<Student’s name>
Nursing and Health Sciences
A healthcare facility is trying to determine whether or not to serve coffee in the waiting rooms for their patients. Since many similar facilities do serve coffee, tea, and water, they want to determine if there is sufficient evidence to show that coffee increases their heart rate. They test 15 patients in the waiting room one day to see if their heart rate increases. They would like to see if there is any supporting data from national studies that support the results of their study. Please present a report that includes your data and national data for the healthcare facility.
My hypothesis is the following: The heart rate increases after drinking coffee.
H0: μ1=μ2Ha: μ1<μ2
- The significance level is being set at a level of 5%. α=0,05. The test is one-tailed. If we would state the hypothesis that “the heart rate before coffee is significantly differ from the heart rate after coffee”, it could be 2-tailed test. But according to our understanding of coffee properties, we can check more accurate hypothesis with one-tailed test.
- We have to compare means using two-sample t-test:
t=x1-x2s12+s22n
It is comfortably to use Excel Tools or Minitab Software, or SPSS to perform this test. Also it can be calculated by hand, according to the formula above. I prefer to use Minitab, here is the output:
Two-sample T for Heart Rate Before Coffee vs Heart Rate After Coffee
N Mean StDev SE Mean
Heart Rate Before Coffee 15 78,3 11,4 2,9
Heart Rate After Coffee 15 84,0 10,9 2,8
Difference = mu (Heart Rate Before Coffee) - mu (Heart Rate After Coffee)
Estimate for difference: -5,73
95% upper bound for difference: 1,22
T-Test of difference = 0 (vs <): T-Value = -1,40 P-Value = 0,086 DF = 27
- Make a decision
As we can see, the p-value is 0.086 which is higher than alpha level. Since p>alpha, we have no enough evidence to state, that there is a significant difference between heart rate before coffee and heart rate after coffee at 5% level of significance. The null hypothesis is NOT rejected.
Now let’s calculate the correlation coefficient r. According to the Excel function “CORREL”, we obtain, that:
r≈0.946
We have obtained, that there is no significant difference between heart rate before and after coffee.
- What information might lead you to a different conclusion?
The correlation coefficient shows, that there is a strong positive association between the variables. And, hence, there is a good trend line can be constructed (with a big R-square value). Indeed, we can construct the regression line and look on the regression equation:
Here x variable is a heart rate before coffee, y variable is a heart rate after coffee. We can see, that the regression shows us that the heart rate after coffee is 0.9048*(heart rate before coffee) plus some positive constant (13.185). This may lead us to the different conclusion. The regression line shows us that the heart rate increases (as 13.185 is added). But in fact we shouldn’t believe this information.
There are many possible variables could be missed in this research. We need more information about people, who were involved in the test. Their sex, age, heart problems, etc.
If we would have more detailed information about persons, we could divide them by some similar groups, and perform tests within these groups. Also it is very important to increase the size of the sample. We know the central limit theorem, which says in fact, that the point estimators are better, when the sample is larger and representative.
Sources
R. A. Fisher (1925).Statistical Methods for Research Workers, Edinburgh: Oliver and Boyd, 1925
