Consider a distribution defined by a probability density function which has the form
A single value y is observed from this distribution. Derive the formula for p, a suitable p-value for testing the null hypothesis $\theta$ = k against the alternative $\theta$> k. Then calculate q, the p-value for when y = 4 and k = 5. Also find z, the value of y which would result in a p-value of 0.1 when k = 5.
As I am relatively new to the world of hypothesis testing, can I know how to find the p-value in this question?
In previous cases, we used a test statistic( eg. x-k/s(x)) and we identified that it will follow a t-distribution. the p-value is usually the upper 5% (if it's a one sided test) probability values. But I cannot link the previous concept here. Any help would be greatly appreciated.
The hypothesis to be tested is $$H_0 : \theta = k \quad \text{vs.} \quad H_1 : \theta > k.$$ If you observe a single value $Y = y$, this furnishes information about $\theta$ in the sense that we must have $\theta \ge y$.
For instance, suppose $k = 5$ as stated in the question, but you observe $Y = 10$. This would actually provide incontrovertible evidence to reject $H_0$, since your observation proves that $\theta \ge 10 > 5$ and the $p$-value of the test is $0$. But what if you had observe $Y = 1$? What does this say about the plausibility of $\theta = 5$ versus $\theta > 5$? Clearly, our previous situation suggests that the larger the value of $Y$, the more plausible is $H_1$; i.e., we want to reject $H_0$ when $Y > y_{\text{crit}}$ for some critical value $y_{\text{crit}}$. Just like in the $t$-test case, we want to choose the critical value in such a way as to control the Type I error of the test; i.e., $$\Pr[Y > y_{\text{crit}} \mid H_0] \le \alpha$$ for some desired significance level $\alpha$. Correspondingly, the $p$-value of the test is $$p = \Pr[Y > y_{\text{obs}} \mid H_0],$$ the probability of erroneously rejecting the null hypothesis when $H_0$ is true, where $y_{\text{obs}}$ is the observed outcome of $Y$.
So for $k = 5$ and $Y = 1$, we would have $$p = \Pr[Y > 1 \mid \theta = 5] = \int_{y=1}^5 f_Y(y) \, dy.$$ Note that you will first have to calculate the appropriate value of $c$, and you will have to compute the integral shown above for the specific choice of parameter $\theta = 5$. I leave this as an exercise.
Then, for your question where $Y = 4$, you would compute $$p = \Pr[Y > 4 \mid \theta = 5] = \int_{y=4}^5 f_Y(y) \, dy.$$
As for the final part of your question, the calculation must be done in reverse: you want to find $y_\text{obs}$ that would give you $p = 0.1$, i.e. solve $$\int_{y=y_\text{obs}}^5 f_Y(y) \, dy = 0.1.$$ Throughout this, keep in mind that the larger the observed value, the smaller the value of $p$, which makes sense since the length of the interval of integration decreases as the lower limit of integration approaches $5$.
Why is this question so different from your "usual" understanding of hypothesis testing? The reason is because here, you are dealing with a single observation from this distribution, rather than a sample; and as a result, the test statistic in this case is $Y$ itself. There is no need to perform data reduction because the single observation is all you have, and the distributional properties of $Y$ have already been completely characterized.