correct. Discussion In a hypothesis test for the population mean (or any parameter), the null hypothesis typically represents the 'status quo' belief; it states that things are the way they have always been. The null hypothesis is usually denoted H0 and is typically listed first. The alternative hypothesis, which usually motivates the test itself, reflects a belief that the status quo may have changed. The alternative hypothesis is usually denoted Ha and listed second. In this question, the status quo belief is that the mean temperature is 77 °F. However the climate researcher has a suspicion that the mean temperature has increased. Therefore, the null hypothesis states that the mean temperature is 77 °F while the alternative hypothesis states that the mean temperature is actually higher than 77 °F. That is, the hypotheses for the test are: Ho:p = 77°F Ha: [I > 77°F 10 of 10 ID: MST.HT.TM.03.0020 A statistician would like to construct a hypothesis test of the mean using the P-value approach. Rank the steps that the statistician can follow to carry out this hypothesis test at a level of significance of a. Note that there may be more than one correct answer. Define the null and alternative hypotheses n Collect sample data Calculate the P-value n Define the distribution of the test statistic hﬂp:/Mww.perdisco.comlel ms/qsam/htm llqsam .ast 12/14 10/2/2015 Graded homework I Calculate the test statistic l 4 v \ Feedback You are correct. Discussion Define the null and alternative hypotheses The first step in carrying out a hypothesis test is to determine the null and alternative hypotheses. The hypotheses must be constructed before the data is collected and analyzed. It is not reasonable to collect the data, determine what value of the parameter should be tested such that the null will be rejected, and then simply test that value. To do so would be to defeat the purpose of hypothesis testing because it would be possible to fix the results. In essence you would no longer be carrying out a test. You would simply be calculating a point estimate for the mean and verifying that point estimate to be valid. Define the distribution of the test statistic Once the null and alternative hypotheses have been defined, the corresponding sampling distribution of the mean can be determined. Which sampling distribution is relevant depends on what parameter is being tested. For a hypothesis test of the mean, the most common sampling distribution of the mean is the normal distribution. The sampling distribution of the mean is then used to determine the distribution of the test statistic. The most common distributions of the test statistic are the standardized normal distribution and the Student's t distribution. The standardized normal distribution is used if the value of the population standard deviation is known. The Student's t distribution is used if the value of the population standard deviation is unknown. Collect sample data Once the hypotheses have been determined, the sample data to be analyzed can be collected. For the results of the hypothesis test to be valid, this data must be collected from a random sample and any bias must be minimized. Calculate the test statistic Using the sample data that has been collected, the test statistic can be calculated. For a hypothesis test of the mean, the calculation of the test statistic requires the preliminary calculation of the sample mean. If the population standard deviation is unknown, the sample standard deviation will also need to be calculated. Calculate the P-value Once the test statistic has been calculated, the P-value can be calculated as the probability that a similar test statistic (or one more extreme) is obtained assuming that the null hypothesis is true. This P-value is obtained using the defined distribution of the test statistic. Draw a conclusion Once the test statistic and P-value have been calculated, a conclusion can be made as to whether the null hypothesis will be rejected or not rejected. If the P-value is less than the level of significance, the null hypothesis will be rejected. Otherwise, the null hypothesis will not be rejected. hﬂp:/Mww.perdisco.comlel ms/qsam/htm llqsam .ast 13/14 10/2/2015 Graded homework Perdisco/ latin /, v., to learn thorough/y © 2005 Perdisco - Terms Of Use | Privacy Policy | Saturday, October 03, 2015, 10:48 e r- d II http ://www.perdisco.com.a u hﬂp:/Mww.perdisco.comlel ms/qsam/htm llqsam .ast 14/14 ...

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