# How to assess whether a coin tossed 900 times and comes up heads 490 times is biased?

A coin is tossed 900 times and heads appeared 490 times. Does the result support the hypothesis that the coin is unbiased?

Here the natural null-hypothesis $H_0$ is that the coin is unbiased, that is, that the probability $p$ of a head is equal to $1/2$. The most reasonable alternate hypothesis $H_1$ is that $p\ne 1/2$, though one could make a case for the one-sided alternate hypothesis $p>1/2$.

We need to choose the significance level of the test. That’s up to you. Two traditional numbers are $5$% and $1$%.

Suppose that the null hypothesis holds. Then the number of heads has *binomial distribution with mean $(900)(1/2)=450$, and standard deviation $\sqrt{(900)(1/2)(1/2)}=15$.

The probability that in tossing a fair coin the number of heads differs from $450$ by $40$ or more (in either direction) is, by symmetry,

This is not practical to compute by hand, but Wolfram Alpha gives an answer of roughly $0.008419$.

Thus, if the coin was unbiased, then a number of heads that differs from $450$ by $40$ or more would be pretty unlikely. It would have probability less than $1$%. so at the $1$% significance level, we reject the null hypothesis.

We can also use the normal approximation to the binomial to estimate the probability that the number of heads is $\ge 490$ or $\le 410$ under the null hypothesis $p=1/2$. Our normal has mean $450$ and variance $15$ is $\ge 490$ with probability the probability that a standard normal is $\ge 40/15$. From tables for the normal, this is about $0.0039$. Double to take the left tail into account. We get about $0.0078$, fairly close to the value given by Wolfram Alpha, and under $1$\%. So if we use $1$\% as our level of significance, again we reject the null hypothesis $H_0$.

Comments: $1$. In the normal approximation to the binomial, we get a better approximation to the probability that the binomial is $\ge 490$ by calculating the probability that the normal is $\ge 489.5$. If you want to look it up, this is the continuity correction. If we use the normal approximation with continuity correction, we find that the probability of $490$ or more or $410$ or fewer heads is about $0.008468$, quite close to the “exact” answer provided by Wolfram Alpha. Thus we can find a very accurate estimate by, as in the bad old days, using tables of the standard normal and doing the arithmetic “by hand.”

$2$. Suppose that we use the somewhat less natural alternate hypothesis $p>1/2$.
If $p=1/2$, the probability of $490$ or more is about $0.00421$. Thus again at the $1$% significance level, we would reject the null hypothesis, indeed we would reject it even if we were using significance level $0.005$.

Setting a significance level is always necessary, for it is possible for a fair coin to yield say $550$ or more heads in $900$ tosses, just ridiculously unlikely.