I have a basic question on p-tests (it is the first time I use them). Imagine having a basket with 70 orange balls and 30 blue balls. The probability to pick to pick an orange ball is 0.7 and the probability to get a blue ball is 0.3.
Now imagine I have a second basket and by picking 10 balls I have to determine wether or not the second basket has the same proportion of orange and blue balls. I could define a random variable such that X(blue) = 0 and X(orange) = 1, the probability distribution of this random variable has $\mu = 0.7$ and $\sigma = 0.483$.
The outcome of the test is 10 orange balls.
1) The probability of picking 10 orange balls in a row is $0.7^{10}=0.028$. This gives a good indication that the second basket has more orange balls than the first.
2) Using p-test: The sample mean of my sample is $\mu_s = 0$. I could make the null hypothesis ($H_0$) that the baskets are the same, hence $\bar \mu_s = \mu$ and $\sigma_s = \sigma$. The sample mean random variable has a standard deviation $\sigma_\mu = \sigma/\sqrt{10}$ (using CLT). Hence, the probability of getting mean $\mu_s$ under $H_0$ is given by:
$$ 2\phi\left(\frac{\mu_s-\bar \mu_s}{\sigma_\mu}\right) = 2\phi(-4.58)=4.6 10^{-6}$$
This p-value rejects the null hypothesis and therefore the baskets are different.
The question is: how are these two methods and probabilities related?
Is it wrong to estimate the standard deviation of my sample as being $\sigma$ in method (2)? I am doing this way because the std of my sample is zero.
Any better method to solve this problem?
Comment. Thanks for the clarification. But this is still somewhat of a tangled mess. Below, I am not necessarily trying to give answers, as much as I am trying to draw some of the distinctions that need to be made before you try to compute answers of your own. I hope some of my comments are helpful.
Let $X$ be the number of orange balls seen when $n = 10$ balls are drawn at random, with replacement, from an urn with 70% orange balls. Then $X \sim \mathsf{Binom}(n = 10, p = .7)$ and $P(X = 10) = .7^{10} = 0.028$ as you say. The distribution of $X$ is shown by the heights of the black balls in the figure below. If you tried to approximate this using a normal distribution then the 'best fitting' formal distribution has $\mu = np = 7$ and $\sigma = \sqrt{npq} \approx 1.449.$ The denisty finction of $\mathsf{Norm}(7, 1.449)$ is also shown.
If you use the normal curve, with continuity correction) to approximation $P(X = 10)$ you would get the area under the normal curve to the right of $x = 9.5$ (shown as a vertical dashed line). As you can see, the fit is not perfect. The normal approximation gives $P(X = 10) \approx 0.042 > 0.028.$
Note: If sampling is without replacement, then you should use a hypergeometric distribution. If you assume that the second basket also has 100 balls (70 orange), then you would obtain $P(X = 10) = 0.023 \ne 0.028.$
I believe your last computation is an attempt to test whether $p = 0.7,$ based the proportion $\hat p = X/n,$ where $X = 0$ orange balls. This seems inappropriate, and I am not surprised it gives nowhere near the same numerical result as above.