Respuesta :
Answer:
Based on statistical analysis the claim is of 3% of patients developing adverse reaction is valid and there is a drug problem
Step-by-step explanation:
Here we have
Percentage of patients that develop nausea em= 153/5783 *100 = 2.65 %
To test the hypothesis, we put our null hypothesis as
H₀: μ = 3%
Alternative hypothesis is then
Hₐ: μ < 3%
Therefore the test statistic is
[tex]t=\frac{\hat{p}-p_0}{\sqrt{\frac{ p_0( 1- p_0)}{n}}}[/tex]
P₀ = 0.03
[tex]\hat p[/tex] = 0.0265
n = 5783
From the above we have our test statistic as -1.579 from which gives our p value as P = 0.057.
Therefore, as we have our P value less than the confidence level of 0.05, we therefore fail to reject the null hypothesis because there is sufficient statistical evidence that suggest that the proportion of patients who would have adverse reaction to the drug = 3%.
Answer:
Based on the statistical solution one can conclude that 3% of users develop adverse reaction of nausea and it appears a drug problem.
Step-by-step explanation:
Given information:
Number of patient = 5783
Adverse reaction developed = 153
Significance level = 0.05
Percentage of patients that develop Nausea
[tex]= (153*5783)/100[/tex]
[tex]=2.65[/tex] %
Now, to test hypothesis
Put the null hypothesis as [tex]H_{0} = 3%[/tex] %
So, alternative hypothesis will be [tex]\leq 3[/tex] %
Therefore ,
The test statistic is
[tex]t=(p-p')/\sqrt{\frac{(p'-(1-p'))}{n} }[/tex]
Where,
[tex]p=0.03\\p'=0.0265\\n=5783\\[/tex]
On solving above equation we get
[tex]t=-1.579\\[/tex]
From which we conclude that
[tex]p=0.057[/tex]
As we have our value less than the confidence level of 0.05%
We can conclude that ,nausea appear to be a problematic adverse reaction.
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