Can the experts help me with this statistics question please?

Question

Your boss promotes you to a senior position for your excellence in statistical analysis. Shortly after, he brings you a set of data comparing patient satisfaction rates between different hospitals. How would you tell whether your hospital is performing better, same or worse than the other hospitals? Please indicate what statistical tests (t-test, z-test, ANOVA, Chi-square, Mann-Whitney) you would perform to make your conclusion? Thank you very much

Answer 1

t-tests, z-tests and the Mann-Whitney are used for testing for a difference between two samples. Since I assume you have more than two hospitals then you should not use these tests because you'll have to conduct multiple tests. you can correct for errors induced by multiple tests by using Bonferroni's correction or Fisher's LSD.

If the data collected is parametric then an ANOVA will be the best test for you. If the data is non-parametric then the Chi-Square test.

Let's explain way.

let's assume that you have four hospitals. Let's also assume you have parametric data. if you do a t/z - test or the Mann-Whitney then you have to conduct 6 different hypothesis tests.

H0: μ1 = μ2
H0: μ1 = μ3
H0: μ1 = μ4
H0: μ2 = μ3
H0: μ2 = μ4
H0: μ3 = μ4

if you are testing at the α level then the probability of a Type I error is bounded between 0 and α.

The probability of not committing a Type I error is (1 - α)

for 6 tests the probability of not committing a Type I error is (1 - α) ^6. For example, testing at the 5% level means the probability of not committing a Type I error is 0.7350919. The probability of committing a Type I error is 0.2649081. This shows that for finding a difference between the means of four groups you have a probability of being wrong going from 5% to over 26%.

Using ANOVA will correct for this as ANOVA is designed to test the hypothesis:

H0: μ1 = μ2 = μ3 = μ4

If the data is non-parametric then you should use a non-parametric test and for the multiple groups you should use the chi-square test.

How do we know if the data is parametric or non-parametric?

Well, since the data is from a satisfaction survey it is most likely non-parametric. If you have a survey with responses such as "strongly disagree," "disagree," "neutral," "agree," and "strongly agree" then you have non-parametric data. Many people will map the responses to an integer value scale such as 1, 2, 3, 4, 5. this will allow you to find a mean and treat the data as parametric. However, we do not know if the scale for the mapping is correct. perhaps the mapping should be 1, 4, 4.5, 4.7, 5.6. The selection of the scale can impact the results of the parametric tests.

Using the Chi-square will look at the proportion of responses for each option in each group and will provide you with more reliable results.

Answer 2

Let's assume you have enough patients surveyed that your data are normally distributed so you can use a parametric test.

Only use a z-test if you are dealing with two whole population groups (such as census data). Because these tests often deal in specific populations (i.e., hospital patients) t-tests are more appropriate for comparing group means. If it's two hospitals, you'd use a t-test. But if you have more than two hospitals to compare, you'd have to use ANOVA.

Suppose the patients surveyed were asked "Did you enjoy our sevice? Yes or No." Then your data would be dichotomous. Now suppose you were comparing that data between two hospitals. Then you would have to do a Chi-square test becuse your variable are dichotomous.