91Ó°ÊÓ

It is given that the proportion \(\theta \)of defective items in a large shipment is unknown and a prior distribution of\(\theta \) is equal to the beta distribution with parameters \(\alpha = 5\) and\(\beta = 10\)

Also, it is given that 20 items are selected randomly from the shipment and each one is found to be defective. That is, n=20 and\(x = 1\) .

Here,

\(n = 20\) and \(x = 1\)

Also,

The prior distribution of \(\theta \) is the Beta distribution with parameters \(\alpha = 5\) and\(\beta = 10\)

Since the Bayes estimate of\(\theta \)for the squared loss function is given by,

\(\widehat \theta = \frac{{\alpha + x}}{{\alpha + \beta + n}}\)

Therefore,

\(\begin{array}{c}\widehat \theta = \frac{{5 + 1}}{{5 + 10 + 20}}\\ = \frac{6}{{35}}\\ = 0.1714\end{array}\)

Therefore, the Bayes estimate of \(\theta \)based on the squared error loss function is\(\frac{6}{{35}}\) .