DillonHughes-HS345-2304A-Unit 6
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Drug Therapy Trial for Lowering Heart Rate
Dillon S. Hughes
Department of Health Care Administration, Purdue University Global
HS 345: Biostatistics
Unit 5 Assignment
Professor Kito Barrow
November 22, 2023
Drug Therapy Trial for Lowering Heart Rate
We will examine the results from a new innovative drug to help lower a patient’s heart
rate. We will first exam the raw numbers form each patient group. Then we will use a statistical
software to produce a t-test analysis. Finally, we will describe all of the results of the analysis.
We will determine whether or not the new innovative drug was effective or not.
The process for this is fairly simple. The first thing you do is collect all of your data. Next,
you will run a T-Test, and write the report. The next step is to summarize the research scenario
and then to justify the reason for running a t-test. At this point you would identify your
hypothesis. Here you interpret the t-test results. Finally, you would provide recommendation.
Our patient groups are broken up into two different groups of twenty-three. The first
group, which is Group 1 was assigned a placebo. The second group, which is Group 2 was
given the new innovative drug. See the table below for the raw data for each patient’s heart rate
after they received their dose of medication.
Group 1
Group 2
81
92
110
91
111
108
100
90
90
108
100
99
95
80
91
90
86
97
82
92
90
105
90
95
96
84
90
86
95
107
85
100
90
100
97
101
91
84
81
92
97
101
104
91
101
99
With this raw data, we produced a t-test analysis. This provided us with very valuable
information. The mean for Group 1 was 93.61 and the mean for Group 2 was 95.30. The
standard deviation for Group 1 was 8.29 and 7.96 for Group 2(2023).
The unpaired t test showed us some interesting results. First, the two-tailed p-value
equals 0.4830. By conventional criteria, this difference is considered to be statistically
significant. When a p-value is smaller that means that it is less likely the results occurred by
random and a stronger chance that the null hypothesis should be rejected(Mclead, 2013). The
difference between the mean of Group 1 and 2 is -1.70. The t value was 0.7075 and the
standard error of difference was 2.397.
With our p-value being as high as it is, this leads to believe that our null hypothesis
should be rejected. When a negative mean between the means in the test groups with group 2
having a lower mean, showed the medication had some effect, but a very low difference in the
patients that received the medication. Any t-value grater than plus two or less than negative 2 is
acceptable. Our t-value being right over 0.7 puts within the area we do not want to be in and
further leading us to believe in rejecting our null hypothesis. Our standard error of difference
was well above the 1.96 to -1.96, does not give us a confidence level of at lest 95%. When
putting all of this data together, it is safe to assume that the null hypothesis will be rejected.
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Reference:
T test calculator
. GraphPad by Dotmatics. (2023). https://www.graphpad.com/quickcalcs/ttest2/
Mcleod, S. (2023, October 13).
P-value and statistical significance: What it is & why it matters
.
Simply Psychology. https://www.simplypsychology.org/p-value.html#:~:text=A%20p
%2Dvalue%20less%20than,≤%200.05)%20is%20statistically%20significant.
Dun & Bradstreet. (2022, December 17).
T-value
. All Business.
https://www.allbusiness.com/barrons_dictionary/dictionary-t-value-4942040-
1.html#:~:text=Definition%20of%20T%2Dvalue&text=Thus%2C%20the%20t
%2Dstatistic%20measures,less%20than%20%2D%202%20is%20acceptable.