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

$ \begingroup $ here the natural null-hypothesis $ H_0 $ is that the mint is indifferent, that is, that the probability $ p $ of a head is equal to $ 1/2 $. The most fair alternate hypothesis $ H_1 $ is that $ p\ne 1/2 $, though one could make a font for the biased alternate hypothesis $ p > 1/2 $ .
We need to choose the meaning level of the quiz. That ‘s up to you. Two traditional numbers are $ 5 $ % and $ 1 $ % .
Suppose that the nothing hypothesis holds. then the number of heads has * binomial distribution with mean $ ( 900 ) ( 1/2 ) =450 $, and standard diversion $ \sqrt { ( 900 ) ( 1/2 ) ( 1/2 ) } =15 $.

The probability that in tossing a fair coin the count 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 $ .
therefore, if the coin was unbiased, then a number of heads that differs from $ 450 $ by $ 40 $ or more would be reasonably improbable. It would have probability less than $ 1 $ %. therefore at the $ 1 $ % meaning level, we reject the null guess.

We can besides use the convention approximation to the binomial to estimate the probability that the total of heads is $ \ge 490 $ or $ \le 410 $ under the nothing guess $ 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 chase into account. We get about $ 0.0078 $, fairly close to the respect given by Wolfram Alpha, and under $ 1 $ \ %. therefore if we use $ 1 $ \ % as our floor of significance, again we reject the nothing 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 convention estimate 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 “ accurate ” answer provided by Wolfram Alpha. Thus we can find a very accurate estimate by, as in the bad previous days, using tables of the standard normal and doing the arithmetic “ by hand. ”
$ 2 $. Suppose that we use the reasonably less natural alternate hypothesis $ p > 1/2 $. If $ p=1/2 $, the probability of $ 490 $ or more is about $ 0.00421 $. frankincense 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 necessity, for it is potential for a bonny coin to yield say $ 550 $ or more heads in $ 900 $ tosses, precisely laughably unlikely .

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