Question

Suppose you measure the number of "success" outcomes n out of N binary (success/failure) trials (e.g. your local oracle predicts tomorrow's weather on N different occasions and you assess the number of correct predictions n). From this sample you determine the proportion p=n/N of the success outcomes. Clearly, we are dealing here with a sample of size N from i.i.d. Bernoulli processes X, and p is an estimator for the underlying probability of success in each Bernoulli trial.

How do we put a confidence interval on the proportion p? In other words, suppose we have a particular sample of N Bernoulli (binary) trials and we compute p=0.58. Is the difference from a random coin toss (which would be p=0.5) significant? Can we calculate a 95% confidence interval, so that we could tell that true underlying proportion (success probability) is likely bounded by, say, [0.48,0.63]?

This is a well known problem and a solution is very easy to find on the internet (you are welcome to use any external resources). However, it is important to not only have the formula but also be able to explain how things work. In particular, this is what I am looking for: an explanation of how things work. Specifically, think about the following:

·        p is a function of the sample drawn, so it is a random variable. How is it distributed (at least as N becomes large), and why? [Hint: represent failure as 0, and success as 1]. What is the distribution of p at any sample size N?

·        What parameter of the estimator's distribution determines the width of the confidence interval and what is the value of that parameter for N Bernoulli trials [Hint: we can easily calculate the variance of a single Bernoulli random variable with values 0,1 and success probability p: the (true!) mean, e.g. the expectation, is μ=0*(1-p)+1*p=p; hence the variance is σ=(1-p)*(0-p)^2+p*(1-p)^2=(1-p)*p^2+(1-p)^2*p=p*(1-p)*(p+1-p)=p*(1-p). What will happen in N trials? You can consult the derivation of the standard error of the mean]

·        Now that we have the estimator p, and we know shape and width of its distribution, how do we form the confidence interval at e.g. 80% confidence level? 90%? Use normal (not t-distribution) as it is what's usually used in conjunction with proportion estimates.

Answer 1

No , P is the proportion of sample drawn. Listen , probability is always same. You just need to understand it & more importantly hypothesis testing. The mean test and the the proportion test is somewhat different. Here you are finding the no of proportion of individual getting success that is 1 from the sample.

Suppose what you are saying that you are computing it with p=0.58 success & yes obvious you can test it at 95% level confidence interval. That is the probability not same with the proportion. that is the chances of occuring the proportion of individual who has got success. ofcourse you will be able to calculate the confidence interval rather you can lower limit of proportion & upper limit of proportion.

In your case p = 0.58 that is 58% of proportion of individual

so at 95% confidence interval it might be the output ( say example)

Lower CI Upper CI

0.57 0.59

57% 59%

Now depends on p-value you can reject your null hypothesis.

Your null hypothesis should be H0 : P0 = 0.58

alternative hypothesis HA : P0 not equal to 0.58

In bernolli's case random variable that only take two values 0 and 1.

Here it is discrete random variable.

We do mean t test with continuous random variable which follows normal distribution.

Here we dont need to do mean test rather wehave done proportion test.

And any chances of happening can be measured by probability & define by statistics.