MATH 225N Week 7 Discussion: Hypothesis Testing
Initial Post Instructions
Describe a hypothesis test study that would help your work or conclusions in
... [Show More] some way. Describe what variable would be tested and what would be your guess of the value of that variable. Then include how the result, if the null were rejected or not, might change your conclusions or actions in some way.
NB: 2 Answers Displayed
Answer 1
We interpret data by assuming a specific structure or outcome and use statistical methods to confirm or reject the assumption. According to Opens tax, the process begins with a model, is build. This theory, if it has validity, will lead to predictions; what we call hypothesis. (Openstax, pg. 381-382) There are two types of statistical hypotheses. Null hypothesis, denoted by H0, is usually the hypothesis that sample observations result purely from chance. Alternative Hypotheses, is denoted by H1 or Ha, is the hypothesis that sample observations and are influenced by some non-random cause. (Star Trek)A hypothesis test can result in one of two way, either rejecting the null hypothesis or claim or by not rejecting the null hypothesis.
The variable that I feel would help my work in some way would be intramuscular antibiotics shorten length of illness to 3 days. For my chosen variable, the null hypothesis is represented as H0 = 3 and the alternative hypothesis, intravenous antibiotics do not shorten illness length to 3 days, as Ha ≠ 3. I chose this variable and value based on my husband’s recent experience with an intramuscular antibiotic used to treat his strep throat. He received an injection on 2-10-2019 and today, 2-12-2019, he states feeling back to normal without the presence of a fever, sore throat, or fatigue.
Considering all the various types of infectious diseases processes there are, my hypothesis would likely result in a rejection of the null hypothesis. Some infections are quite serious and remain present in a person’s body for longer periods of time. There are also some individuals that avoid seeking medical attention when a change in health status appears, resulting in the ability of an infection to colonize for a longer period. If I were to conduct this hypothesis test, I was specifically focus on cases of strep throat and I would also limit my study population to those individuals that sought medical attention when symptoms first presented. By controlling these two areas, there is a higher chance of obtaining accurate results and determining the validity of my claim.
References
What Is Hypothesis Testing? (2019, December 5). Retrieved from http://stattrek.com/hypothesis-test/hypothesis-testing.aspx (Links to an external site.)
Holmes, A., Illowsky, B., & Dean, S. (2017). Introductory Business Statistics. Houston, TX: OpenStax CNX. Retrieved from https://openstax.org/details/books/introductory-business-statistics
Answer 2
I could use what we learned this week to test a hypothesis regarding whether a medication helps to reduce length of hospital stay for patients after cardiac surgery. I reviewed a study from 2006 that researched length of stay after cardiac surgery. In a sample of 1620 patients, the mean LOS for female patients with an age >65 was 12 days. The average length of stay can be drastically altered by many things, most commonly postoperative atrial fibrillation and pleural effusions (Rosborough, 2006). I will use the following data to determine if adding a new medication reduces length of hospital stay. The fictional new medication will be called (Med-X). For statistical example I will use the following fictional values:
Established length of stay: 5.5 Days
SD: 1.5 Days
Established length of stay after Med-X added: 4.7 Days:
Sample Size – 25 Patients
Significance level: 95%
With this data I would write a hypothesis as such:
Null Hypothesis: Mean length of stay = 5.5 Days
Alternative Hypothesis: Mean length of stay < 5.5 Days
The critical value for this hypothesis would be -1.64. I would calculate a test statistic with the following equation: Zc = (4.7 – 5.5) / (1.5 / Square root of 25). This calculates to -2.67. Because this value is less than the critical value (-1.64), and it is a left-tailed test, I could say with 95% confidence that Med-X does reduce average length of stay after open heart surgery. I could then implement Med-X postoperatively for all cardiac surgery patients. This is not to say with absolute certainty that this is correct; a type one error could be made. A type one error would result in rejecting the null hypothesis when in fact it is true. If this were the case I could incorrectly assume average LOS = 4.7 when in fact the null hypothesis is true and average LOS = 5.5. This may result in planning for shorter stays and being inadequately supplied and staffed for the patient’s actual length of stay. [Show Less]