Statistics for Health Management Decision-Making
Part I: [60 points]
Continuation of the “Hospitalizations for Heart Failure in Hamilton Health System”
Recall the hfdata.dta posted under Unit 1 and Assignment 1. Please refer to the Practicing with STATA: Hospitalizations for Heart Failure in Hamilton Health System (PDF) under Data and Statistical Software Resources to remind yourself of this exercise. Recall that the assignment was based on 974 discharges of heart failure patients from the Hamilton Health System (HHS).
You are now provided with a random sample of 100 hospital discharges of patients with heart failure from the HHS. The data (hf100sample) are provided in the data folder for Assignment 2 in STATA and Excel formats. You can answer the questions below using STATA or Excel.
STATA hints: Consider the following commands: ci, ttest, prtest
[10 points] Test the null hypothesis at 95 percent level of confidence that the average length of stay (los) for the population is 3.80 days against the alternative that length of stay is not 3.80 days, and state your conclusion.
Fill in the blanks and the table (up to 3 decimal points) below:
Null hypothesis: H0: _________________
Alternative hypothesis: H1: _________________
Mean los Standard Deviation of los Standard Error of mean los Critical t-value
Tinv () T-Stat Lower Bound of the Confidence Interval Upper Bound of the Confidence Interval
Interpret the confidence interval________________________________
Conclusion: Do you reject the Null hypothesis? Why/Why not? ________
[10 points] Test the same hypothesis for length of stay (los) at the 99 percent level of confidence and state your conclusion. Interpret the confidence interval.
Fill in the blanks and the table (up to 3 decimal points) below:
Null hypothesis: H0: _________________
Alternative hypothesis: H1: _________________
Mean los Standard Deviation of los Standard Error of mean los Critical t-value
Tinv () T-Stat Lower Bound of the Confidence Interval Upper Bound of the Confidence Interval
Interpret the confidence interval________________________________
Conclusion: Do you reject the Null hypothesis? Why/Why not? ________
[10 points] Test the null hypothesis at 95 percent level of confidence that the average charges (totchgs) for the population is $9,500 against the alternative that charges is not $9,500, and state your conclusion. Interpret the confidence interval.
Fill in the blanks and the table (up to 3 decimal points) below:
Null hypothesis: H0: _________________
Alternative hypothesis: H1: _________________
Mean charges Standard Deviation of charges Standard Error of mean charges Critical t-value
Tinv () T-Stat Lower Bound of the Confidence Interval Upper Bound of the Confidence Interval
Interpret the confidence interval________________________________
Conclusion: Do you reject the Null hypothesis? Why/Why not? ________
[10 points] Test the same hypothesis for average charges (totchgs) at the 99 percent level of confidence and state your conclusion. Interpret the confidence interval.
Fill in the blanks and the table (up to 3 decimal points) below:
Null hypothesis: H0: _________________
Alternative hypothesis: H1: _________________
Mean charges Standard Deviation of charges Standard Error of mean charges Critical t-value
Tinv () T-Stat Lower Bound of the Confidence Interval Upper Bound of the Confidence Interval
Interpret the confidence interval________________________________
Conclusion: Do you reject the Null hypothesis? Why/Why not? ________
[10 points] Test the null hypothesis at 95 percent level of confidence that 6% of the discharges result in the patient’s death while in hospital (pat_died). The alternative hypothesis is that the percentage of discharges that result in patient death while in hospital does not equal 6%.
Calculate the Z test statistic
Calculate the p-value (Hint: NORM.S.DIST in Excel)
State whether you reject the null hypothesis or not, and explain why.
[10 points] Test the null hypothesis at 95 percent level of confidence that more than 75% of the patients are admitted through the emergency room (eradmit). The alternative hypothesis is that less than 75% of patients are admitted through the emergency room.
Calculate the Z test statistic
Calculate the p-value (Hint: NORM.S.DIST in Excel)
State whether you reject the null hypothesis or not, and explain why.
Part II: [40 points]
Hospital Compare Exercise
Hospital Compare is a hospital reporting database in the United States (U.S) that was created by the Centers for Medicare & Medicaid Services (CMS) in collaboration with consumers and health care providers including hospitals, physicians and other organizations. The database is publicly available, and includes quality information at over 4,000 Medicare-certified hospitals in the U.S. It is a valuable resource that provides quality transparency to patients, and enables quality monitoring and improvement by hospitals and policy makers. You can read more about Hospital Compare here:
https://www.medicare.gov/hospitalcompare/search.html
Latest set of all Hospital Compare data files can be found here:
https://data.medicare.gov/data/hospital-compare
Your team is given a hospital-level sample on 30-day readmission rate for heart attack (AMI) patients. You can read about the full set of readmission measures of Hospital Compare in detail here:
https://www.medicare.gov/hospitalcompare/Data/30-day-measures.html
In addition to the original data pulled from Hospital Compare, you are also provided a variable that assigns a “readmission grade” to each hospital. Grades range from 1 to 6, with 1 assigned to the hospitals with the highest re-admission rates, and 6 assigned to those with the lowest readmission rates.
Reducing readmission rates has been an important policy initiative nationally in the U.S. You can read more about CMS’s readmission reduction program here: https://www.medicare.gov/hospitalcompare/readmission-reduction-program.html
Consider you are a project lead working with a team of physicians and policy makers at the Minnesota Department of Health. Your task is to investigate:
How select Minnesota hospitals compare to the average Minnesota benchmark in terms of their readmission rate for AMI patients.
How select states on average compare to the average U.S. hospital in terms of their hospitals’ readmission “grade” for AMI patients.
You are provided with a sample of 2,274 U.S. hospitals that reported their 30-day re-admission rate for AMI patients. Below is a description of the hospcompare data available both in STATA and Excel.
——————————————————————————————— variable name type format label variable label
———————————————————————————————
hospital_id str6 %6s hospital’s identifier
hospital_name str50 %50s hospital’s name
hospital_city str20 %20s hospital’s city
hospital_state str2 %2s hospital’s state
sampsize int %10.0g number of discharges used for readmission rate
readm_ami double %10.0g 30 day readmission rate of AMI patients
grade float %9.0g hospital’s grade for readmissions: high grade
means low readmissions
———————————————————————————————
[20 points] You will see that there are 26 Minnesota hospitals in the data. You are interested in studying select Minnesota (MN) hospitals on how they compare to the average MN benchmark in terms of their readmission rate for AMI patients. You are given readmission rate estimate (readm_ami) for each hospital. You are also given the sample size (sampsize) over which the readmission rates are estimated. Perform a test comparing each hospital’s readmission rate to the state of MN benchmark.
What is the MN benchmark readmission rate (average across all hospitals in Minnesota) in 3 decimal points?
STATA hint: summ readm_ami if hospital_state==”MN”
Assume readmission rate is normally distributed, and assume a critical value of 1.96 for the 95% confidence interval. Construct the Z-score to test the readmission rate of each of the hospitals below to the MN benchmark readmission rate you found in 1.1 above.
Hint: This is a test for a single population proportion compared to a benchmark. (Lesson 4, slide 19)
Ho: Hospital’s readmission rate = MN benchmark
2-sided test
Fill in the Table below. Explain briefly why you reject or do not reject the Null hypothesis.
Hospital Name (hospital id) Readmission Rate Sample size Z score Lower CI Upper CI Conclusion:
Reject the Null? Yes/No
NORTH MEMORIAL MEDICAL CENTER (240001)
MAYO CLINIC HOSPITAL ROCHESTER (240010)
ST CLOUD HOSPITAL (240036)
ABBOTT NORTHWESTERN HOSPITAL (240057)
MERCY HOSPITAL (240115)
You can do this exercise either in Excel or STATA.
Here are some STATA hints:
Step 1: generate a new variable for the numerator of the z-score
gen numerator_z=readm_ami-0.167
Step 2: generate a new variable for the denominator of the z-score
gen denominator_z=((0.167)*(1-0.167)/sampsize)^0.5
Step 3: generate a new variable for their ratio (this is the z-score)
gen z_score= numerator_z/ denominator_z
You can now list the hospitals in MN to see their z-score for your hypothesis testing
list hospital_id hospital_name readm_ami sampsize z_score if hospital_state==”MN”
You could also compute confidence intervals by generating new variables.
For the lower bound of the confidence interval:
gen ci_low=readm_ami-(1.96*denominator_z)
For the upper bound of the confidence interval:
gen ci_high=readm_ami+(1.96*denominator_z)
list hospital_id hospital_name ci_low ci_high if hospital_state==”MN”
Once you construct the confidence intervals, then you can check whether the confidence interval includes the MN benchmark MN benchmark readmission rate you found in 1.1 above. If it does include it, then we cannot reject the null hypothesis.
[20 points] Suppose you are interested in studying how each state on average compare to the average U.S. hospital in terms of their readmission “grade” for AMI patients. In particular, you would like to compare readmission grade averages across hospitals in Alabama (AL), California (CA), Massachusetts (MA), Minnesota (MN), Nevada (NV) and Vermont (VT) to the national U.S. average hospital grade.
2.1.What is the U.S. national benchmark hospital grade (average across all hospitals in the U.S) in 3 decimal points?
STATA hint: summ grade
2.2. Calculate average hospital grade for each state, construct the t-statistic to test whether the average hospital grade in the state equals to the U.S. national benchmark grade you found in 2.1 above (at 95% confidence level).
Hint: This is a test for a single population mean (hospital grade is a continuous variable) compared to a benchmark. (Lesson 4, slide 15)
Ho: Hospital’s grade = U.S. national benchmark grade
2-sided test
Fill in the Table below. Explain briefly why you reject or do not reject the Null hypothesis.
State Name (State Abbreviation) Average hospital grade Std. Dev. off hospital grade Number of hospitals in the sample (n) t- statistic Critical value for 95% confidence Lower CI Upper CI Conclusion:
Reject the Null? Yes/No
ALABAMA (AL)
CALIFORNIA (CA)
MASSACHUSETTS (MA)
MINNESOTA (MN)
NEVADA (NV)
VERMONT (VT)
You can do this exercise either in Excel or STATA.
Here are some STATA hints:
Step 1: summarize the hospital grade by state to fill in average hospital grade and the standard deviation of hospital grade.
summ grade if hospital_state==”AL”
Step 2: Calculate the number of hospitals by state
One way to do this is by producing lists of hospitals and counting by state.
list hospital_id hospital_name if hospital_state==”AL”
Another way would be to use the “egen” command in STATA. This command, used together with “count” option would count the number of non-missing observations. For example, we want to count the number of hospitals with non-missing hospital grade information by state.
egen hospcount=count(grade), by( hospital_state)
summ hospcount if hospital_state==”AL”
Step 3: You can go state by state and calculate the t-stat.
t-statistic=(avg.grade-national benchmark grade)/(stddev of grade/√hospcount)
Step 4: Calculate the Confidence Interval for each state.
Hint: use the tinv() function in Excel to find the relevant critical value for the 95% Confidence Interval.
CI:
Alternatively, STATA’s ttest command is the most efficient:
ttest grade==3.43 if hospital_state==”AL”
