phase 5 project

Running head: PROJECT PHASE 4-INFECTIOUS DISEASES 0

Project Phase 4 – Infectious Diseases

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Project Phase 4 – Infectious Diseases

Introduction:

As a healthcare professional, you will work to improve and maintain the health of individuals, families, and communities in various settings. Basic statistical analysis can be used to gain an understanding of current problems. Understanding the current situation is the first step in discovering where an opportunity for improvement exists. This course project will assist you in applying basic statistical principles to a fictional scenario in order to impact the health and wellbeing of the clients being served.

This assignment will be completed in phases throughout the quarter. As you gain additional knowledge through the didactic portion of this course, you will be able to apply your new knowledge to this project. You will receive formative feedback from your instructor on each submission. The final project will be due on week 5.

Scenario:

You are currently working at NCLEX Memorial Hospital in the Infectious Diseases Unit. Over the past few days, you have noticed an increase in patients admitted with a particular infectious disease. You believe that the ages of these patients play a critical role in the method used to treat the patients. You decide to speak to your manager and together you work to use statistical analysis to look more closely at the ages of these patients. You do some research and put together a spreadsheet of the data that contains the following information:

Client number
Infection Disease Status
Age of the patient

You need the preliminary findings immediately so that you can start treating these patients. So, let’s get to work!!!!

Background information on the Data:

The data set consists of 60 patients that have the infectious disease with ages ranging from 35 years of age to 76 years of age for NCLEX Memorial Hospital. Remember this assignment will be completed over the duration of the course.

To begin let’s learn what infectious disease is. Infectious diseases are caused by pathogenic microorganisms, which are bacteria, viruses, parasites or fungi; the diseases can be spread directly or indirectly, through one person to another (WHO, 2017).

This scenario will aim to improve the quality of healthcare services that are provided to individuals, families, and communities at different levels of age. Therefore, the project utilized at NCLEX Memorial Hospital, over the past few days has seen a larger level of infectious disease occurrences. The data set composed was for sixty patients ranging in age from thirty-five to seventy-six.

a) Qualitative infectious: Disease

b) Quantitative: Age

2) Age is a constant variable as it may take any value.

3) A variable is any quantity that can be measured and whose value differs through the

Population and here we see the level of measurement s the age.

a) These measures of center are vital as it analyzes all the data provided in a particular set to understand and the approximate a middle value or average.

b) Variation is defined to mean, range or dispersion to differentiate it from systematic trends or differences. Measurers of variation are either property of a probability distribution or sample estimates of them. While, the range of a data set is among the biggest and littlest value.

5) Mean – 61.82

Median – 61.50

Mode – 69.00

Midrange – 58.00 76 35

Range – 41.00

Variance – 79.64

Standard Deviation- 8.92

By studying the data set we see that patients after the particular age over fifty and more likely over the ages of sixty, to be commonly affected by infectious diseases (Everitt & Skrondal, 2012). Therefore, there should be a prevention plan put in place to lower the amount of infected or more likely to be affected by different viruses.

Infectious Diseases

What are Confidence Intervals?

Confidence intervals are a range of values that have acquired a definition and have a particular probability that the value of parameter is able to attain lies within this range. A confidence interval exists to indicate the level of precision in a measurement made (Cumming, 2012). A point estimate is a single value that exists as an estimate of a population parameter calculated from the sample data of a population.

The best point estimate for the population mean is the mean obtained from the sample calculated as the average. The reason is that, the sample mean has characteristics of being unbiased as an estimate of the mean of the population (Cumming, 2012). Confidence intervals are important because having considered the size of the sample and the variations that lie in this sample potentially; they produce an estimate in which the real answer can be found. Confidence intervals introduce the potential for risks in decision making. Risks are increased when confidence intervals are underestimated. They give a picture of how accurate or precise an estimate is.

From the sample in the excel sheet g310, the total is 3709 while the sample mean is 61.8166 The sample mean is not the same as the population mean but it is a good point estimate for the population mean. The standard deviation of the sample is 8.92433. The Z-value is 0.41953.

Confidence interval

= CI for sample mean with unknown  = x */ z* s/60 = 1.8167 / 0.41953 * 8.92433/60 =3.7588

Following the calculation of the confidence interval for the mean, the values obtained as the mean of the population will lie between the values 61.3334 ± 3.7588. The values give a range in which to expect the value of the mean of the population. The values are just but a risk. This is an expectation of the outcome in case of anything but things may turn out to be different from what has been obtained.

The confidence interval in this case ranges from 57.5752 to 65.0928 in which the sample mean is included. The sample mean is the midpoint of the two values that give the confidence interval. This is part of reason why the sample mean is referred to as the best point estimate of the population mean.

Shifting the confidence intervals from 95% to 99% leads to a change in the standard deviation because all other variables are constant in this case. This change means that there is a reduction in risk as one increase the confidence interval. An increase in confidence interval means a subsequent increase in the standard deviation of the sample and a reduction in risk level. A reduction in confidence interval means a subsequent reduction in the standard deviation and an increment in the level of risk.

Conclusion

The confidence interval is designed to give a range of values where an estimate value is supposed to fall. The mean of the sample is the best value for use as a point estimate. It occurs as the midpoint of the two values that provide the range in which the actual value may fall. Reducing the value of confidence interval increases the risk of falling outside the interval. This means that the likelihood of obtaining the actual value reduces. On the other hand, increasing the confidence interval increases the probability of obtaining the actual value and reduces the risk of not doing so.

Explain the 8 Steps in hypothesis testing.

1. State null hypothesis (H 0) – this is an initial claim that the researcher specifies using previous knowledge. However, the researcher tries to reject or nullify this claim. Karl Popper’s principle or Falsifiability, it not possible to exclusively confirm a hypothesis but it is possible to conclusively negate a hypothesis.

2. Alternative hypothesis (Ha) - This is a hypothesis used in hypothesis testing that is contrary to the null hypothesis and it is accepted if the null hypothesis is rejected. It is stated as not all equal, because despite the fact that it is possible to have not all equal variables it is only one of the many chances. For instance, when comparing effect of infectious disease of the color of urine the alternative hypothesis can state that disease 1 results in tinting of the color of urine to yellow but disease 2, 3… and normal un-infected persons do not differ in the color of urine.

3. Set α- this is the level of significance. This is the probability of a statistical test will reject the data and hypothesis, despite the hypothesis actually being true.

There are two types of errors;

	Reality
decision	H0 is correct	H0 in incorrect
Accept H0	OK	Type II error which is the β equal to possibility of type II error.
H0 rejected	Type I error α=possibility of type I error	OK

4. Data collection- it is important to use valid data collection techniques possibly for this case use observational or experimental method

5. Stating and calculating the statistics for the study- this are the statistic values tested which include the mean, population, sample proportion and the difference in mean and sample proportions.

mean	61.82
median	61.50
mode	69.50
Mid-range	58
range	41
variance	79.64
Standard deviation	8.92

6. Decide on the test to be used- there are basically two types of tests; one tailed and two tailed. The decision is reached depending on the spread of error; two tailed is used when error spread is on two extremes side while one tailed test is used when error is spread on one side in the distribution.

7. Create accept and reject regions- a critical F value is established, you can establish the study F value from the statistical tables it is also called the Fα. It represents the minimum value for the study test statistics which determine which values should be rejected. With the value of F you locate it in the F distribution which forms the location for boundary for acceptance and rejection.

8. Standardize the test statistics to draw a conclusion- using step 5 and 6 you can make some inference on the study, but to make more specific conclusion computation of z-test will help decide on the whether to reject or accept the hypothesis. In such cases p-value lower then α, then null hypothesis H0 is conclusively negated and therefore should accept the alternative hypothesis HA.

In testing a hypothesis using the above eight steps I prefer using critical value.

This method include the coming up with the unlikely or likely which involves testing is the involved test statistics are more extreme than it would be possible if the null hypothesis was to true. It’s a straightforward, if the critical value if greater than the critical value α and therefore, the null hypothesis are incorrect which means the alternative hypothesis is accepted. It is has more valid explanation for accepting or rejecting a null hypothesis using this method.

Performing the hypothesis:

Claim:

The average age of patients admitted to NLEX hospital is less than 65 years is the test claim using α=0.05 in a normally distributed population of unknown standard deviation

Mean= 65 years

Scenario 2

Mean= 62.82 years scenario 1 standard deviation= 8.3 with a sample size of 60.

The H0 will be that the average age of patients admitted to NLEX hospital is 65 years.
The HA is the average age of patients admitted to NLEX hospital is less than $ 65,000
Therefore t=(65-62.82)(8.3/√60)= 2.336
α=0.05 and n-1 degree of freedom which is 60-1=59 in one tiled test.
Tabulated t-value is 1.65.
With a greater computed t-value of 2.336 against tableted t-value of 1.65 the null hypothesis is conclusively negated.

Based on the option selected;

Write the alternative and null hypothesis and indicate which the claim is.

H0 =µ=65 years

HA= µ˂ 65 years

The alternative hypothesis is the claim hypothesis.

Is the test two tailed, left tailed or right tailed? Explain.

It’s left tailed, because a test to find probability/chance of the null hypothesis and the mean is on the left side of the normal distribution curve.

Which test will you use to test the hypothesis test, z-test or t-test? Explain.

To answer the question here, the T-test, due to the data, this is normally distributed in the population of the unknown standard deviation.

What is the value of the test statistic? Show your calculations.

T= 2.336

T= (65-62.82) (8.3/√60)

T= 2.18/1.072

=2.336

What is the p-value? Explain how you determine it.

The value of p=0.05

Determined by finding if Z is beyond the statistic test; looking up in the z-table for the probability that it is greater than alternative hypothesis and subtract from 1.

What is the critical value and how to calculate it?

T (0.5-0.05) d= 0.45

T-value=1.65

Subtracting alpha (0.05) from 0.5=0.45 (in the z-table that is 0.455) the looking up in the z-table for the corresponding value is 1.65.

Decision, whether to reject of accept the null hypothesis.

I reject the null hypothesis, because the computed t-value 2.336 is greater than tabulated t-value of 1.65.

I reject the null hypothesis because it has been conclusively being negated i.e. the average patients admitted in NLEX hospital is less than 65 years.

Conclusion

I reject the null hypothesis because it has been conclusively being negated i.e. the average patients admitted in NLEX hospital is less than 65 years.

References

WHO, (2017). Infectious Diseases.

Retrieved from, http://www.who.int/topics/infectious_diseases/en/

Everitt, B.S., & Skrondal, A., (2010). The Cambridge Dictionary of Statistics. 4th Ed.

Retrieved from, http://www.stewartschultz.com/statistics/books/Cambridge%20Dictionary%20Statistics%204th.pdf

Cumming, G. (2012). Understanding the New Statistics: Effect Sizes, Confidence Intervals, and Meta-Analysis. New York: Routledge.

Lane, D. M. (2002). Steps in Hypothesis Testing.