NURS 701 - Statistical Analysis Questions – Assignment 3 (Due by Midnight – Sunday, September 27h, 2020) For each of the questions below, carry out an appropriate analysis to answer the research...

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Three partial questions, two full questions. (1-3 are partially complete...looking to have findings summarized)


NURS 701 - Statistical Analysis Questions – Assignment 3 (Due by Midnight – Sunday, September 27h, 2020) For each of the questions below, carry out an appropriate analysis to answer the research questions. The JMP files for some of the questions are located on the Datasets page on the course website. Most of the problems required you to enter the data yourself. When possible, include the relevant JMP output. 1. Improving control of blood glucose levels is an important motivation for the use of insulin pumps by diabetic patients. However, certain side effects have been reported with pump therapy. The table below comes from the paper “Acute complications associated with insulin pump therapy: Report of experience with 161 patients” by Mecklenberg et al. published in JAMA 252 (23), 3265-3269. The table provides data on the occurrence of diabetic ketoacidosis (DKA) in patients before and after pump therapy. Before pump therapy After pump therapy No DKA DKA Row Totals No DKA 128 7 135 DKA 19 7 26 Column Totals 147 14 n = 161 Conduct an appropriate test to determine if the rate of diabetic ketoacidosis (DKA) is higher following diabetic pump therapy. Summarize your findings. (5 pts.) 2. Schoenbaum et al. looked at risk factors for human immunodeficiency virus (HIV) infection among intravenous drug users enrolled in a methadone program in their paper “Risk factors for human immunodeficiency virus infection in intravenous drug users” published in the New England Journal of Medicine, 321 (13), 874-879. The data table below presented the HIV antibody status among 120 non-Hispanic white subjects by total family income. Family Income Level (ordinal) HIV Status A - < $10k="" b="" –="" $10k="" –="" $20k="" c="" -=""> $20k Row Totals HIV Positive 13 5 2 20 HIV Negative 59 21 20 100 Column Totals 72 26 22 120 Conduct a test to determine if total family income is related in a consistent manner to percentage of HIV-positive? Summarize your findings. (5 pts.) 3. In a study to investigate the potential relationship between age at first birth and the development of breast cancer the following results were obtained. Age at First Birth (ordinal) Case-Control status 1 < 20="" 2="" 20="" –="" 24="" 3="" 25="" –="" 29="" 4="" 30="" –="" 34="" 5=""> 35 Row Totals Case 320 1206 1011 463 220 3220 Control 1422 4432 2893 1092 406 10245 Column Total 1742 5638 3904 1555 626 n = 13,465 Is there evidence of an increasing trend in the proportion of breast cancer cases as a age at first birth increases? Conduct appropriate statistical test to answer this question and summarize the results. (5 pt.) 4. A case-control study was carried out to look at the potential risk for myocardial infarctions associated with oral contraceptive use. In addition to case-control status and current OC use, the age of the subject was also recorded. The variable age group described below is a ordinal variable created from the ages of the subjects. The data-file OC-Age-MI.JMP contains these data and the variables in this file and their coding are defined below. Case-Control Status 1 = Case (Myocardial Infarction (MI)) 2 = Control Oral Contraceptive Use? 1 = Yes 2 = No Age Group 1 = 25 - 29 yrs., 2 = 30 – 34 yrs., 3 = 35 – 39 yrs., 4 = 40 – 44 yrs., 5 = 45 – 49 yrs. a) Ignoring age group, estimate the risk for MI associated with OC use. Provide both a point estimate and an associated confidence interval for this measure of risk. Discuss. In doing this in JMP it will be best to use the OC coded and Case-Control coded so the risk measure is calculated in the preferred way. The appropriate 2x2 contingency table is shown below so you can easily check your calculation by hand. (5 pts.) b) What type of confounder would you expect age to be, positive or negative? Explain your reasoning by considering the relationship between age, oral contraceptive use, and myocardial infarctions. (3 pts.) c) The tables on the following page were obtained by stratifying on age group. Below each calculate the associated OR. Does age appear to be a confounder? What type given these results? Explain. (5 pts.) Age Group: 1 = 25 - 29 yrs., 2 = 30 – 34 yrs., 3 = 35 – 39 yrs., 4 = 40 – 44 yrs., 5 = 45 – 49 yrs. Age Group = 1 Age Group = 2 Age Group = 3 Age Group = 4Age Group = 5 d) Conduct an appropriate test to determine if a significant relationship exists between OC use and MI adjusting for the age group classification of the individual. Summarize your findings. (4 pts.) e) We can obtain a better estimate of the odds ratio than the crude one from the combined table in part (a) by combining the results from the age group stratified tables. The formula for this combined OR is given by where k = number of strata, (note: k = 5 in this case) and are the usual cells from strata i. Remember cell where risk factor and disease are present. This estimate of the odds ratio is known as Cochran-Mantel-Haenszel OR or . Find the for this study and compare it to the crude OR from part (a). What does the CMH odds ratio suggest about the risk OC use presents when considering myocardial infections? (5 pts.) 5. Rosenberg et al. (1980) studied the relationship between coffee drinking and myocardial infarction in young women, aged 30-49 years. This retrospective study including 487 cases hospitalized for the occurrence of a myocardial infarction (MI). Nine hundred eighty controls hospitalized for an acute condition (trauma, acute cholecystitis, acute respiratory diseases and appendicitis) were selected. The measured variables in this study are defined below. These data are contained in the file Coffee-MI.JMP. Case-Control Status 1 = Case (Myocardial Infarction (MI)) 2 = Control Drink Coffee? 1 = Yes (5 cups of coffee or more) 2 = No (< 5 cups of coffee) smoker group 1 = never., 2 = former, 3 = 1 – 14 cigarettes, 4 = 15 – 24 cigarettes, 5 = 25 – 34 cigarettes, 6 = 35 – 44 cigarettes, 7 = 45+ cigarettes a) ignoring smoking group, estimate the risk for mi associated with drinking coffee as defined above. provide both a point estimate and an associated confidence interval for this measure of risk. discuss. in doing this in jmp it will be best to use drink coffee? and case-control as coded above so the risk measure is calculated in the preferred way. the appropriate 2x2 contingency table is shown below so you can easily check your calculation by hand. (5 pts.) b) what type of confounder would you smoking status to be, positive or negative? explain your reasoning by considering the relationship between smoking status, coffee use as defined, and myocardial infarctions. (3 pts.) c) the tables below were obtained by stratifying on smoking status. below each calculate the associated or. does smoking status appear to be a confounder? what type given these results? explain. (7 pts.) never former 1 – 14 cigarettes 15 – 24 cigarettes 25 – 34 cigarettes 35 – 44 cigarettes 45+ cigarettes d) conduct an appropriate test to determine if a significant relationship exists between coffee use as defined and mi adjusting for the smoking status classification of the individual. summarize your findings. (4 pts.) e) we can obtain a better estimate of the odds ratio than the crude one from the combined table in part (a) by combining the results from the smoking status stratified tables. the formula for this combined or is given by where k = number of strata, (note: k = 7 in this case) and are the usual cells from strata i. remember cell where risk factor and disease are present. this estimate of the odds ratio is known as cochran-mantel-haenszel or or . find the for this study and compare it to the crude or from part (a). what does the cmh odds ratio suggest about the risk coffee use as defined presents when considering myocardial infections? (5 pts.) 8 = r o ˆ = r o ˆ = r o ˆ = r o ˆ = r o ˆ = r o ˆ = r o ˆ = r o ˆ = r o ˆ = r o ˆ = r o ˆ = r o ˆ = r o ˆ 5="" cups="" of="" coffee)="" smoker="" group="" 1="Never.," 2="Former," 3="1" –="" 14="" cigarettes,="" 4="15" –="" 24="" cigarettes,="" 5="25" –="" 34="" cigarettes,="" 6="35" –="" 44="" cigarettes,="" 7="45+" cigarettes="" a)="" ignoring="" smoking="" group,="" estimate="" the="" risk="" for="" mi="" associated="" with="" drinking="" coffee="" as="" defined="" above.="" provide="" both="" a="" point="" estimate="" and="" an="" associated="" confidence="" interval="" for="" this="" measure="" of="" risk.="" discuss.="" in="" doing="" this="" in="" jmp="" it="" will="" be="" best="" to="" use="" drink="" coffee?="" and="" case-control="" as="" coded="" above="" so="" the="" risk="" measure="" is="" calculated="" in="" the="" preferred="" way.="" the="" appropriate="" 2x2="" contingency="" table="" is="" shown="" below="" so="" you="" can="" easily="" check="" your="" calculation="" by="" hand.="" (5="" pts.)="" b)="" what="" type="" of="" confounder="" would="" you="" smoking="" status="" to="" be,="" positive="" or="" negative?="" explain="" your="" reasoning="" by="" considering="" the="" relationship="" between="" smoking="" status,="" coffee="" use="" as="" defined,="" and="" myocardial="" infarctions.="" (3="" pts.)="" c)="" the="" tables="" below="" were="" obtained="" by="" stratifying="" on="" smoking="" status.="" below="" each="" calculate="" the="" associated="" or.="" does="" smoking="" status="" appear="" to="" be="" a="" confounder?="" what="" type="" given="" these="" results?="" explain.="" (7="" pts.)="" never="" former="" 1="" –="" 14="" cigarettes="" 15="" –="" 24="" cigarettes="" 25="" –="" 34="" cigarettes="" 35="" –="" 44="" cigarettes="" 45+="" cigarettes="" d)="" conduct="" an="" appropriate="" test="" to="" determine="" if="" a="" significant="" relationship="" exists="" between="" coffee="" use="" as="" defined="" and="" mi="" adjusting="" for="" the="" smoking="" status="" classification="" of="" the="" individual.="" summarize="" your="" findings.="" (4="" pts.)="" e)="" we="" can="" obtain="" a="" better="" estimate="" of="" the="" odds="" ratio="" than="" the="" crude="" one="" from="" the="" combined="" table="" in="" part="" (a)="" by="" combining="" the="" results="" from="" the="" smoking="" status="" stratified="" tables.="" the="" formula="" for="" this="" combined="" or="" is="" given="" by="" where="" k="number" of="" strata,="" (note:="" k="7" in="" this="" case)="" and="" are="" the="" usual="" cells="" from="" strata="" i.="" remember="" cell="" where="" risk="" factor="" and="" disease="" are="" present.="" this="" estimate="" of="" the="" odds="" ratio="" is="" known="" as="" cochran-mantel-haenszel="" or="" or="" .="" find="" the="" for="" this="" study="" and="" compare="" it="" to="" the="" crude="" or="" from="" part="" (a).="" what="" does="" the="" cmh="" odds="" ratio="" suggest="" about="" the="" risk="" coffee="" use="" as="" defined="" presents="" when="" considering="" myocardial="" infections?="" (5="" pts.)="" 8="R" o="" ˆ="R" o="" ˆ="R" o="" ˆ="R" o="" ˆ="R" o="" ˆ="R" o="" ˆ="R" o="" ˆ="R" o="" ˆ="R" o="" ˆ="R" o="" ˆ="R" o="" ˆ="R" o="" ˆ="R" o="">
Answered 3 days AfterSep 21, 2021

Answer To: NURS 701 - Statistical Analysis Questions – Assignment 3 (Due by Midnight – Sunday, September 27h,...

Swapnil answered on Sep 23 2021
127 Votes
91689/Solution.docx
        1
        To test if the rate of diabetic ketoacidosis (DKA) is higher following diabetic pump therapy.
Given, Before pump therapy.
        After pump therapy
        No DKA
        DKA
        Row total
        No DKA
        128
        7
        135
        DKA
        19
        7
        26
        Column total
        147
        14
        161
The test statistic for the given data,
X^2 = n (ΣΣ f^2ij/fi0fj0 – 1)
Where, f​​​​​ij is the frequency of (i, j) th cell. And f​​​​​​i0 and f​​​​​​0j are marginal frequency of the ith row and jth column respectively.
Therefore, n = 161.
= 1.0805
Now, x^2 = 161* (1.0805 -1) = 12.9605
with degrees of freedom =
(1) *(1) = 1
X^2 = 0.05,
1 = 3.841.
Here, observed x^2 > tabulated x^2 0.05,1.
So basically it can give us the result significant.
And there is no homogeneity that can gives you the homogeneity for the pump therapy after and before.
So the rate of diabetic ketoacidosis can be gives the higher rate for the diabetic pump therapy. Hence, the rate of diabetic ketoacidosis is higher than the diabetic pump therapy.
        2
        We are basically used the chi square test fir the independence of two variables.
Null hypothesis: H0: Total family income is not related in a consistent manner to percentage of HIV positive.
Alternative hypothesis: Ha: Total family income is related in a consistent manner to percentage of HIV positive.
We assume level of significance = α = 0.05
Test statistic formula is given as below:
Chi square = ∑ [(O – E) ^2/E]
Where, O is observed frequencies and E is expected frequencies.
E = row total * column total / Grand total
We are given
Number of rows = r = 2
Number of columns = c = 3
Degrees of freedom = df = (r – 1) *(c – 1) = 2
α = 0.05
Critical value = 5.9914
Calculation tables for test statistic are given as below:
        Observed Frequencies
        
        Family income level
        
        HIV Status
        A
        B
        C
        Total
        Positive
        13
        5
        2
        20
        Negative
        59
        21
        20
        100
        Total
        72
        26
        22
        120
        Expected Frequencies
        
        Family income level
        
        HIV Status
        A
        B
        C
        Total
        Positive
        12
        4.3333
        3.6666
        20
        Negative
        60
        21.6666
        18.3333
        100
        Total
        72
        26
        22
        120
        (O - E)
        1
        0.6666
        -1.6666
        -1
        -0.6667
        1.6666
        (O - E)^2/E
        0.0833
        0.1025
        0.7575
        0.0166
        0.0205
        0.1515
Chi square = ∑ [(O – E) ^2/E] = 1.1321
P-value = 0.5677
P-value > α = 0.05
So, we do not reject the null hypothesis
There is insufficient evidence to conclude that Total family income is related in a consistent manner to percentage of HIV positive.
        3
        The investigation for the potential relationship between age at the first birth and the development for the breast cancer will give the following table.
        Case control Status
        1
<20
        2
20-40
        3
25-29
        4
30-34
        5
>= 35
        Row Totals
        Case
        320
        1206
        1011
        463
        220
        3220
        Control
        1422
        4432
        2893
        1092
        406
        10245
        Column Total
        1742
        5638
        3904
        1555
        626
        n = 13,465
To increase the trend in the proportion of breast cancer cases as the age at first
The expected values for the row and columns that can computed to the formula. In fact, the formula is Eij=Ri*Ci/T
Where = Ri = Sum elements in Row I, Ci = Sum elements in column j, T = grand total.
Expected values =
R1 = 1742*3220/13465 = 416
C1 = 1742*10245/13465 = 1325
R2 = 5638*3220/13465 = 1348
C2 = 5638*10245/13465 = 4289
R3 = 3904*3220/13465 = 933
C3 = 3903*10245/13465 = 2969
R4 = 1555*3220/13465 = 371
C4 = 1555*10245/13465 = 1183
R5 = 626*3220/13465 = 149
C5 = 626*10245/13465 = 476
R Total = R1+R2+R3+R4+R5/5 = 416+1348+933+371+149/5 = 643
C Total = C1+C2+C3+C4+C5/5 = 1325+4289+2969+1183+476/5 = 2048
So next we got the increasing trend for the proportion of the breast cancer in the above calculation. Based on the expected values we can squared to the computed distances that can give the following formula:
(O-E) ^2 / E
So the squared distance shown below
(643) ^2/13465 = 30.7054
(2048) ^2/13465 = 311.4967
So the statistical test that can be done by the investigation to the potential relationship.
Which gives us the result 311.50 which will give the stat for the breast cancer cases.
        4A
        While estimating the age group the risk can be MI associated with the OC use. The interval for this age group between
1 = 25 - 29 yrs.
2 = 30 – 34 yrs.
3 = 35 – 39 yrs.
4 = 40 – 44 yrs.
5 = 45 – 49 yrs.
Can give you the associated confidence interval for the measure risk.
A point here is to estimation for the single value parameters. So the simple estimation point can be gives the population mean. An interval can be used to give the range between these age group values where the parameters can be lied itself in the JMP.
And if we work with the JMP software then the best use for the OC code can be added to the controlling the risk. The following table can give us the risk measure calculation:

        Count Row %
        1
        2
        
        1
        29
12.39
        205
87.51
        234
        2
        135
7.75
        1607
92.25
        1742
        
        164
        1812
        1976
Here the control codec can give the estimation for the age group confidence interval. And the OC can be used for the contraceptive code that can give the control for the parameters.
        4B
        The positive confounder would be expected age.
Age: The age of 16 is considered for the teenagers that can basically start the contraceptive. The cycle starts at this age only.
Oral contraceptive use: the risk is basically used in the oral contraceptive can be included to the six fold that can be increased to the incidence of the venous thromboembolism for the apparent. The Oral contraceptive is a method basically include the spontaneity and it can be used for the controlling the hormonal methods.
Myocardial Infarctions: Myocardial Infarctions can be called as the heart attack which can be occurred when the blood flow can be decreased to the part of the heart. The main cause of myocardial Infarctions involves the blockage into the heart or the coronary arteries....
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