MPH 515 Week 6 Hypothesis Testing Exercise Example

Epidemiology and Biostatistics for Public Health Application I Chamberlain University Free custom sample in 24 to 48h

Whether a county's rate of first nonfatal opioid overdose visits changed between 2023 and 2024 is the question worked through in this MPH 515 Week 6 sample paper. In Chamberlain MPH 515 (MPH-515), Epidemiology and Biostatistics for Public Health Application I, Master of Public Health (MPH) students complete a hypothesis testing exercise step by step. The sample states the hypotheses about the population before any data, sets the significance level first, chooses a two-rate comparison, shows every step of the working, reaches a decision, writes a statistical and a practical conclusion, ties the possible error to the decision and checks the approximation it used.

CourseMPH 515 Epidemiology and Biostatistics for Public Health Application I
Week6
Paper typeHypothesis testing exercise
LengthAbout 1,130 words, 5 pages
FormatAPA 7 student paper
SchoolChamberlain University
ProgramMaster of Public Health (MPH)
UpdatedOctober 2026

Free sample paper for MPH 515 Week 6

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Hypothesis Testing Exercise: Did the Rate of First Nonfatal Opioid Overdose Visits Among Bellamy County Residents Change Between 2023 and 2024?

Student Name

Chamberlain University

MPH 515: Epidemiology and Biostatistics for Public Health Application I

Instructor Name

Month Day, Year

What this page is doingThe title asks the question about the population and both years, which is where a hypothesis test has to start before any data are examined.
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Hypothesis Testing Exercise: Did the Rate of First Nonfatal Opioid Overdose Visits Among Bellamy County Residents Change Between 2023 and 2024?

The Question and the Data

Bellamy County's health department reported fewer first emergency department visits for nonfatal opioid overdose among residents in 2024 than in 2023. Its board asked whether the decline was real or could reflect chance variation. This exercise tests whether the incidence rate in the county's population changed between the two years. The data are 655 new cases among about 412,000 residents in 2023 and 586 new cases among about 415,000 residents in 2024, with person-time approximated by midyear population.

Step 1: Hypotheses About the Population

The hypotheses concern the true underlying rates in the county's population, not the observed counts.

Null hypothesis: the true incidence rate of first nonfatal opioid overdose visits among county residents was the same in 2024 as in 2023.

Alternative hypothesis: the true incidence rate in 2024 differed from that in 2023, in either direction.

A two-sided alternative was chosen before looking at the data, because an increase would have been as important to the board as a decrease.

What this page is doingStating both hypotheses about the population, and the choice of a two-sided test before seeing the data, secures the marks the exercise assigns to the first step.
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Step 2: Significance Level Set First

The significance level was set at 0.05 before calculation. This means the analysis accepts a 5% probability of concluding that the rate changed when it did not, if the null hypothesis is true.

Step 3: The Test and Why

The outcome is a count of new cases over a known amount of person-time in each year, so a comparison of two Poisson rates is appropriate. With several hundred cases each year, a z test based on the normal approximation to the Poisson distribution is adequate. The rate ratio with its 95% confidence interval is calculated alongside the test.

Step 4: Statistic With Working Shown

Rate in 2023: 655 divided by 412,000, or 159.0 per 100,000 person-years.

Rate in 2024: 586 divided by 415,000, or 141.2 per 100,000 person-years.

Difference in rates: 159.0 less 141.2, or 17.8 per 100,000 person-years.

Standard error of the difference: the square root of the sum of each year's cases divided by its person-time squared, which for these data is 8.5 per 100,000 person-years.

Test statistic: the difference divided by its standard error, 17.8 divided by 8.5, gives z = 2.09.

Two-sided p value for z = 2.09: 0.037.

Rate ratio, 2024 compared with 2023: 141.2 divided by 159.0, or 0.89, with a 95% confidence interval of 0.79 to 0.99, calculated on the log scale from the two case counts.

Table 1

First Nonfatal Opioid Overdose Visits by Year, Bellamy County Residents

YearFirst visitsResidents, midyearFirst visits per 100,000 residents per year
2023655412,000159.0
2024586415,000141.2

Note. Note. Annual counts are composite figures built for this exercise; person-years use midyear population.

Step 5: Decision Against the Criterion

Since 0.037 falls under the preset 0.05 threshold, equal rates in the two years are rejected as the explanation.

Step 6: Two Conclusions

Statistical conclusion: the data provide evidence, at the 0.05 level, that the incidence rate of first nonfatal opioid overdose visits among county residents was lower in 2024 than in 2023.

Practical conclusion: the rate fell by about 11%, or about 18 fewer first visits per 100,000 residents a year. The interval shows that the decline could be as small as about 1% or as large as about 21%. That is a modest change. It is consistent with real improvement, but it is also close to the boundary, and the board should not treat the decline as settled or attribute it to any single program on this evidence alone. Statistical significance here does not establish that the decline was large or that any particular cause produced it (Greenland et al., 2016).

What this page is doingWriting a separate practical conclusion, with the size of the change and its interval, prevents the common error of letting a p value stand in for importance.
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Step 7: Errors Tied to This Decision

Because the null hypothesis was rejected, the error that could have occurred is a Type I error: concluding the rate changed when in truth it did not. The probability of this error was set at 5%. A Type II error, failing to detect a real change, is not possible for this decision, because the null was rejected. Had the result been p = 0.08, the right conclusion would not have been that the rate was unchanged, but that this analysis did not detect a change; failing to reject a null hypothesis is not evidence that it is true (Wasserstein & Lazar, 2016).

What the Test Cannot Address

The test treats the two years as random samples of an underlying rate, but surveillance data also change for nonstatistical reasons. A change in how emergency departments code chief complaints, a shift of patients to a hospital across the county line, or more overdoses being reversed by bystanders without an emergency visit could each lower the count without any change in overdose occurrence. Before the board concludes that fewer residents overdosed, the health department should check whether coding rules, hospital use patterns or naloxone distribution changed between the two years, and should compare the trend in overdose deaths over the same period.

Checking the Approximation

The z test uses a normal approximation to the Poisson distribution, which works well when counts are large. With 655 and 586 cases, both counts are far above the usual minimum of about 10 to 20 cases per group, so the approximation is sound. With small counts, such as a comparison for one planning area with a few dozen cases, an exact test based on the binomial distribution of cases between the two years, given their total, would be preferred.

A Second Way to See the Same Result

The confidence interval for the rate ratio gives the same decision as the test. The interval, 0.79 to 0.99, excludes 1.0, the value that would mean no change, which agrees with rejecting the null hypothesis at the 0.05 level. The interval adds information the p value does not: it shows the plausible size of the change, which is the information the board needs. Because the upper limit is close to 1.0, a decline of only about 1% remains compatible with the data.

Reporting It to the Board

A sentence for the board would read: the rate of first overdose visits among residents fell from 159.0 to 141.2 per 100,000 between 2023 and 2024, an 11% decline (rate ratio 0.89, 95% CI 0.79 to 0.99); chance by itself is an unlikely explanation for the drop, but the drop is modest and may partly reflect changes in coding or in where residents seek care.

What this page is doingShowing that the interval and the test agree, and translating both into a sentence for decision makers, completes the exercise beyond the arithmetic.
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References

Greenland, S., Senn, S. J., Rothman, K. J., Carlin, J. B., Poole, C., Goodman, S. N., & Altman, D. G. (2016). Statistical tests, P values, confidence intervals, and power: A guide to misinterpretations. European Journal of Epidemiology, 31(4), 337-350. https://doi.org/10.1007/s10654-016-0149-3

Wasserstein, R. L., & Lazar, N. A. (2016). The ASA statement on p-values: Context, process, and purpose. The American Statistician, 70(2), 129-133. https://doi.org/10.1080/00031305.2016.1154108

The Chamberlain MPH 515 Week 6 instructions

The sixth week asks you to complete a hypothesis test in a fixed sequence. State the null and alternative hypotheses about the population, not the sample, and choose one-sided or two-sided before seeing the data. Set the significance level in advance. Choose the test and justify it from the type of data. Calculate the test statistic, showing your working. Compare the result with your criterion and state the decision. Write two conclusions, one statistical and one practical, explaining the size of the effect. Identify the type of error that could have occurred given your decision. Do not describe a nonsignificant result as proof that no difference exists, and check any approximation you relied on.

How this MPH 515 Week 6 example is built

The exercise uses 655 first visits among 412,000 residents in 2023 and 586 among 415,000 in 2024. It states two-sided hypotheses about the true rates, fixes alpha at 0.05 and explains why a comparison of two Poisson rates fits counts over person-time. Each step is shown: rates of 159.0 and 141.2 per 100,000, a difference of 17.8, a standard error of 8.5 and z = 2.09 with p = 0.037. The rate ratio is 0.89, interval 0.79 to 0.99. The practical conclusion calls the decline modest and warns against attributing it to any program. Type I error is named, and sections check the normal approximation and translate the result for the board.

MPH 515 Week 6 rubric: what full points look like

The defining loss on this exercise is a conclusion that claims proof, such as accepting the null hypothesis or saying a result proves no difference exists. Instructors mark it heavily because the whole exercise exists to correct it. Hypotheses written about the sample, a significance level chosen after the result, and working left out also cost points. Conclusions that stop at statistical significance without saying how large the change was lose the practical marks. Strong exercises state population hypotheses first, fix alpha in advance, justify the test, show every step, decide against the criterion, give statistical and practical conclusions and name the error that could follow from the decision. Leaving out the confidence interval removes the size of the effect from view.

MPH 515 Week 6 help from the desk

Hypothesis testing exercises follow a strict sequence, and most lost marks come from steps done out of order or stated loosely. If you share the problem your instructor set, we can help you write population hypotheses, choose and justify the test, lay out the working and phrase both conclusions correctly. We can also help you identify which error type applies to your decision. The arithmetic and the conclusions are yours to own. When a section expects a particular test, such as a chi-square or t test rather than a rate comparison, tell us so the custom sample follows that test's steps. We can also help you check whether a normal approximation is acceptable for your counts or whether an exact test is needed.

Get a MPH-515 Week 6 example written to your instructions

Send the Week 6 instructions with the data and the steps your section requires, and a custom example is worked through in that sequence and returned inside 24 to 48 hours. The first one is free. The computation itself is what the exercise is assessing.

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MPH 515 Week 6 questions, answered

Why can I not say the null hypothesis is proven?

Because a test that fails to detect a difference cannot distinguish between there being none and the study being too small to find one. The correct language is failing to reject, and the correct addition is a comment on power or sample size. This is the single most heavily marked point in the exercise and the easiest to lose after doing everything else right.

Does the significance level have to be the conventional one?

Follow whatever your section sets, which is usually conventional. What matters is that it is fixed before the analysis rather than adjusted afterward. Where an assignment asks you to justify the level, the argument turns on the cost of each error type in that particular context, which is a public health question rather than a statistical one.

What is the difference between statistical and practical significance?

A large study can detect a difference too small to matter, and a small study can miss one that matters a great deal. Statistical significance says a difference was detected; practical significance asks whether it would change anything. Assignments frequently plant a case where they diverge, and answering both separately is what the rubric rewards.

Where can I find a free MPH 515 Week 6 sample paper?

This page holds a complete MPH 515 Week 6 hypothesis test comparing overdose visit rates in two years, with every step shown.

What practical conclusion does the MPH 515 Week 6 sample reach?

That the rate fell by about 11%, a modest decline that could be as small as about 1%, so it should not be attributed to any single program.