91Ó°ÊÓ

To derive the cumulative distribution function (CDF) of the given pdf, integrate the pdf from the lower limit 0 to x. This yields \(F(x) = \int_0^x f(t) dt = \int_0^x \frac{3t^2}{\theta^3} dt = \frac{t^3}{\theta^3} \Bigg|_{t=0}^{t=x} = \frac{x^3}{\theta^3}\). Thus, \(F(x) = \frac{x^3}{\theta^3}\) for \(0 < x < \theta\).

As \(\frac{Y_n}{\theta}\) follows the same distribution as \(X\), we can also write \(P(c < \frac{Y_n}{\theta} < 1) = P(c<X<1) = F(1) - F(c) = 1 - c^3\). Since there are \(n\) number of samples, the probability for the nth order statistic is \(P(c < \frac{Y_n}{\theta} < 1) = 1 - c^{3n}\).

First, we set \(P(c < \frac{Y_n}{\theta} < 1) = 1 - c^{3n}\) equal to 0.025 (since the confidence level is 95%, the two-tailed alpha level is 0.05, which means we have 0.025 in each tail of the distribution). Solving this equation for \(c\), we get \(c = (1 - 0.025)^{\frac{1}{3n}}\). Plugging in the observed value of \(Y_4 = 2.3\) and solving for \(\theta\), we can create the 95% confidence interval for \(\theta\) using the equation \(\theta = \frac{Y_4}{c}\). The resultant 95% confidence interval for \(\theta\) will be the range \((\frac{Y_4}{c_{upper}},\frac{Y_4}{c_{lower}})\) where \(c_{upper} = (1 - 0.025)^{\frac{1}{12}}\) and \(c_{lower} = (1 - 0.975)^{\frac{1}{12}}\). Since this is a one-sided interval, the upper bound for theta will be \(\infty\).