5. A new drug designed to reduce fever (and relieve aches and pains) is being te
ID: 3053154 • Letter: 5
Question
5. A new drug designed to reduce fever (and relieve aches and pains) is being tested for efficacy and side effects. Ten patients entering a hospital with a high fever were selected at random. The temperature (in °F) of each patient was measured, the drug was administered, and two hours later the temperature was measured again The data are given below. Assuming normality, test at the 5% significance level to see if the new drug did reduce fevers. (17 points) 102.6 99.2 102.3 101.1 102.7 102.6 100.5 103.5 105.7 104.3 Before Drug After Drug 99.8 98.8 97.5 100.3 99.6 102.8 99.0 101.8 97.1 99.2 He Ha Test Statistic Rejection Region Conclusion P-value Based on your conclusion, what type of error have you possibly subjected yourself to and explain the error in terms of this particular situation?Explanation / Answer
Requirements
The dependent variable is normally distributed and is continuous (measured on ratio/interval scale of measurement).
Independent variable is categorical and two samples are dependent on each other. Data is paired; for example, it is in form of before, after etc.
Ho: there is no significant difference in the mean fever before and after drug. U1 = u2
H1: new drug produces fever. U1 > u2
Test statistic
Mean_diff= 2.86
sd_diff= 2.680
n= 10
t= (mean_diff)/(sd_diff/sqrt(n))
t= 2.86/(2.680049750782/sqrt(10))
t= 3.375
rejection region
t(a,n-1) =t(0.05,10-1) =abs(F.INV(0.05,10-1)) = 1.833
Reject the null hypothesis if test statistic t is greater than 1.833
Conclusion
I reject the null hypothesis at 5% level of significance and conclude that new drug produces fever. U1 > u2
P value
p-value= P(T>|t|
P(T>(3.37460679804269)
T.DIST.rT(3.37460679804269,10-1)
0.004098412
Type of error
They are chances of conducting Type 1 error since I reject the null hypothesis. Type 1 error is defined as rejection of null hypothesis when it is true.
But the probability of Type 1 error is controlled by fixing the level of significance, Alpha at a predetermined value. I assume alpha, level of significance 5%
before after di 102.6 99.8 2.8 99.2 98.8 0.4 102.3 97.5 4.8 101.1 100.3 0.8 102.7 99.6 3.1 102.6 102.8 -0.2 100.5 99 1.5 103.5 101.8 1.7 105.7 97.1 8.6 104.3 99.2 5.1Related Questions
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