Due 03/18/2019 Though data analysis occurs after the study has completed a data collection stage, the researcher needs to have in mind what type of analysis will allow the researcher to obtain an answer to a research question. The researcher must understand the purpose of each method of analysis, the characteristics that must be present in the study for the design to be appropriate and any weaknesses of the design that might limit the usefulness of the study results. Only then can the researcher select the appropriate design. Choosing the appropriate design enables the researcher to claim the data that is potential evidence that provides information about the relationship being studied. Notice that it is not the statistical test which tells us that research is valid, rather, it is the research design. Social workers must be aware of and adjust any limitations of their chosen design that may impact the validity of the study. To prepare for this Discussion, review the handout, A Short Course in Statistics and pages 210–220 in your course text Social Work Evaluation: Enhancing What We Do. If necessary, locate and review online resources concerning internal validity and threats to internal validity. Then, review the “Social Work Research: Chi Square” case study located in this week’s resources. Consider the confounding variables, that is, factors that might explain the difference between those in the program and those waiting to enter the program.Please use 4 APA Refrences, Be detailed in responses Post an interpretation of the case study’s conclusion that “the vocational rehabilitation intervention program may be effective at promoting full-time employment.” Describe the factors limiting the internal validity of this studyExplain why those factors limit the ability to draw conclusions regarding cause and effect relationships. Reference Dudley, J. R. (2014). Social work evaluation: Enhancing what we do. (2nd ed.) Chicago, IL: Lyceum Books. Chapter 9, “Is the Intervention Effective?” (pp. 226–236: Read from “Determining a Causal Relationship” to “Outcome Evaluations for Practice”) Plummer, S.-B., Makris, S., & Brocksen S. (Eds.). (2014b). Social work case studies: Concentration year. Baltimore, MD: Laureate International Universities Publishing. [Vital Source e-reader]. Read the following section: “Social Work Research: Chi Square” (pp. 63–65) Document: Week 4: A Short Course in Statistics Handout (PDF) Document: Week 4: Handout: Chi-Square findings (PDF)
single_subject_social_work_case_studies__concentration_year__wk3_.pdf
social_work_case_studies__chi_square.pdf
handout_chi_square_findings.pdf
shortcourseinstatisticshandout.pdf
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Social Work Research: Single Subject
Chris is a social worker in a geriatric case management program located in a midsize Northeastern town. She has an MSW and is part of a team of
case managers that likes to continuously improve on its practice. The team is currently using an approach that integrates elements of geriatric case
management with short-term treatment methods, particularly the solution-focused and task-centered models. As part of their ongoing practice, the
team regularly conducts practice evaluations. It has participated in larger scale research projects in the past.
The agency is fairly small (three full-time and two part-time social work case managers) and is one of several providers in a region of
approximately 50,000 inhabitants. Strengths of the agency include a strong professional network and good reputation in the local community as well
as the team of experienced social workers. Staff turnover has been almost nonexistent for the past 3 years. The agency serves about 60–70 clients at
any given time. The clients assisted by the case management program are older adults, ranging from their early 60s to over 100 years of age, as well
as their caregivers.
To evaluate its practice approach, the team has decided to use a multiple-baseline, single-subject design. Each of the full-time case managers will
select one client new to the caseload to participate in the study. The research project is explained to clients by the respective case manager and
informed consent to participate is requested.
George was identified by Chris as a potential candidate for the evaluation. As a former science teacher who loved to do research himself, he
agreed to participate in the project. George is 87 years old, and although he is not as physically robust as he once was, at 5 feet 9 inches tall, he has a
strong presence. He has consistent back pain and occasional flare-ups of rheumatoid arthritis. His wife of 45 years passed away two summers ago
after a long fight with cancer. After his initial grief, he has managed fairly well to adapt to life on his own. George entered the program after being
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hospitalized for fainting while at the grocery store. A battery of medical tests was conducted, but no specific cause of his fainting attack could be
found. However, the physicians assessed signs of slight cognitive impairments/dementia and suggested a geriatric case management program.
An initial assessment by the case manager showed the need for assistance in the following areas: 1) personal care, 2) decrease in mobility, and 3)
longer-term planning around living arrangement and home safety. The case manager also thought that George could benefit from setting up advance
directives, which he did not want to discuss at that time. They agreed that the case manager could bring this topic up again in the future.
As part of the practice process, the case manager used clinical rating scales that were adapted from the task-centered model. At the beginning of
each client contact, case manager and client collaboratively evaluated how well the practice steps (tasks) undertaken by client and/or case manager
were completed using a 10-point clinical scale. Concurrently, they evaluated changes to the respective client problems, also using a 10-point clinical
scale. George was able to actively participate in the planning and implementation of most care-related decisions. During the course of their
collaborative work, most needs were at least partially addressed. Two tasks were completed regarding personal care, two regarding mobility, and three
addressing home safety issues. Only personal mobility was still a minor problem and required some additional work.
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After finishing the reassessment at 3 months, Chris completed gathering and evaluating the data for the single-subject design (SSD). As promised, she
also provided George with the finished SSD findings. The following is an overview of the data that was collected for this case:
TABLE 1. TASK COMPLETION SCORES
WEEK:
1
Area:
Personal Care
Mobility
Home Safety
2
3
7
10
10
4
5
6
2
10
N/A
10
10
7
TABLE 2. PROBLEM CHANGE SCORES
WEEK:
Area:
Personal Care
Mobility
Home Safety
1
2
3
4
5
6
7
3
5
4
3
5
4
8
2
4
5
5
9
8
7
10
9
7
10
9
7
10
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Week 4 Handout: Chi-Square Findings
The chi square test for independence is used to determine whether there is a relationship between
the two variables that are categorical in the level of measurement. In this case, the variables are:
employment level and treatment condition. It tests whether there is a difference between groups.
The research question for the study is: Is there a relationship between the independent variable,
treatment, and the dependent variable, employment level? In other words, is there a difference in
the number of participants who are not employed, employed part-time and employed full-time in
the program and the control group (i.e., waitlist group)?
The hypotheses are:
H0 (The null hypothesis): There is no difference in the proportions of individuals in the three
employment categories between the treatment group and the waitlist group. In other words, the
frequency distribution for variable 2 (employment) has the same proportions for both categories
of variable 1 (program participation).
** It is the null hypothesis that is actually tested by the statistic. A chi square statistic
that is found to be statistically significant, (e.g. p< .05) indicates that we can reject the
null hypothesis (understanding that there is less than a 5% chance that the relationship
between the variables is due to chance).
H1 (The alternative hypothesis): There is a difference in the proportions of individuals in the
three employment categories between the treatment group and the waitlist group.
** The alternative hypothesis states that there is a difference. It would allow us to say
that it appears that the treatment (voc rehab program) is effective in increasing the
employment status of participants.
Assume that the data has been collected to answer the above research question. Someone has
entered the data into SPSS. A chi-square test was conducted, and you were given the following
SPSS output data:
Week 4: A Short Course in Statistics Handout
This information was prepared to call your attention to some basic concepts underlying
statistical procedures and to illustrate what types of research questions can be
addressed by different statistical tests. You may not fully understand these tests without
further study. However, you are strongly encouraged to note distinctions related to type
of measurement used in gathering data and the choice of statistical tests. Feel free to
post questions in the “Contact the Instructor” section of the course.
Statistical symbols:
µ mu (population mean)
α alpha (degree of error acceptable for incorrectly rejecting the null hypothesis,
probability that results are unlikely to occur by chance)
≠ (not equal)
≥ (greater than or equal to)
≤ less than or equal to)
ᴦ (sample correlation)
ρ rho (population correlation)
t r (t score)
z (standard score based on standard deviation)
χ2 Chi square (statistical test for variables that are not interval or ratio scale, (i.e.
nominal or ordinal))
p (probability that results are due to chance)
Descriptives:
Descriptives are statistical tests that summarize a data set.
They include calculations of measures of central tendency (mean, median, and mode),
and dispersion (e.g., standard deviation and range).
Note: The measures of central tendency depend on the measurement level of the
variable (nominal, ordinal, interval, or ratio). If you do not recall the definitions for these
levels of measurement, see
http://www.ats.ucla.edu/stat/mult_pkg/whatstat/nominal_ordinal_interval.htm
You can only calculate a mean and standard deviation for interval or ratio scale
variables.
For nominal or ordinal variables, you can examine the frequency of responses. For
example, you can calculate the percentage of participants who are male and female; or
the percentage of survey respondents who are in favor, against, or undecided.
Often nominal data is recorded with numbers, e.g. male=1, female=2. Sometimes
people are tempted to calculate a mean using these coding numbers. But that would be
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Page 1 of 5
meaningless. Many questionnaires (even course evaluations) use a likert scale to
represent attitudes along a continuum (e.g. Strongly like … Strongly dislike). These too
are often assigned a number for data entry, e.g. 1–5. Suppose that most of the
responses were in the middle of a scale (3 on a scale of 1–5). A researcher could
observe that the mode is 3, but it would not be reasonable to say that the average
(mean) is 3 unless there were exact differences between 1 and 2, 2 and 3 etc. The
numbers on a scale such as this are ordered from low to high or high to low, but there is
no way to say that there is a quantifiably equal difference between each of the choices.
In other words, the responses are ordered, but not necessarily equal. Strongly agree is
not five times as large as strongly disagree. (See the textbook for differences between
ordinal and interval scale measures.)
Inferential Statistics:
Statistical tests for analysis of differences or relationships are Inferential,
allowing a researcher to infer relationships between variables.
All statistical tests have what are called assumptions. These are essentially rules that
indicate that the analysis is appropriate for the type of data. Two key types of
assumptions relate to whether the samples are random and the measurement levels.
Other assumptions have to do with whether the variables are normally distributed. The
determination of statistical significance is based on the assumption of the normal
distribution. A full course in statistics would be needed to explain this fully. The key point
for our purposes is that some statistical procedures require a normal distribution and
others do not.
Understanding Statistical Significance
Regardless of what statistical test you use to test hypotheses, you will be looking to see
whether the results are statistically significant. The statistic p is the probability that the
results of a study would occur simply by chance. Essentially, a p that is less than or
equal to a predetermined (α) alpha level (commonly .05) means that we can reject a null
hypothesis. A null hypothesis always states that there is no difference or no relationship
between the groups or variables. When we reject the null hypothesis, we conclude (but
don’t prove) that there is a difference or a relationship. This is what we generally want to
know.
Parametric Tests:
Parametric tests are tests that require variables to be measured at interval or ratio
scale and for the variables to be normally distributed.
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These tests compare the means between groups. That is why they require the data to
be at an interval or ratio scale. They make use of the standard deviation to determine
whether the results are likely to occur or very unlikely in a normal distribution. If they are
very unlikely to occur, then they are considered statistically significant. This means that
the results are unlikely to occur simply by chance.
The T test
Common uses:
To compare mean from a sample group to a known mean from a population
To compare the mean between two samples
o The research question for a t test comparing the mean scores between
two samples is: Is there a difference in scores between group 1 and group
2? The hypotheses tested would be:
H0: µgroup1 = µgroup2
H1: µgroup1 ≠ µgroup2
To compare pre- and post-test scores for one sample
o The research question for a t test comparing the mean scores for a
sample with pre and posttests is: Is there a difference in scores between
time 1 and time 2? The hypotheses tested would be :
H0: µpre = µpost
H1: µpre ≠ µpost
Example of the form for reporting results: The results of the test were not statistically
significant, t (57) = .282, p = .779, thus the null hypothesis is not rejected. There is not a
difference in between pre and post scores for participants in terms of a measure of
knowledge (for example).
An explanation: The t is a value calculated using means and standard deviations and a
relationship to a normal distribution. If you calculated the t using a formula, you would
compare the obtained t to a table of t values that is based on one less than the number
of participants (n-1). n-1 represents the degrees of freedom. The obtained t must be
greater than a critical value of t in order to be significant. For example, if statistical
analysis software calculated that p = .779, this result is much greater than .05, the usual
alpha-level which most researchers use to establish significance. In order for the t test
to be significant, it would need to have a p ≤ .05.
ANOVA (Analysis of variance)
Common uses: Similar to the t test. However, it can be used when there are more than
two groups.
The hypotheses would be
H0: µgroup1 = µgroup2 = µgroup3 = µgroup4
H1: The means are not all equal (some may be equal)
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Correlation
Common use: to examine whether two variables are related, that is, they vary together.
The calculation of a correlation coefficient (r or rho) is based on means and standard
deviations. This requires that both (or all) variables are measured at an interval or ratio
level.
The coefficient can range from -1 to +1. An r of 1 is a perfect correlation. A + means that
as one variable increases, so does the other. A – means that as one variable increases,
the other decreases.
The research question for correlation is: “Is there a relationship between variable 1 and
one or more other variables?”
The hypotheses for a Pearson correlation:
H0: ρ = 0 (there is no correlation)
H1: ρ ≠ 0 (there is a real correlation)
Non-parametric Tests
Nonparametric tests are tests that do not require variable to be measured at
interval or ratio scale and do not require the variables to be normally distributed.
Chi Square
Common uses: Chi square tests of independence and measures of association and
agreement for nominal and ordinal data.
The research question for a chi square test for independence is: Is there a relationship
between the independent variable and a dependent variable?
The hypotheses are:
H0 (The null hypothesis) There is no difference in the proportions in each category of
one variable between the groups (defined as categories of another variable).
Or:
The frequency distribution for variable 2 has the same proportions for both categories of
variable 1.
H1 (The alternative hypothesis) There is a difference in the proportions in each category
of one variable between the groups (defined as categories of another variable).
The calculations are based on comparing the observed frequency in each category to
what would be expected if the proportions were equal. (If the proportions between
observed and expected frequencies are equal, then there is no difference.)
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See the SOCW 6311: Week 4 Working With Data Assignment Handout to explore the
Crosstabs procedure for chi square analysis.
Other non-parametric tests:
Spearman rho: A correlation test for rank ordered (ordinal scale) variables.
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