disc0222

Statistical Inferences" Please respond to the following:

· Choose one (1) of the statistical tests presented in Chapter 2, and provide a real-world example in which it is applicable. Explain why the test is appropriate for the example.

· Conclude why reporting results with a margin of error is more informational than just the point estimate, such as the sample mean. Provide support for your rationale.

LEARNING OBJECTIVE 2: TO BE ABLE TO COMPARE DIFFERENT TYPES OF DATA USING STATISTICAL INFERENCE AND HYPOTHESIS TESTING

Bivariate Analysis

The second primary function of managerial statistics is to be able to compare two or more variables within a set of data. This can mean comparing a variable measured at two points in time, such as the number of births from one year to the next, or in two locations, such as comparing births between hospitals. It can also mean comparing two different types of data, such as the number of emergency department (ED) visits over a time period with the number of lab tests performed during that same period. Each type of analysis again requires knowing what types of data you are comparing. Like descriptive statistics, there are certain types of analyses that you will perform and others we can set aside depending on how the data are measured. Before doing so, however, we first need to review the need for hypothesis development and testing.

Hypothesis Testing

Students may recall from an introductory statistics course, that comparisons of variables are best tested using hypotheses. These are simply statements of association that are first stated and then, using analysis, either supported or refuted. The reason for doing this is that most data are simply representations of phenomena that exist in real life. The problem is that it is usually unrealistic or impossible to measure all phenomena completely. Think about measuring population. We may want to know the actual number of people in the United States, but our ability to measure this is limited. Realistically, we cannot find and count everyone in the United States without missing some people, and the number of people changes daily because of births and deaths, so that by the time we were done measuring, the “real” answer would have already changed. Yet, we also know that for a given point in time there is a “real” measurement; we just are unable to observe it. We can, however, estimate within some level of certainty, whether or not the measurement we observe, or the data comparison we make, is likely to be representative of what is “real” at that point in time.

This is why we create hypotheses and then use statistical tests to either support or refute them. Two primary types of hypotheses are used in statistical analysis. The first is the null hypothesis, which is always the hypothesis of no association or difference. The second is the alternative hypothesis, which is most often the converse of the null, but that can be directional, such as two data elements having a positive or negative association. Both are examined here briefly.

Managers in healthcare settings are often assessing data for comparative purposes, and often they use samples of data taken at a point in time. The question managers should be concerned with is not only if there is an observable difference or association in the data, but with what level of confidence can you believe it to be true and not because of chance. Otherwise stated, if the manager collected another sample of data, could the association or difference reverse itself or would the data be reflecting the same pattern? These are the foundational questions behind hypothesis testing. For example, consider a manager who has collected data on the number of safety protocol violations within two units of the hospital. In summarizing these violations, she finds the mean number of violations of unit A to be 21 over a 1-year period, and 27 in unit B over the same period. In real terms, unit B does report more violations than unit A. The question is whether this trend is a “real” trend, or simply a result of chance. Thus, the question we ask is how likely are we to record or see a difference as big as we have when the difference is actually zero in reality.

Often, the stating of hypotheses is unduly confusing. In fact, it is quite simple. The null hypothesis always states that there is no difference or association between the two things being observed (i.e., data variables). For example, we could hypothesize that the mean number of violations in site A is actually no different than the mean number of violations at site B “in reality” if we were to continue to measure over time. The alternative is that there is a real difference between the two things being observed. But here we have an option. We can say that we think, for example, that the mean number of violations at site B is simply different than site A, whether that is higher or lower. When direction doesn’t matter, we are conducting a two-tailed hypothesis test. Our second alternative is to say that one will be higher than the other or lower than the other. In this case we might say the alternative hypothesis is that the mean number of violations at site B is higher than the mean number of violations at site A. In this case we are conducting a one-tailed hypothesis test. The difference occurs primarily with respect to interpretation of the tests and is explored later in the chapter. The stated null hypothesis, abbreviated H0, or that of no difference, for this example would be:

H0: There is no difference in the mean number of safety violations between site A and site B over the 1-year period.

The stated alternative hypothesis, abbreviated Ha, is the converse of this and, assuming a two-tailed test, would be:

Ha: There is a difference in the mean number of safety violations between site A and site B over the 1-year period.

The analysis we perform will allow us to either reject or fail to reject our null hypothesis. The rule here is that we never accept a hypothesis. Why? Because we can never be 100% certain what the relationship between two things is “in reality” at a given point in time, for reasons stated earlier in this section. Instead, we use hypothesis testing and statistics to make probabilistic inference into the relationship between two sets of measured data or observations. Interpreting our hypotheses now requires the use of statistics, and also a brief introduction to theoretical probability distributions, otherwise thought of as why we can be certain we are at least partially certain.

Probability Distributions

If someone were to ask you what the probability of flipping a normal coin and having it come up heads, you would no doubt say that it is a 50/50 chance, or 50% of the time. Yet you would also likely agree that it is quite possible that you could flip a coin and heads would come up three times in a row. How can this be? Two reasons. One is that each coin flip is not dependent on the previous one. There are two sides of the coin, so you only have two possible outcomes. Each time you flip, they are equally likely to come up (if the coin is balanced and not a trick coin). The second is that we know if you continue to flip over and over again, the number of heads and the number of tails will start to equal out. In statistical language, we would say the probability of heads grows closer to 0.5 as your n (number of flips) increases. Suffice it to say that flipping a coin has a known probability. Could we observe 37 heads in a row? Sure, but it is highly unlikely.

Most phenomena in the world have a distribution of measurement, whether height, weight, income, hair length, etc. Consider height. There are a range of heights of individuals throughout the world. Some are quite tall, and others are not. If, for example, we see someone who is 8 feet tall, we might think that it is unusual, but not impossible. (Seeing is believing.) But how do we test this statistically?

Here we offer a non-statistical explanation of a statistical occurrence. As we stated before, at a point in time, all phenomena are theoretically measurable. If we are examining a data element that is continuous in nature, such as height, then at a point in time there is also a “true” mean of the observed data—right now there is a “true” mean height of all people in the world. Similarly, if we were to measure all persons, there would also be a “true” standard deviation around that mean. Some measurements will be close to the mean and others further away. We would expect that observations that were further from the mean would be less likely to occur, as with our 8-foot friend. From statistics we know how likely certain data will be to occur in relation to its mean by measuring how far those observations are from the mean in units of standard deviation. The reason for this is that many types of data are distributed normally, or in a fashion in which there is a mean and a symmetrical distribution of values on either side in the shape of a bell curve ( Figure 2-3A ). For example, if we were to know that the “true” mean of heights in the United States for men is 68 inches, and the “true” standard deviation is 2 inches, someone who is 8 feet tall (96 inches) would be 14 standard deviations above the mean or (96–68)/2. And because all data can be examined by how far in standard deviations they are from the mean, we can construct a theoretical normal distribution in which the mean is zero and the area under the curve represents units of standard deviation, called z-values. Why is the mean zero? If the distances under the curve are measures of standard deviation, then how many standard deviations away from the mean is the mean? Zero. Not only does this allow us to assess the probability of occurrence for certain data, it allows us to compare any type of data because the units are the same (standard deviations) (see  Figure 2-3 ,  A , B ).

Figure 2-3A Normal Distribution Showing Mean, Standard Deviations, and Percentage of Observations Falling Within the Standard Deviations

Figure 2-3B Distribution of Height (M = 68″ and SD = 2″)

Further, we know that the areas under standard normal distributions have known probabilities. The 68-95-99.7 rule states that approximately 68% of observations fall within one standard deviation of its mean, approximately 95% of observations fall within two standard deviations of the mean, and approximately 99.7% of observations fall within three standard deviations of the mean. In addition, each z-value under the curve has a known probability of occurrence. There are also other distributions that do not quite follow the symmetrical shape of a z-distribution (standard normal distribution), but nonetheless have known probabilities under them, such as t-distributions and f-distributions to name only two. For our purposes, it is not important to know what these distributions look like, but that they have known probabilities, and so any observations we may make can be tested to see what the likelihood of its occurrence is. All we need to know is what test to run to for what type of data, and then how to interpret the results we get from our computer analysis. We next examine this for a number of data comparisons.

Comparing Continuous Data

There are different analytic techniques for comparing a continuous data variable measured at different points in time or across locations, and two different continuous variables to one another. First, let us examine comparing two different continuously measured variables, lab tests and ED visits.

Correlation

In this instance, we look to statistics to provide a measure of association between two differently measured phenomena. Because they are measured in increments of equal distance, respectively, we can assess how unit changes in one variable are correlated to unit changes in the other. This becomes an algebraic relationship, in which if we label one variable x and the other y, we can express y as being some function of x. Examine  Table 2-3 , which measures the number of both ED visits and lab tests for a sample period during the month of September. If they were perfectly correlated, these variables would have a one-to-one relationship. In this example, a perfectly positive correlation would mean each additional ED visit would result in the same additional number of lab tests. Similarly, if there was a perfectly negative relationship, for every ED visit, lab tests would consistently decrease by a set amount. To do this we calculate the linear correlation coefficient (r), which will indicate the associative, but not causal relationship between the number of ED visits and the number of lab tests performed per day. The correlation coefficient will also indicate the strength of the linear association between the two variables.

Statistics indicates that a correlation coefficient (r) of +1.00 indicates a perfectly positive correlation (an increase in X is always associated with a parallel increase in Y) and that a negative correlation coefficient of –1.00 indicates a perfectly negative correlation. By definition, correlation coefficients can only range from +1.00 to –1.00. For the data in  Table 2-3 , the correlation coefficient is 0.977, which indicates a strong positive correlation between ED visits and lab tests. Although this seems to be indication of a powerful correlation, the association cannot yet be said to be statistically significant or not one because of random chance.

Table 2-3 Observational Data for ED Visits and Lab Tests

Date	Ed Visits	Lab Tests
15-Sep	5	12	Correlation Coefficient r = .977 Critical value of r = .532 with n = 14
16-Sep	8	12
17-Sep	9	24
18-Sep	12	36
19-Sep	14	48
20-Sep	2	0
21-Sep	4	0
22-Sep	8	12
23-Sep	7	12
24-Sep	12	36
25-Sep	14	48
26-Sep	6	12
27-Sep	18	60
28-Sep	12	36

Here we have collected a sample of data based on 14 days of observation. The question we must ask is whether the observed phenomenon could be owing simply to chance rather than some real association. Stating our hypotheses is helpful in doing this. Here, our null and alternative hypotheses are:

Ho: There is no relationship between the number of ED visits and the number of lab tests.

Ha: There is a relationship between the number of ED visits and the number of lab tests.

To address our hypotheses, we must now determine a critical value of r to assess the likelihood of the relationship being because of chance. Although some computer programs give the actual probability, or likelihood of the relationship with a p-value, others, such as Excel, do not. Here we present a table of critical r-values, shown in  Table 2-4 , which gives the critical values of r at various sample sizes (n) at the alpha of 0.05. Given our example, we would use the critical value of r for n = 14, which is 0.532. If r-calculated > r-critical, we can be 95% confident that the association is not because of random chance. Note that analysis programs such as Excel will compute the correlation coefficient r, but do not provide the critical value of r, or the probability associated with r. Here, r-calculated (0.979) is greater than r-critical (0.532), so we can say that there is a statistically significant positive correlation between ED visits and lab tests. For each additional ED visit (1 unit), we would expect lab test volume to increase by 0.977 units.

Table 2-4 Critical Values of the Correlation Coefficient r for Various Sample Sizes n

n	r	n	r
5	0.878	18	0.468
6	0.811	19	0.456
7	0.754	20	0.444
8	0.707	22	0.423
9	0.666	24	0.404
10	0.632	26	0.388
11	0.602	28	0.374
12	0.576	30	0.361
13	0.553	40	0.312
14	0.532	50	0.279
15	0.514	60	0.254
16	0.497	80	0.220
17	0.482	100	0.196

T-Tests

A second common analysis is to examine a continuously measured variable at two points in time, or in two locations. For example, say we wish to compare the number of births as Port City Hospital with other U.S. hospitals of similar size. To do so we collect data over 12 months, shown in  Table 2-5 .

Examining the data we see that for all months, Port City Hospital performs more births than the average hospital of similar size in the United States and that the mean number of births over the period was 37 at Port City and 29 at other hospitals. Our question is whether the data we see here for 1 year represent the “true” relationship between Port City and other similar size hospitals. Because this is only a sample of data from 1 year, we must use statistics to assess this. First, however, we should state our hypotheses.

Ho: There is no difference between the mean number of births at Port City Hospital and other U.S. hospitals of similar size.

Ha: There is a difference between the mean number of births at Port City Hospital and other U.S. hospitals of similar size.

To test the difference between two means requires the use of a t-test. T-tests are used to compare means between groups. These groups can be paired, as would be the case with a group who is measured on some variable, for example, blood pressure, undergoes some intervention, for example, an exercise routine, and re-measured. The groups can also be different, as is the case with our comparison of mean births at Port City Hospital with other hospitals. What cannot be compared are means for different variables, such as the mean average length of stay compared with the mean number of births. The means must be measured in similar units for comparison with a t-test.

Table 2-5 Comparative Monthly Births

	Port City Hospital	U.S. for Similar Size Hospitals
January	24	22
February	25	21
March	33	26
April	35	27
May	37	31
June	38	25
July	41	36
August	35	27
September	45	39
October	39	35
November	42	34
December	50	23
Mean	37	29

Most analytic software including Excel can calculate a number of t-tests. What is important to note is that the different types of t-tests (paired, assuming equal variances, and assuming unequal variances) revolve, as the names suggest, around variation of the data. For our purposes, we will assume that variances are unequal in cases other than paired data. In practice, this difference in variances would be analyzed with an f-test, and some programs will provide output for both equal and unequal variances assumed. Because Excel does not do this, assume unequal variances. Rarely will the interpretation differ between the two, but it can. The t-test output for our data is shown in  Table 2-6 .

Table 2-6 Excel Output t-Test: Two-Sample Assuming Unequal Variances

	Port City	U.S.
Mean	37	28.8
Variance	56	36.0
Observations	12	12
Hypothesized Mean Difference	0
df	21
t Stat	2.9499
P(T≤t) one-tail	0.0038
t Critical one-tail	1.7207
P(T≤t) two-tail	0.0076
t Critical two-tail	2.0796

In  Table 2-6  we are given a number of analytic outputs. The first is the mean of the data for both Port City and the United States. We are also given the variance and the number of observations. The hypothesized mean difference is simply the null hypothesis restated. Examining the lower half of the table we are given the t-statistic, the probability of t for both one-sided and two-tailed tests, and the critical value of t that cuts off the upper or lower 2.5% of the distribution (one-tailed) and the value of t that cuts off the upper and lower 2.5% of the distribution (two-tailed). Thus, values of the t-statistic that lie beyond the critical value are statistically significant (different) at the 95% level of confidence. The p-value gives the exact probability that our means are truly different and that the observed difference is not because of random chance. Here, we would use a two-tailed p-value because our hypothesis was not directional. That is, our null stated that the mean number of births was different, but not in which direction (greater than or less than). Doing so would require a one-tailed test, because we would only be interested in values at one end of the distribution. Here it doesn’t matter, so we conduct and interpret the two-tailed test. Interpreting the t-statistic we see that:

T Stat (2.95) > t Critical two tail (2.07) or p(T ≤ t)(0.0007) < 0.05

In either case, we would reject the null hypothesis.

Otherwise stated, our hypotheses ask what is the likelihood of seeing a difference in observed means as large as the one we did (8.2, or 37–28.8) if in fact the real difference were zero (the null hypothesis). Examining our t Stat relative to the critical values tells us the likelihood is less than 5% of the time. Examining our p-value tells us the exact likelihood, or less than 0.7% of the time (0.007). So, if we continue to collect samples of data, the means would likely only be the same in 0.7% of samples.

Comparing Categorical Data

Table 2-7 2 × 2 Contingency Table

	Group 1	Group 2	Total
variable 1	a	b	a + b
variable 2	c	d	c + d
Total	a + c	b + d	a + b + c + d

Finally, we may often be interested in analysis of two variables that are measured categorically through either rates or proportions. Examples of these types of data include: What type of insurance does the patient have; what is the gender; and were they satisfied with their visit? Summarizing these involves creating counts and percentages. However, often we wish to compare how two groups of categories compare with one another. We do this using the chi-square statistic (χ2), which compares the observed differences in proportions with what would be expected if proportions were equal. For example, if we were to examine the satisfied/unsatisfied percentages of 40 men and 40 women on a satisfaction questionnaire, we would expect that if they were equal, 20 would say satisfied and 20 would not in each gender category (or 50% for each). When we observe actual data, however, we often see different results. The basic null and alternative hypotheses hold true here. The question we are asking is what is the chance of seeing a difference of the magnitude observed in the collected data if in fact there is no true difference (all proportions are equal) in the population. The chi-square statistic and its associated probability allow us to test these hypotheses. The simplest form of chi-square analysis is of two variables using a 2 × 2 contingency table, shown in  Table 2-7 .