Hypothesis Testing for Shape Parameter of Gamma Distribution

$H_0:k=5$
vs.
$H_1:k>5$

where k is the shape parameter of the Gamma distribution from which a sample of 50 points is taken.

• •

  For gamma distribution (GM, AM) is a minimal sufficient statistic for (k,$\beta$).

• •

  Based on the paper by Keating et al.[1] we use the test statistic w:

  	$GM$	$={({\displaystyle \prod _{i=1}^n}{T}_i)}^{1/n}$
  	$AM$	$={\displaystyle \sum _{i=1}^n}{T}_i/n$
  	$w$	$=GM/AM$

  where $T_i$ are the values from a gamma sample.

• •

  (w, AM) is also minimally sufficient for (k, $\beta$) and the pair is stochastically independent.

• •

  For our sample, we obtain the value w = 0.909968

• •

  This is the region within which if obtained value of w lies, we reject the null-hypothesis.

• •

  $w_c=w_{(\alpha ,k,n)}$ is the critical value for $\alpha$ confidence level.

• •

  Critical region $CR=\{w:w>w_c\}$

• •

  For $n>30$ we approximate the distribution to Beta distribution by employing technique by Patnaik (1949).

• •

  In this technique we equate the first two moments of W and Beta distribution to find the parameters $\alpha$ and $\beta$ of Beta distribution.

• •

  Equations to solve :

  	$\alpha /(\alpha +\beta )$
  	$={[\mathrm{\Gamma }(k+1/n)/\mathrm{\Gamma }(k)]}^{n-1}\times {\displaystyle \prod _{i=1}^{n-1}}[\mathrm{\Gamma }(k+i/n)/\mathrm{\Gamma }(k+1/n+i/n)]$

  	$\alpha (\alpha +1)/[(\alpha +\beta )(\alpha +\beta +1)]$
  	$={[\mathrm{\Gamma }(k+2/n)/\mathrm{\Gamma }(k)]}^{n-1}\times {\displaystyle \prod _{i=1}^{n-1}}[\mathrm{\Gamma }(k+i/n)/\mathrm{\Gamma }(k+2/n+i/n)]$

• •

  For $k=5$ and $n=50$ with level of significance 0.05 we get $w_c=0.932226$

• •

  We see that . Hence for our sample we accept the null hypothesis.

• •

  Also $w_{(0.5,5,50)}=0.904888$ , the median for null distribution is quite close to our observed w.

• •

  For 1000 simulations, we found our hypothesis to be rejected only 55 times i.e. 0.055 in fraction.

• •

  The MU estimator for $k^{\ast }$ is such that $w=w_{(0.5,k^{\ast },n)}$.

• •

  By solving the above equations taking $\alpha ,\beta ,k$ as variables, we obtain $k^{\ast }=5.28736$, which we see to be quite close to value used for sample generation.

• •

  Solving for $k^{\prime }$ in $w_{(0.05,k^{\prime },n)}=w$, using same method as above, we obtain the $95\%$ confidence bound.

• •

  For all $\kappa$ above this value, the null hypothesis $H_0:k=\kappa$ is not rejected for the alternative $H_1:k>\kappa$

• •

  For our sample $\kappa =3.7599$