Suppose a child plays with a die and the mean obtained after $25$ rolls is equal to $4.5$. Perform a statistical test to check if the die is fair at significance level $\alpha=0.05$.
Suppose we perform hypothesis testing with null hypothesis $\mu_0 = 4.5$. If we have a fair die the average number of points after throwing it $25$ times is just the same as after throwing it once so $E(\bar{X})=\frac{1+2+3+4+5+6}{6}=3.5$. The alternative I guess is that $\mu_1=3.5$ and we want to show that the die is unfair. I am not sure if I am on the right track though? Or do we just convert this to probabilities, and say that under the null $p=\frac{1}{6}$. Thanks for any help.
I managed to solve the exercise myself after the suggestion in the comments. We suppose that $\mu_0=3.5$. Then we use the $z$-score: $$\frac{\bar{X}-\mu_0}{\sigma / \sqrt{n}}.$$
In this case $\bar{X}=4.5$, $\mu_0=3.5$ and the standard deviation can be calculated as the square root of the variance divided by the number of rolls $\sigma / \sqrt{n}=\sqrt{0.1166}=0.34$
So $$\frac{\bar{X}-\mu_0}{\sigma / \sqrt{n}}=\frac{1}{0.34}=2.94.$$ Then $p=P(Z\geq 2.94)=0.0033$ so we reject the null hypothesis.