Week 6 Healthcare Assignment | Complete Solution

Week 6 Healthcare Assignment
Your Name:_________________________________
Instructions: Please everything carefully, follow any noted directions, and fill all answers into this document.
In this document, I include hints and tutorials. Be sure to use them 
Next, in this document, I note places where you will answer in BOLD and in YELLOW- be sure to place answers there and to answer all questions noted.
Submit this paper with your answers, etc. to the classroom by entering Week 6, choosing Assignments, and clicking on the name of this Assignment. There you will see a place to upload your submission.

Part I
A researcher believes a risk exists between the strain to human papillomavirus (HPV) infection and the risk of cervical cell abnormalities. 
In a random sample of 130 biopsies. 
So n = 130
Of these 130 people, 20 are negative, 44 are CIN I, 32 are CIN II, 24 are CIN III, and 10 are CIS. 
Also, of these 130 people, 1 person is HPV negative (no strain), 92 are HPV positive/low-risk strain, and 37 are HPV positive/high-risk strain.
Note: Cervical abnormalities are labeled as:
None, CIN I (Mild Abnormality on cervix surface), CIN II (Moderate Abnormality on cervix surface), CIN III (Severe Abnormality on cervix surface), CIS (cervical carcinoma).

1.    Identify the null hypothesis.
Hints: The null is the hypothesis that represents no significant difference. Here, the researcher believes that there IS a sig difference between the strain of HPV and the risk of cell abnormalities. The null would prove the researcher to be incorrect. 
Ho:  PLACE NULL HERE….

2.    The study reports the association between the type of HPV infection and cervical cell abnormalities has a p-value of 0.06. If the alpha for the study is set at 0.05, what should the researcher conclude regarding the null hypothesis? Why? 
ANSWER HERE
What does this say about the relationship between HPV infections and cervical cell abnormalities?
    ANSWER HERE
3.    If the researchers decided to set this pilot study’s alpha at 0.10 instead of 0.05, what would the researcher conclude about the null hypothesis (p = 0.06)? 
ANSWER HERE
How does the change in the alpha value alter your explanation on the relationship between HPV infection and cervical cancer?
    ANSWER HERE

Part II
Data from 155 patients at a diabetes clinic include measures of their HbAlc (a measure of how blood sugar is controlled). The mean HbAlc value is 8.2, with a standard deviation of 1.3.
1.    Compute a 95% confidence interval using the above data.
What formula will you use?
ANSWER HERE
Tutorial to Help: 
Visual Tutorial Step by Step: Dr Ami Gates MOOT Intro Stats Confidence Interval for Population Mean
http://www.mathandstatistics.com/learn-stats/confidence-intervals/confidence-intervals-for-means

What is E? (show your work)
ANSWER HERE
What is the final Confidence interval answer? (Please show work)
ANSWER HERE

2.    What does the result of the confidence interval say about the HbAlc values of the diabetes patients at this clinic?
ANSWER HERE (2 -4 sentences)

Part III
Assume 41% of people have Group O blood. Your hospital is conducting an urgent blood drive because its supply of Group O blood is low, and it needs at least 136 people to donate Group O blood. Three hundred thirty (330) volunteers donate blood.
1.    Estimate the probability that the number with Group O blood is at least 136. Include your work and calculations.
Hints and Help – read everything I have here and then answer where I note.
We know that the population proportion for Type O is .41 (as given in the question)
First, let’s think about the question here. We know that we have 330 volunteer donors. We also know that 41% of people in the population are thought to have Type O. However, because we are only sampling, we cannot be certain that exactly 41% of our specific sample will have Type O. 
We can also directly calculate that 41% of 330 is .41 * 330 = 135.3 (or 135 people).
This means that we expect that each sample will have ** approximately ** 135 people with Type O. But again, this number might be larger or smaller depending on the specific sample we take. 
This is a very important statistical concept. Remember, just because we know that 41% of the “population” is Type O, this does not mean that exactly 41% of every sample we take will be Type O – right? 
Let’s use hypothesis testing to determine the p value here: 
Call the population p as “pop p” and the sample p as “sam p”
I will get you started…
Ho: sam p  <= pop p  
Ha: sam p > pop p
Remember, we need our sample proportion to be larger than the population proportion. With the population proportion of 41%, and a sample of 330 people, we have 135 people (not enough). 
Let’s use alpha = .05 here and let’s run this hypothesis test to see if our sample is likely to have at least 136 people or more….
Step 1: You must get the std dev:
std dev = sqrt[ pop p * ( 1 – pop p ) / n ]
stde dev = ……ANSWER HERE (keep at least 4 decimal places)
Then get the z –test value
z = (sam p – pop p) / std dev
The z value is ANSWER HERE

Once you have the z value, use the z table to get the p value
The p value is ANSWER HERE
The p value here from the z tale is …..

Next, compare the z value to your alpha of .05
Compare HERE….what is the result? 
Do you reject Ho or accept Ho? 
What does that mean?

2.    What would you tell your hospital manager at the end of the day regarding the blood drive’s success (before blood typing has been done)? Do you need to plan an additional blood drive tomorrow? Why or why not? Explain using statistical data.
Hint: Use what you calculated above to answer this question. Again, you will need to run the z test and review the result to answer.
Your paper should be double spaced, 12 point font, and all references and citations should be APA formatted. All responses should include proper spelling and grammar, and questions should be answered using complete sentences.