A study was performed to assess the rate of adverse events l

A study was performed to assess the rate of adverse events lambda n for a new surgical method. It is known that the rate of adverse events for the standard surgical method is lambda s = 5.2 per 1000 operations. Let the random variable X denote the number of adverse events observed on 723 operations with the new surgical method. Suggest a distribution for X assuming lambda n = lambda s. Substantiate your answer. Suppose that x = 0 was observed. Test H0 : lambda n = lambda s versus H1 : lambda n lambda s at the 5% significance level. Interpret your result. Provide a 95% confidence interval for the rate of adverse effects for the new surgical method.

Solution

a) as the sample size is very large...we can assume a normal distribution with sample proportion = 0.00052....

b) = sqrt[ P * ( 1 - P ) / n ]
so, s.d = 0.00085.....

test statistic= (0.00052 - 0) / 0.00085 = 0.62...

p-value = 0.5352578 which is greater than 0.05...
so, the null hypothesis is true and there is no evidence to reject that...

c) upper limit = 0.00052 + ( 1.96 * 0.00085) = 0.002186...
lower limit =  0.00052 - ( 1.96 * 0.00085) = - 0.001146.....
confidence interval is= [ - 0.001146 ,  0.002186 ]