Can anyone guide me through this question?
From what I gather, I have to do a hypothesis test and get a number of consecutive tails such that I can reject H0. H0: The coin is unbiased H1: The coin is biased
I’m unfamiliar with what I should do. Should I use a geometric distribution for this case? X ~ Geo (0.5), X = number of consecutive heads up to and including first tail
Can I approximate this to a normal distribution? E(X) = 1/p and Var(X) = (1-p)/p^2 X = N(2,2)
And then do P(X>n-2) < 0.025 in order for H0 to be rejected? Where n is the number of throws until first tailed is obtained?
I’m not sure why I thought of approximating to a normal distribution, but we are not allowed to use calculators/ not given any formula table etc in the test where this question came from, so I assume it has to be simple.
The hypothesis test we're performing is $$H_0:p=1/2$$ $$H_a:p\neq 1/2$$ where $p$ is the (true) probability of landing on heads.
Suppose you flip the coin $X$ times until observing your first tails. Then $N=X-1$ and the pmf of $X$ becomes $$p_{X}(x)=(0.5)^x:x=1,2,3,...$$ The cdf of $X$ is $$P_{X}(x)=1-(.5)^x$$
In finding our rejection region for a two$-$tailed test we typically solve the equations $P_X(x)=0.025$ and $P_X(x)=0.975$ to identify the ${2.5}^{\text{th}}$ and ${97.5}^{\text{th}}$ percentiles of $X$ given $H_0$. But these respective solutions are $\approx 0.04,5.32$ neither of which belong in the support of $X$. The smallest possible outcome of $X$ larger than our critical value $0.04$ is $1$, but including the outcome of $1$ in our rejection region would lead to a $p-$value that's at least $0.50$. In other words, we need only identify the upper $2.5$% to find the critical region. Since $P(X\geq 6)= 0.03125$ and $P(X\geq 7)\approx 0.016$ we see flipping the coin at least $7$ times lands us in the upper $2.5$%. This corresponds to observing at least $N=7-1=6$ consecutive heads.