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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