Overview

As part of continuing your work on your evidence-based project proposal, identify one to three outcomes of interest. That is,what you hope to change or improve based on the implementation of your project (your project’s dependent variable). For example, if you are designing an intervention to reduce hospital re-admission rates for patients with heart failure, the outcome is re-admission rates. Another example would be a mindfulness-based intervention for critical care nurses to reduce burnout; in this case, burnout is the outcome of interest.

Once you have identified the outcomes of interest in your project, you need to determine how each outcome will be measured. Consider the examples above. How might you measure re-admission rates? (For example, you might measure the percentage of patients with heart failure who are re-admitted with a diagnosis of heart failure within 90 days of being discharged). How about measuringburnout in critical-care nurses? (Would you use the Maslach Burnout Inventory, or some other tool?). As you can see, there are different ways to measure outcomes.

The Maslach Burnout Inventory is an example of a measurement tool. Many tools such as this exist to measure a variety of phenomena such as resilience, moral distress, self-efficacy, and many others! These tools can include surveys or questionnaires that have been used in the literature to evaluate similar evidence-based practice projects. Many tools may be available to you depending on your topic. This assignment involves you searching the databases to learn about how your topic has been evaluated in the past.

To find tools and determine how outcomes can be measured, start by reading the literature. What tools are frequently used to assess the variables or outcomes of interest? Some are very commonly used. When you find a tool, you’ll want to review the original primary source —the published (or revised and updated) book or article where the tool was first described. Evaluate how the tool was developed and if it was found to be reliable and valid. It is very interesting to read instrument-development articles, so please do if you get the chance! Often the titles of these articles contain the terms “development and psychometric testing of the _ scale/tool.” Please note when actually conducting research that there are many considerations for the selection and use of measurement tools, including permission to use the tool from the researcher.

Assignment Instructions

For this assignment, select one to three outcomes and identify how each outcome will be measured. How many outcomes you have depends on your individual project.

As discussed, you might need to determine exactly how certain outcomes will be measured (such as re-admission rate). It is also possible that there is no tool available to measure an outcome of interest in your study. For example, if I wanted to assess “knowledge” of some topic, I would need to create questions to obtain data about this outcome. In either case, include the following information:

Clearly state the outcome and how, specifically, you will measure it.
Describe why you selected the measurement method and how you plan to use it in the project.

If you are able to identify an appropriate measurement tool that already exists (such as the Maslach Burnout Inventory) for an outcome of interest, include the following information:

The outcome and the name of the tool that will be used to measure the outcome.
A brief description the tool. How many items are there? How are items scored? What do scores mean?
An explanation of why you selected this tool and how you plan to use it in your project.
The validity and reliability of the tool.

Here is an example:

Self-Efficacy. Self-efficacy will be measured using the General Self-Efficacy Scale (GSE) (Schwarzer & Jerusalem, 1995).

The scale was developed in 1979 and subsequently revised and adapted to 26 languages, and consists of 10 items, scored on a scale of 1 (not at all true) to 4 (exactly true), with a score range of 10 to 40 (where lower scores indicate lower self-efficacy and higher scores indicate higher self-efficacy) (The General Self-Efficacy Scale, n.d.).

In a sample of 747 early career nurses, the scale had a Cronbach’s alpha of 0.884 (Wang et al., 2017). This establishes reliability in that sample (early career nurses).

The GSE scale was selected because it offers a general overview of the concept of self-efficacy and is not specific to nursing practice. In the proposed study, the GSE scale will be given to participants before and after the intervention.

After you identify between two and five peer-reviewed tools, in a Microsoft Word document, describe in 300 to 500 words why you have selected them and how you plan to use them in your proposal. Include the validity and reliability of the tools (which is found in journal articles). Submit the names of the tools along with your 300- to 500-word justification and ensure that you use APA format.

Please refer to the Grading Rubric for details on how this activity will be graded.

A study of an association between which ear is used for cell phone calls and whether the subject is​ left-handed or​ right-handed began with a survey​ e-mailed to 5000 people belonging to an otology online​ group, and 717 surveys were returned.​ (Otology relates to the ear and​ hearing.) What percentage of the 5000 surveys were​ returned? Does that response rate appear to be​ low? In​ general, what is a problem with a very low response​ rate?

Constructing a Confidence Interval for the Difference between Two Population Proportions
In order to determine if a new instructional technology improves students' scores, a professor wants to know if a larger percentage of students using the instructional technology passed the class than the percentage of students who did not use the new technology. Records show that 45 out of 50 randomly selected students who were in classes that used the instructional technology passed the class and 38 out of 51 randomly selected students who were in classes that did not use the instructional technology passed the class. Construct a 95%

confidence interval for the true difference between the proportion of students using the technology who passed and the proportion of students not using the technology who passed.

Solution

We are going to show how to construct the confidence interval first without a TI-83/84 Plus calculator and then with one.
Step 1: Find the point estimate.

First, we'll let Population 1 be those students who used the new technology and Population 2 be those students who did not. Next, we need to calculate the sample proportions. The sample proportion for Sample 1 (using instructional technology) is calculated as follows.

pˆ1=x1n1=4550=0.9

The sample proportion for Sample 2 (without the instructional technology) is found as follows.

pˆ2=x2n2=3851≈0.745098

Now that we have the sample proportions, we can calculate the point estimate.

pˆ1−pˆ2=0.9−0.745098=0.154902

Step 2: Find the margin of error.

Notice that the samples are indeed independent of one another. Because they are two separate groups of students, they are not connected in any way. We can assume that the other necessary conditions are met to allow us to use the standard normal distribution to calculate the margin of error. The level of confidence is c=0.95
, so the critical value is zα2/=z0.052/=z0.025=1.96

. Substituting the values into the formula gives us the following.

E=zα2/pˆ1(1−pˆ1)n1+pˆ2(1−pˆ2)n2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√=1.960.9(1−0.9)50+0.745098(1−0.745098)51⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√≈0.145675

Step 3: Subtract the margin of error from and add the margin of error to the point estimate.

Subtracting the margin of error from the point estimate and then adding the margin of error to the point estimate gives us the following endpoints of the confidence interval.

Lower endpoint: (pˆ1−pˆ2)−E=0.154902−0.145675≈0.009Upper endpoint: (pˆ1−pˆ2)+E=0.154902+0.145675≈0.301

Thus, the 95%
confidence interval for the difference between the two population proportions ranges from 0.009 to 0.301

. The confidence interval can be written mathematically using either inequality symbols or interval notation, as shown below.

0.009<p1−p2<0.301

or

(0.009,0.301)

Therefore, we are 95%
confident that the percentage of students who passed the class is between 0.9% and 30.1% higher for the population of students who used the new instructional technology (Population 1) than for the population of students who did not use the technology (Population 2). Thus, with 95%

confidence, the professor can conclude that the new instructional technology improves students' scores.

To calculate the confidence interval for the difference between two proportions on the calculator, we don't need to find the individual sample proportions; we just need to enter the number of successes and the sample size for each sample, as well as the level of confidence. Press STAT , scroll to TESTS, and then choose option B:2-PropZInt. x1 is the number of successes from the first sample and n1 is the first sample's size. Similarly, x2 is the number of successes from the second sample and n2 is the second sample's size. As usual, C-Level is the confidence level, which must be entered as a decimal. The data should be entered as shown in the first screenshot below. After you select Calculate and press ENTER , the results will be displayed on the screen as shown in the second screenshot below.
2-PropZInt data entry screen with x_1 equal to 45, n_1 equal to 50, x_2 equal to 38, n_2 equal to 51, and C-Level equal to .95. 2-PropZInt results screen shows ( .00923 , .30057 ), p hat_1 equal to .9 , p hat_2 equal to .7450980392, n_1 equal to 50, and n_2 equal to 51.

Notice that the calculator gives the same interval but with more decimal places. The interpretation of the confidence interval is still the same. The proportion of students passing the class was higher for the population of students who used the new instructional technology than for the population of students who did not use the technology.

In this section we will turn our attention to comparing two population proportions. Once again, there are times when we aren't necessarily focused on the exact proportion, but rather how proportions from two populations compare, that is, if they are equal, or if one is larger than the other.

When we were comparing population means, we constructed a confidence interval for the difference between the two population means. Similarly, when comparing two population proportions, we use a confidence interval for the difference between the population proportions. The best point estimate for the difference is pˆ1−pˆ2

. In this section we will restrict our discussion to comparing two population proportions when the following conditions are met. Notice that the conditions are similar to those discussed for estimating a single population proportion.

All possible samples of a given size have an equal probability of being chosen; that is, simple random samples are used.

The samples are independent.

The conditions for a binomial distribution are met for both samples.

The sample sizes are large enough to ensure that n1pˆ1≥5

, n1(1−pˆ1)≥5, n2pˆ2≥5, and n2(1−pˆ2)≥5

.

When these conditions are met, we can apply the Central Limit Theorem to the sampling distribution of the differences between the sample proportions for two independent samples. This means that we will use the standard normal distribution to calculate the margin of error of a confidence interval for the difference between two population proportions. You can assume that the necessary criteria are met for all examples and exercises in this lesson.
Memory Booster

Population Proportion

p=xN=# of successespopulation size

Sample Proportion

pˆ=xn=# of successessample size

Properties of a Binomial Distribution

    The experiment consists of a fixed number, n, of identical trials.

    Each trial is independent of the others.

    For each trial, there are only two possible outcomes. For counting purposes, one outcome is labeled a success, and the other a failure.

    For every trial, the probability of getting a success is called p. The probability of getting a failure is then 1−p

.

The binomial random variable, X, counts the number of successes in n trials.

If there are n pairs of data values and the population distribution of the paired differences is approximately normal, then the sampling distribution for the sample statistic d⎯⎯ follows a t-distribution with n, n−1 degrees of freedom. Hence, the formula for the margin of error is as follows. This is the same formula that is used when estimating a single population mean when σ is unknown. This is because we use the paired differences as a single set of sample data rather than using the data from the two samples separately when working with paired data.
Margin of Error of a Confidence Interval for the Mean of the Paired Differences for Two Populations ( σ Unknown, Dependent Samples)

When both population standard deviations are unknown, the samples taken are dependent, simple random samples of paired data, and either the number of pairs of data values in the sample data is greater than or equal to 30

or the population distribution of the paired differences is approximately normal, the margin of error of a confidence interval for the mean of the paired differences for two populations is given by

E=(tα2/)(sdn⎯⎯√)

where tα2/
is the critical value for the level of confidence, c=1−α such that the area under the t-distribution with n−1 degrees of freedom to the right of tα2/ is equal to α2

.

sd

is the sample standard deviation of the paired differences for the sample data, an

n is the number of paired differences in the sample data.