Let X1 X2 Xm be a rs from Nmux sigmax2 and let y1 y2 yn be

Let X_1, X_2, ... X_m be a r.s. from N()mu_x, sigma_x^2) and let y_1, y_2, ... y_n be a r.s. from N (mu_y, sigma_y^2) where both mu_x and mu_y are known Use theorem on X^2, t and F dist\'s show that n/m summation I = 1 m (x_1 - mu_x)^2 / summation I = 1 m (y_i - mu_y) sigma_y^2 / sigma_x^2 ~ F (m, n) Assume the two r.s. are independent. Find a pivotal quantity for sigma_y^2 / sigma_x^2 and use it to derive a (100(1 - alpha)%) confidence interval estimator for sigma_y^2 / sigma_x^2. Assume the 2 r.s. are independent.

Solution

If X1,X2,.....Xn be a random sample from N(,²), then by definition of chi-square=[summation over i=1ton((X_i-)/)²) follows chi-square with n degrees of freedom. And if X and Y are two independent chi-square variates with n and m degrees of freedon respectively, then F=(X/n)/(Y/m) follows F(n,m) degrees of freedom.

Given X1,X2,.....Xm be a random sample from N(_x,²_x) and Y1,Y2,.....Yn be a random sample from N(_y,²_y).

(a): Consider F={[summation over i=1tom((X_i-_x)²/²_x)]/m}/{[summation over j=1ton((Y_j-_y)²/²_y)]/n}=(n/m)*[summation over i=1tom((X_i-_x)²)/summation over j=1ton((Y_j-_y)²]*(²_y/²_x) follows F(m,n). Hence proved

(b):Pivotal quantity for (²_y/²_x) is {[(²_y/²_x)]*((n-1)/(m-1))*[summation over i=1tom-1((X_i-_x)²)/summation over j=1ton-1((Y_j-_y)²] follows F(n-1,m-1).

Then (100(1-alpha))% confidence interval estimator for (²_y/²_x) is [{(²_y/²_x)/F_{n-1,m-1,(1-alpha/2)}},{(²_y/²_x)/F_{n-1,m-1,(alpha/2)}}].