Discussion: Statistical Concepts in Public Health

Discussion: Statistical Concepts in Public Health ORDER NOW FOR CUSTOMIZED AND ORIGINAL ESSAY PAPERS ON Discussion: Statistical Concepts in Public Health Need help with a statistics homework assignment. I have attached it for review. One of the documents provided contains the multiple choice options. I have also included a document with my answers highlighted, but it is only ~70% correct, and I’m trying to figure out which ones I got wrong. Discussion: Statistical Concepts in Public Health Covers all aspects of stat from theory to confidence intervals. statistical_concepts_in_public_health_1___homework.pdf stat_concepts_1_hw4_2020.doc stat_concepts_1_hw4_2020.pdf statistical_concep Homework 4 Part A 1. This be g ins Exercise 1 which consistes of questions 1 – 7 Two investigators are interested in studying the health of men 18-24 years old in Baltimore City. Investigator A plans to take a random sample of 100 men age 18-24 from Baltimore City.Investigator B will take a random sample of 1,000 such men. Both investigators will measure the waist size of men in their samples.- Which investigator will get a smaller (in value) sample standard deviation (s) for the waist sizes of the men in his/her sample? Investigator A Investigator B It is not possible to predict which investigator will get a smaller sample standard 1 point 2.Two investigators are interested in studying the health of men 18-24 years old in Baltimore City. Investigator A plans to take a random sample of 100 men age 18-24 from Baltimore City.Investigator B will take a random sample of 1,000 such men. Both investigators will measure the waist size of men in their samples.- What is quantified by the sample standard deviation (s)? The variability in individual sample values (waist sizes) around the sample The variability in sample means from multiple samples of the same taken from the same population 1 point 3.Two investigators are interested in studying the health of men 18-24 years old in Baltimore City. Investigator A plans to take a random sample of 100 men age 18-24 from Baltimore City.Investigator B will take a random sample of 1,000 such men. Both investigators will measure the waist size of men in their samples.- Which investigator will (most likely) get a smaller (in value) estimated standard error for the sample mean waist size? Investigator A Investigator B It is not possible to predict which investigator will get a smaller estimated standard. 1 point 4.Two investigators are interested in studying the health of men 18-24 years old in Baltimore City. Investigator A plans to take a random sample of 100 men age 18-24 from Baltimore City.Investigator B will take a random sample of 1,000 such men. Both investigators will measure the waist size of men in their samples.- What is quantified by the standard error of the sample mean? The variability in individual sample values (waist sizes) around the sample 1 point The variability in sample means from multiple samples of the same taken from the same population. 5.Two investigators are interested in studying the health of men 18-24 years old in Baltimore City. Investigator A plans to take a random sample of 100 men age 18-24 from Baltimore City.Investigator B will take a random sample of 1,000 such men. Both investigators will measure the waist size of men in their samples.- Which investigator is likely to estimate a narrower 95% confidence interval for the true mean waist size of all men 18-24 in Baltimore City? Investigator A Investigator B It is not possible to predict which investigator will get a narrower confidence 1 point 6.Two investigators are interested in studying the health of men 18-24 years old in Baltimore City. Investigator A plans to take a random sample of 100 men age 18-24 from Baltimore City.Investigator B will take a random sample of 1,000 such men. Both investigators will measure the waist size of men in their samples.- Which investigator will definitely get the sample mean closest to the true population mean among all male 18-24 year olds in Baltimore City? Investigator A Investigator B This relative closeness of these two individual sample mean estimates to the true population mean cannot be definitely 1 point 7.Two investigators are interested in studying the health of men 18-24 years old in Baltimore City. Investigator A plans to take a random sample of 100 men age 18-24 from Baltimore City.Investigator B will take a random sample of 1,000 such men. Both investigators will measure the waist size of men in their samples.- For which sample size: n=100 or n=1,000 will sample mean estimates tend to be closer on average to the true population mean? n=100 n=1,000 1 point 8. This be g ins Exericise 2 which consistes of questions 8- 13 In most “real-life research”, do researchers have access to the population level data (i.e data on the entire population or populations of interest? Yes No 1 point 1 point n most “real-life research” How many samples will be taken from each population of interest? One Between 3 and 17 More than 100 10.In most “real-life research” , how many 95% confidence intervals for the true population quantity of interest (mean, difference in proportions, incidence rate ratio, etc..) will be computed? One Two Ninety-five 1 point 11.In most “real-life research”, will the resulting 95% confidence interval definitely contain the value of the population quantity of interest being estimated with the study results? Yes No 1 point 12.What does the 95% refer to in the phrase “95% confidence interval” for a population level quanitity (population mean, mean difference, rleative risk etc..)? A 95% chance that the study results are A 95% chance the null hypothesis in not For 95% of studies that are based on a random sample(s) from a population(s), the 95% percent confidence interval will contain the true value of the population level quantity, For 95% of studies that are based on a random sample(s) from a population(s), the 95% percent confidence interval will contain the value of study based estimate (sample mean difference, sample relative risk etc..) of the population level 1 point 13.Will the corresponding hypothesis test for a population level quantity definitely reject the null hypothesis if the alternative hypothesis is in fact true? 1 point 14. This be g ins Exericise 14 which consistes of questions 14 – 20 Recall, from Homework #3, the following abstract taken from a September, 2012 article appearing in American Journal of Psychiatry.1 point In this study, the outcome of interest was the change in women’s responses on the Hamilton Depression Rating Scale between the treatment (antidepressant + oral creatinine supplementation) and control (antidepressant + oral placebo supplementation)Because of randomization, the Hamilton Depression Rating Scales (HSRD) were nearly identical on average between the treatment and control groups. As such, the Hamilton scores at the end of the eight-week follow-up period can be compared to assess differences, if any, in the average Hamilton Score change. In Homework #3, you reported sample mean difference in Hamilton Score changes between the Treated and Control groups of -4.4, with a 95% CI of (-6.5 to -2.3).The group specific means and sd are as follows:Group n mean standard deviationTreatment 17 5.4 3.0Control 22 9.8 3.5- Suppose you plan to perform a two-sample t-test for this study. What is the null hypothesis for this test? The mean HDRS scores are different in the treatment and control populations The mean difference in HDRS scores between the treatment and control populations is The mean HDRS scores are equal in the treatment and control The mean difference in HDRS scores between the treatment and control populations is not 1 point Recall, from Homework #3, the following abstract taken from a September, 2012 article appearing in American Journal of Psychiatry. In this study, the outcome of interest was the change in women’s responses on the Hamilton Depression Rating Scale between the treatment (antidepressant + oral creatinine supplementation) and control (antidepressant + oral placebo supplementation)Because of randomization, the Hamilton Depression Rating Scales (HSRD) were nearly identical on average between the treatment and control groups. As such, the Hamilton scores at the end of the eight-week follow-up period can be compared to assess differences, if any, in the average Hamilton Score change. In Homework #3, you reported sample mean difference in Hamilton Score changes between the Treated and Control groups of -4.4, with a 95% CI of (-6.5 to -2.3).The group specific means and sd are as follows:Group n mean standard deviationTreatment 17 5.4 3.0Control 22 9.8 3.5-Based on the resulting confidence interval only, what can you ascertain about the p-value from the two-sample t-test from item 14? . As the 95% CI for the mean difference does not include 0, the p-value from the two-sample t-test is greater than 05. As the 95% CI for the mean difference does not include 0, the p-value from the two-sample t- test is less than 05. As the 95% CI for the mean difference includes 1, the p-value from the two-sample t-test is greater than 05. As the 95% CI for the mean difference includes 1, the p-value from the two-sample t-test is less than 05 16.Recall, from Homework #3, the following abstract taken from a September, 2012 article appearing in American Journal of Psychiatry.1 point In this study, the outcome of interest was the change in women’s responses on the Hamilton Depression Rating Scale between the treatment (antidepressant + oral creatinine supplementation) and control (antidepressant + oral placebo supplementation)Because of randomization, the Hamilton Depression Rating Scales (HSRD) were nearly identical on average between the treatment and control groups. As such, the Hamilton scores at the end of the eight-week follow-up period can be compared to assess differences, if any, in the average Hamilton Score change. In Homework #3, you reported sample mean difference in Hamilton Score changes between the Treated and Control groups of -4.4, with a 95% CI of (-6.5 to -2.3).The group specific means and sd are as follows: Group n mean standard deviation Treatment 17 5.4 3.0 Control 22 9.8 3.5 – How far is the observed (sample) mean difference in Hamilton Depression scores between the treatment and control groups from what the expected difference is under the null hypothesis from item 14, in units of standard errors? The result is 4.2 standard errors above the expected null mean difference of The result is 4.2 standard errors below the expected null mean difference of 0 17.Recall, from Homework #3, the following abstract taken from a September, 2012 article appearing in American Journal of Psychiatry.1 point In this study, the outcome of interest was the change in women’s responses on the Hamilton Depression Rating Scale between the treatment (antidepressant + oral creatinine supplementation) and control (antidepressant + oral placebo supplementation)Because of randomization, the Hamilton Depression Rating Scales (HSRD) were nearly identical on average between the treatment and control groups. As such, the Hamilton scores at the end of the eight-week follow-up period can be compared to assess differences, if any, in the average Hamilton Score change. In Homework #3, you reported sample mean difference in Hamilton Score changes between the Treated and Control groups of -4.4, with a 95% CI of (-6.5 to -2.3).The group specific means and sd are as follows: Group n mean standard deviation Treatment 17 5.4 3.0 Control 22 9.8 3.5 – The resulting (two-sided) p-value from the two-sample t-test is 0.0002. What is the interpretation of this p-value (i.e. what probability does this p-value quantify)? The probability that the null hypothesis is The probability that the null hypothesis is The probability of getting a sample mean difference of -4.4 or something more extreme (farther from the null value of 0) if the population mean difference in HDRS scores is greater than The probability of getting a sample mean difference of -4.4 or something more extreme (farther from the null value of 0) if the population mean difference in HDRS scores is equal to 18.Recall, from Homework #3, the following abstract taken from a September, 2012 article appearing in American Journal of Psychiatry.1 point In this study, the outcome of interest was the change in women’s responses on the Hamilton Depression Rating Scale between the treatment (antidepressant + oral creatinine supplementation) and control (antidepressant + oral placebo supplementation)Because of randomization, the Hamilton Depression Rating Scales (HSRD) were nearly identical on average between the treatment and control groups. As such, the Hamilton scores at the end of the eight-week follow-up period can be compared to assess differences, if any, in the average Hamilton Score change. In Homework #3, you reported sample mean difference in Hamilton Score changes between the Treated and Control groups of -4.4, with a 95% CI of (-6.5 to -2.3).The group specific means and sd are as follows:Group n mean standard deviationTreatment 17 5.4 3.0Control 22 9.8 3.5- Suppose you compute the study results in the opposite direction, i.e. the mean difference in Hamilton Score changes between the Control and Treatment groups. What would be the resulting sample mean difference? -4.4 4.4 There is not enough information to answer this question. 19.Recall, from Homework #3, the following abstract taken from a September, 2012 article appearing in American Journal of Psychiatry.1 point In this study, the outcome of interest was the change in women’s responses on the Hamilton Depression Rating Scale between the treatment (antidepressant + oral creatinine supplementation) and control (antidepressant + oral placebo supplementation)Because of randomization, the Hamilton Depression Rating Scales (HSRD) were nearly identical on average between the treatment and control groups. As such, the Hamilton scores at the end of the eight-week follow-up period can be compared to assess differences, if any, in the average Hamilton Score change.In Homework #3, you reported sample mean difference in Hamilton Score changes between the Treated and Control groups of -4.4, with a 95% CI of (-6.5 to -2.3).The group specific means and sd are as follows: Group n mean standard deviation Treatment 17 5.4 3.0 Control 22 9.8 3.5 – Suppose you compute the study results in the opposite direction, i.e. the mean difference in Hamilton Score changes between the Control and Treatment groups. What would be the resulting 95% CI for the population level mean difference? (-6.5, -2.3) (2.3, 6.5) There is not enough information to answer this question 20.Recall, from Homework #3, the following abstract taken from a September, 2012 article appearing in American Journal of Psychiatry.1 point In this study, the outcome of interest was the change in women’s responses on the Hamilton Depression Rating Scale between the treatment (antidepressant + oral creatinine supplementation) and control (antidepressant + oral placebo supplementation)Because of randomization, the Hamilton Depression Rating Scales (HSRD) were nearly identical on average between the treatment and control groups. As such, the Hamilton scores at the endof the eight-week follow-up period can be compared to assess differences, if any, in the average Hamilton Score change. In Homework #3, you reported sample mean difference in Hamilton Score changes between the Treated and Control groups of -4.4, with a 95% CI of (-6.5 to -2.3).The group specific means and sd are as follows: Group n mean standard deviation Treatment 17 5.4 3.0 Control 22 9.8 3.5 – Suppose you compute the study results in the opposite direction, i.e. the mean difference in Hamilton Score changes between the Control and Treatment groups. What would be the resulting p-value for the two –sample t-test comparing the population means? . -0.0002 0.0002 0.05 There is not enough information given to answer this question 21. This be g ins Exericise 4 which consistes of questions 21 – 27 Recall from Homework #3 the JAMA article reports the results of a study of treatment outcomes for children with mild gastroenteritis who were given oral rehydration therapy. Enrolled children were randomized to received either rehydration with diluted apple juice (DAJ), or an electrolyte maintenance solution (EMS). As per the study authors:“The primary outcome was a composite of treatment failure defined by any of the following occurring within 7 days of enrollment: intravenous rehydration, hospitalization, subsequent unscheduled physician encounter, protracted symptoms, crossover, and 3%or more weight loss or significant dehydration at in-person follow-up. Secondary outcomes included intravenous rehydration, hospitalization, and frequency of diarrhea and vomiting.”Of the 323 children randomized to DAJ, 54 experienced treatment failure. (17%). Of the 324 children randomized to EMS, 81 experienced treatment failure. (25%)In Homework #3, you reported a sample relative risk of treatment failure for the DAJ group compared to the EMS group of 0.68 with a 95% CI of 0.49 to 0.93. – What is the null hypothesis for this test in terms of the risk difference (RD)? RD=1 RD not equal to 1 RD = 0 RD not equal to 0 1 point 22.Recall from Homework #3 the JAMA article reports the results of a study of treatment outcomes for children with mild gastroenteritis who were given oral rehydration therapy. Enrolled children were randomized to received either rehydration with diluted apple juice (DAJ), or an electrolyte maintenance solution (EMS). As per the study authors:1 point“The primary outcome was a composite of treatment failure defined by any of the following occurring within 7 days of enrollment: intravenous rehydration, hospitalization, subsequent unscheduled physician encounter, protracted symptoms, crossover, and 3%or more weight loss or significant dehydration at in-person follow-up. Secondary outcomes included intravenous rehydration, hospitalization, and frequency of diarrhea and vomiting.”Of the 323 children randomized to DAJ, 54 experienced treatment failure. (17%). Of the 324 children randomized to EMS, 81 experienced treatment failure. (25%)In Homework #3, you reported a sample relative risk of treatment failure for the DAJ group compared to the EMS group of 0.68 with a 95% CI of 0.49 to 0.93.- What is the null hypothesis for this test in terms of the relative risk (RR)? RR=1 RR not equal to 1 RR = 0 RR not equal to 0 23.Recall from Homework #3 the JAMA article reports the results of a study of treatment outcomes for children with mild gastroenteritis who were given oral rehydration therapy. Enrolled children were randomized to received either rehydration with diluted apple juice (DAJ), or an electrolyte maintenance solution (EMS). As per the study authors:“The primary outcome was a composite of treatment failure defined by any of the following occurring within 7 days of enrollment: intravenous rehydration, hospitalization, subsequent unscheduled physician encounter, protracted symptoms, crossover, and 3%or more weight loss or significant dehydration at in-person follow-up. Secondary outcomes included intravenous rehydration, hospitalization, and frequency of diarrhea and vomiting.”Of the 323 children randomized to DAJ, 54 experienced treatment failure. (17%). Of the 324 children randomized to EMS, 81 experienced treatment failure. (25%)In Homework #3, you reported a sample relative risk of treatment failure for the DAJ group compared to the EMS group of 0.68 with a 95% CI of 0.49 to 0.93. What is the null hypothesis for this test in terms of the odds ratio (OR)? OR = 1 OR not equal to 1 OR = 0 OR not equal to 0 1 point 24.Recall from Homework #3 the JAMA article reports the results of a study of treatment outcomes for children with mild gastroenteritis who were given oral rehydration therapy. Enrolled children were randomized to received either rehydration with diluted apple juice (DAJ), or an electrolyte maintenance solution (EMS). As per the study authors:“The primary outcome was a composite of treatment failure defined by any of the following occurring within 7 days of enrollment: intravenous rehydration, hospitalization, subsequent unscheduled physician encounter, protracted symptoms, crossover, and 3%or more weight loss or significant dehydration at in-person follow-up. Secondary outcomes included intravenous rehydration, hospitalization, and frequency of diarrhea and vomiting.”1 pointOf the 323 children randomized to DAJ, 54 experienced treatment failure. (17%). Of the 324 children randomized to EMS, 81 experienced treatment failure. (25%)In Homework #3, you reported a sample relative risk of treatment failure for the DAJ group compared to the EMS group of 0.68 with a 95% CI of 0.49 to 0.93. Based on the resulting confidence interval for the relative risk only, what can you ascertain about the p-value from the hypothesis test for items 21-23? As the 95% CI for the relative risk includes 0, the p-value is greater than 05. As the 95% CI for the relative risk does not include 0, the p-value is less than 05. As the 95% CI for the relative risk includes 1, the p-value is greater than 05. As the 95% CI for the relative risk does not include 1, the p-value is less than 05. 25.Recall from Homework #3 the JAMA article reports the results of a study of treatment outcomes for children with mild gastroenteritis who were given oral rehydration therapy. Enrolled children were randomized to received either rehydration with diluted apple juice (DAJ), or an electrolyte maintenance solution (EMS). As per the study authors:“The primary outcome was a composite of treatment failure defined by any of the following occurring within 7 days of enrollment: intravenous rehydration, hospitalization, subsequent unscheduled physician encounter, protracted symptoms, crossover, and 3%or more weight loss or significant dehydration at in-person follow-up. Secondary outcomes included intravenous rehydration, hospitalization, and frequency of diarrhea and vomiting.”Of the 323 children randomized to DAJ, 54 experienced treatment failure. (17%). Of the 324 children randomized to EMS, 81 experienced treatment failure. (25%)In Homework #3, you reported a sample relative risk of treatment failure for the DAJ group compared to the EMS group of 0.68 with a 95% CI of 0.49 to 0.93. Why is one test (and hence one p-value) sufficient for testing all 3 null hypotheses from items 21-23? One test is NOT sufficient for testing all 3 null All 3 null hypotheses are differing ways stating the same basic underlying equivalency, p_DAJ=p_EMS. 1 point 26.Recall from Homework #3 the JAMA article reports the results of a study of treatment outcomes for children with mild gastroenteritis who were given oral rehydration therapy. Enrolled children were randomized to received either rehydration with diluted apple juice (DAJ), or an electrolyte maintenance solution (EMS). As per the study authors:“The primary outcome was a composite of treatment failure defined by any of the following occurring within 7 days of enrollment: intravenous rehydration, hospitalization, subsequent unscheduled physician encounter, protracted symptoms, crossover, and 3%or more weight loss or significant dehydration at in-person follow-up. Secondary outcomes included intravenous rehydration, hospitalization, and frequency of diarrhea and vomiting.”Of the 323 children randomized to DAJ, 54 experienced treatment failure. (17%). Of the 324 children randomized to EMS, 81 experienced treatment failure. (25%)In Homework #3, you reported a sample relative risk of treatment failure for the DAJ group compared to the EMS group of 0.68 with a 95% CI of 0.49 to 0.93.1 point The p-value from a chi-square test was 0.01. What is the interpretation of this p-value (i.e. what probability does this p-value quantify)? The probability of getting a sample result (for example a relative risk of 68) as extreme or more extreme (as far or farther from the null value) if the alternative hypothesis is true is 0.01 The probability of getting a sample result (for example a relative risk of 68) as extreme or more extreme (as far or farther from the null value) if the null hypothesis is true is 0.01. The probability that the null hypothesis is true is 01. he probability that the alternative hypothesis is true is 01. 27.Recall from Homework #3 the JAMA article reports the results of a study of treatment outcomes for children with mild gastroenteritis who were given oral rehydration therapy. Enrolled children were randomized to received either rehydration with diluted apple juice (DAJ), or an electrolyte maintenance solution (EMS). As per the study authors:“The primary outcome was a composite of treatment failure defined by any of the following occurring within 7 days of enrollment: intravenous rehydration, hospitalization, subsequent unscheduled physician encounter, protracted symptoms, crossover, and 3%or more weight loss or significant dehydration at in-person follow-up. Secondary outcomes included intravenous rehydration, hospitalization, and frequency of diarrhea and vomiting.”Of the 323 children randomized to DAJ, 54 experienced treatment failure. (17%). Of the 324 children randomized to EMS, 81 experienced treatment failure. (25%)In Homework #3, you reported a sample relative risk of treatment failure for the DAJ group compared to the EMS group of 0.68 with a 95% CI of 0.49 to 0.93.- Assuming an alpha level of 0.05, what decision should the researchers make based on the p- value from item 26? The researchers should “fail to reject the null hypothesis”. The researchers should “reject the null hypothesis”. The researchers should question whether the study was well designed because of this p-value. The researchers should feel confident that the study results are1 point 28. This be g ins Exercise 5, which consists of questons 28 – 31. A 2013 article in the American Journal of Public Health presents the results of study of the relationship between social isolation and mortality.This1 pointanalysis is based on data collected in the Third National Health and NutritionalExamination Survey (NHANES), and the National Death Index. The following snippet comes from the article abstract: The authors chose to present the results of their analyses separately for females and males. This homework exercise will use only the results for females.The following Kaplan-Meier curve shows the time to death after participation in NHANES 3, when the factors included in computing the social isolation index were measured. The curves are presented separately for the four social index score groups. The authors report that “Among women, social isolation was associated with mortality (log rank p <0.001)”- What is the null hypothesis for the log-rank test performed by the authors? All four survival curves are equivalent at the population At lease two of the four curves differ at the population 29. A 2013 article in the American Journal of Public Health presents the results of study of the relationship between social isolation and mortality.1 pointThis analysis is based on data collected in the Third National Health and NutritionalExamination Survey (NHANES), and the National Death Index. The following snippet comes from the article abstract: The authors chose to present the results of their analyses separately for females and males. This homework exercise will use only the results for females.The following Kaplan-Meier curve shows the time to death after participation in NHANES 3, when the factors included in computing the social isolation index were measured. The curves are presented separately for the four social index score groups The authors report that “Among women, social isolation was associated with mortality (log rank p <0.001)” What is the alternative hypothesis for the log-rank test performed by the authors? All four survival curves are equivalent at the population At least two of the four curves differ at the population30. A 2013 article in the American Journal of Public Health presents the results of study of the relationship between social isolation and mortality.This analysis is based on data collected in the Third National Health and NutritionalExamination Survey (NHANES), and the National Death Index. The following snippet comes from the article abstract: The authors chose to present the results of their analyses separately for females and males. This homework exercise will use only the results for females.The following Kaplan-Meier curve shows the time to death after participation in NHANES 3, when the factors included in computing the social isolation index were measured. The curves are presented separately for the four social index score groups.1 pointThe authors report that “Among women, social isolation was associated with mortality (log rank p <0.001)” What is the interpretation of the p-value from the log-rank test (i.e. what probability does this p-value quantify) The probability that the null hypothesis is The probability that the alternative hypothesis is The probability of getting the observed study results (or results even less likely) if the null hypothesis is he probability that the study is not 31. A 2013 article in the American Journal of Public Health presents the results of study of the relationship between social isolation and mortality.This analysis is based on data collected in the Third National Health and NutritionalExamination Survey (NHANES), and the National Death Index. The following snippet comes from the article abstract: The authors chose to present the results of their analyses separately for females and males. This homework exercise will use only the results for females.The following Kaplan-Meier curve shows the time to death after participation in NHANES 3, when the factors included in computing the social isolation index were measured. The curves are presented separately for the four social index score groups.1 pointThe authors report that “Among women, social isolation was associated with mortality (log rank p <0.001)” Based on the log-rank results and the Kaplan-Meier curve, what general conclusion can be made about the association between social isolation and mortality? There is no association between social isolation and Greater social isolation is associated with greater Greater social isolation is associated with lower . Discussion: Statistical Concepts in Public Health Get a 10 % discount on an order above $ 100 Use the following coupon code : NURSING10

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