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

**Answer**

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,

2\sum_{k=490}^{900} \binom{900}{k}\left(\frac{1}{2}\right)^{900}.

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.

**Attribution***Source : Link , Question Author : Sanu , Answer Author : Community*