Computing a confidence interval in Bernoulli trials

Question

A dice is thrown 9000 times and a throw of 3 or 4 is observed 3240 times.Show that the dice cannot be regarded as a unbiased one and find the limits between which the probability of a throw of 3 or 4 lies?

To show the dice is not unbiased, suppose the dice is unbiased.

Then probability(getting 3 or 4)=1/3

no.of times that can get 3 0r 4 =9000*(1/3)=3000
since 3000 !=3240,the observed value it can be concluded that the dice is not unbiased.

To get the limit I constructed a confidence interval for probability of success p. As sample probability of success is 3240/5000=0.36 The 95% confidence interval I get is (0.35008,0.3699) But to get the limit should I have constructed a confidence interval and if so since the significance level is not given at what level of significance should I test?

Answer 1

Our hypothesis is that the dice is unbiased, which means that $p=1/3$ in the Bernoulli trial of throwing the dice and regarding event $A = [3\text{ or } 4\text{ is observed}]$ as success and the complement event $A^c=[1,2,5\text{ or } 6\text{ is observed}]$ as failure.

In $n=9000$ trails $k=3240$ successes have been observed. Under our hypethesis that $p=1/3$, we expect to see $3000$ successes "on average" (that is if we keep repeating this experiment of throwing the dice 9000 times). The observed value of successes $k=3240$ seems to be high, which gives us a basis for rejecting our hypothesis that $p=1/3$.

What you should do is to calculate the probability (under the hypothesis that $p=1/3$) of observing $3240$ or more successes in $9000$ trials. If this probability is less than $5$%, you can reject the hypothesis at $95$% confidence level. Note that in this calculation, the parameter $p$ is an unknown constant, an only the number of successes is a random variable.

What you attempted to do, that is to calculate probabilities about $p$, is a very different approach, that is called Bayesian statistics.

Answer 2

In principle we have two problems, the first a hypothesis-testing problem, and the second a confidence interval problem.

For the hypothesis testing, the null hypothesis is that the probability of "success" (a $2$ or a $4$) is $\frac{1}{3}$. Under that hypothesis, the number $X$ nof successes in $9000$ independent trials has binomial distribution, $n=9000$, $p=\frac{1}{3}$.

This binomial has mean $np=3000$ and variance $np(1-p)=2000$. Since $n$ is very large, and $p$ is not close to either $0$ or $1$, probabilities for $X$ are well-approximated by probabilities for the normal distribution, mean $3000$, variance $2000$. So the standard deviation is about $44.72$. Our actual result of $3240$ is about $\frac{3240-3000}{44.72}$ stahdard deviation units up from the mean under the null hypothesis. Calculation shows that the result is about $5.36$ standard deviation units up from the mean. That is an enormous number of standard deviation units: A result of $3240$ or more successes for all practical purposes cannot happen with fair dice.

For the confidence interval for $p$, the probability of success, again we use the normal approximation.

The sample mean $\frac{X}{9000}$ can be assumed to be normally distributed, with standard deviation $$s=\frac{a(1-a)}{\sqrt{9000}},\tag{1}$$ where $a=\frac{3240}{9000}$.

The number obtained using (1) is not necessarily the true standard deviation of $\frac{X}{9000}$, but it is good enough, particularly since $\sqrt{x(1-x)}$ is quite insensitive to smallish changes in $x$ when $x$ is not very far from $\frac{1}{2}$.

Unless there is good reason not to, one might as well go for a $95\%$ confidence interval. This will be the interval $$\frac{3240}{9000}-1.96 s \le p \le \frac{3240}{9000}+1.96 s,$$ where $s$ is the number obtained in (1).