Find a $100(1-\alpha)\%$ confidence interval from a population $\sim$ Gamma$(5,\theta)$

For $$f(x|\theta)=\frac{1}{24}x^4\theta^{-5}\exp^{-\frac{x}{\theta}}I_{(0,\infty)}(x)\hspace{0.5cm}\theta\in\mathbb{R}^+$$

I want to find a $$100(1-\alpha)\%$$ confidence interval.

I know $$T(X_1,\dots,X_n)=\sum\limits_{i=1}^nX_i\sim\text{Gamma}(5n,\theta)$$ and T is sufficient

So $$f_T(t)=\frac{t^{5n-1}e^{-t/\theta}}{\Gamma(5n)\theta^{5n}}$$ Then I used $$t=y\theta$$ so

$$f_Y(y)=\frac{y^{5n-1}e^{-y}}{\Gamma(5n-1)}$$ Then I should find $$\gamma_1,\gamma_2$$ such that $$P(\gamma_1 So I need $$F_Y(\gamma_1)$$ and $$F_Y(\gamma_2)$$ and solve for $$\gamma_1$$ and $$\gamma_2$$ but I can't do it.

• Add the self-study tag if this is homework. – StubbornAtom May 23 '20 at 18:36

You have a random sample of $$n$$ observations $$X_1, X_2, \dots, X_n$$ from a population with distribution $$\mathsf{Gam}(\mathrm{shape}=5, \mathrm{scale}=\theta),$$ where $$\theta$$ is unknown.

Suppose you wish to use these data to make a 95% confidence interval for $$\theta.$$ Let $$T = \sum_{i=1}^n X_i,$$ which has $$E(T) = 5n\theta.$$

You propose using the pivotal quantity $$Y = T/\theta,$$ which has distribution $$\mathsf{Gam}(\mathrm{shape}=5n, \mathrm{scale}=1)$$ and $$E(Y) = 5n.$$ [Note: Your density function for the distribution of $$Y$$ needs to have denominator $$\Gamma(5n),$$ not $$\Gamma(5n-1).]$$

If $$L$$ and $$U$$ cut probability 0.025 from the lower and upper tails, respectively, of $$\mathsf{Gam}(5n,1),$$ then with $$n = 30,$$ we can use R statistical software to find that $$L=126.9562, U= 174.9372.$$ [Note: In R, the second parameter is the rate $$\lambda$$, which is the reciprocal of the scale. But here both rate and scale are $$1.]$$

a = 5; n = 30
LU = qgamma(c(.025,.975), a*n, 1);  LU
[1] 126.9562 174.9372


Then

$$0.95 = P(L < Y = T/\theta < U) = P(T/U < \theta < T/L),$$

so that a 95% confidence interval of $$\theta$$ is $$(T/U,\, T/L).$$

Example: As an example, we let $$\theta = 9$$ and use R to get a random sample of size $$n = 30$$ from $$\mathsf{Gam}(\mathrm{shape} = \alpha = 5, \mathrm{scale} = 9).$$

a = 5;  th = 9
x = rgamma(n, a, 1/th)
summary(x)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
11.92   28.20   45.19   46.48   58.94   99.61
t = sum(x);  t
[1] 1394.351


The resulting 95% CI for $$\theta$$ is $$(7.97, 10.98).$$

t/qgamma(c(.975,.025), a*n, 1)
[1]  7.970578 10.982932


So my random sample was among the 'lucky' 95% of samples that produces a CI covering $$\theta = 8.$$

Note: If you're allergic to software, you can express the distribution $$\mathsf{Gam}(5n, 1)$$ as an equivalent chi-squared distribution and use printed chi-squared tables to find lower and upper boundaries.

qchisq(c(.025,.975), 300)/2
[1] 126.9562 174.9372
qgamma(c(.025,.975), 150, 1)
[1] 126.9562 174.9372

• Thank you for your answer and how can I can express the distribution Gam(5n,1) as an equivalent chi-squared distribution? – Jhon Knows May 23 '20 at 18:20
• I will leave that for you. If it isn't clear from your text, you can look at Wikipedia articles on gamma and chi-squared under Related distributions. (Erlang is a Gamma with integer shape parameter.) ) – BruceET May 23 '20 at 18:36