98% confidence interval
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A clinical trial of n=600 infants using Drug A found that 40% were cured of sepsis. What is 98% confidence interval for the drug's effectiveness? A previous study showed that the proportion of subjects under a placebo is 36%. Based on the answer from a, is there enough evidence to suggest that Drug A does significantly better?
a)
sample size, n=600 sample proportion, p = 0.4 we use normal approximation, for this we check that both np and n(1p) > 5. Since n*p = 240 > 5 and n*(1p) = 360 > 5, we can take binomial random variable as normally distributed, with mean = p = 0.4 and std deviation = root( p * (1p) /n ) = 0.02 
For constructing Confidence interval, Margin of Error (ME) = z x SD = 0.0465 98% confidence interval is given by: Sample Mean +/ (Margin of Error) 0.4 +/ 0.0465 = (0.3535 , 0.4465) 
b)
We have n = 600, sample proportion p = 0.4 and P = 0.36 we want to test whether hypothesis: H0: true P <= 0.36 (null hypothesis) Ha: true P > 0.36 (alternative hypothesis) now, calculating statistics, (test statistic) sigma s = sqrt(P*(1P)/n) = 0.0196 zscore = (pP)/s = 2.0408 this is right tailed test, hence pvalue = P( z > 2.0408) = 0.0413 
this is less than 0.05, so we reject null, and yes this is significantly beter
Thank you! Can you help me with this last question? A random sample of 100 preschool children in Tokyo revealed that only 60 had been vaccinated. Provide a 90% confidence interval for the proportion of the vaccinated city.
Thank you! Can you help me with this last question? A random sample of 100 preschool children in Tokyo revealed that only 60 had been vaccinated. Provide a 90% confidence interval for the proportion of the vaccinated city.
sorry dear .i dont know
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