Solution

First, we need to write down the null and alternative hypotheses. In this case

$H_0$: The coin is not biased.

$H_1$: The coin is biased in favour of tails.

The important thing to note here is that we only need a one-tailed test as the alternative hypothesis says “in favour of tails”. A two-tailed test would be the result of an alternative hypothesis saying “The coin is biased”.

We need to calculate more than just the probability that it lands on heads $6$ times. If it landed on heads fewer than $6$ times, that would be even more evidence that the coin is biased in favour of tails. Consequently we need to add up the probability of it landing on heads $1$ time, $2$ times, $\ldots$ all the way up to $6$ times. Although a calculation is possible, it is much quicker to use the cumulative binomial distribution table. This gives $\mathrm{P}[X\leq 6] = 0.058$.

We are asked to perform the test at a $5$% significance level. This means, if there is less than $5$% chance of getting less than or equal to $6$ heads then it is so unlikely that we have sufficient evidence to claim the coin is biased in favour of tails. Now note that our $p$-value $0.058>0.05$ so we do not reject the null hypothesis. We don't have sufficient evidence to claim the coin is biased.

But what if the coin had landed on heads just $5$ times? Again we need to read from the cumulative tables for the binomial distribution which shows $\mathrm{P}[X\leq 5] = 0.021$, so we would have had to reject the null hypothesis and accept the alternative hypothesis. So the point at which we switch from accepting the null hypothesis to rejecting it is when we obtain $5$ heads. This means that $5$ is the critical value.