91影视

Donate with\(\theta \)

\(\theta = {\alpha _i} - \alpha _i^\prime \)

Estimator of \({\alpha _i}\) is \({\bar X_{i \ldots }} - {\bar X_ \ldots }\)and of \({\bar X_{i' \ldots }} - {\bar X_ \ldots }\)thus, the estimator of the \(\theta \)is

\(\hat \theta = {\bar X_{i..}} - {\bar X_{i' \ldots }}\)

The following holds

For\(i \ne i'\), random variables \({\bar X_{i' \ldots }}\)and $\\({\bar X_{i' \ldots }}\)are independent for every j and every k. Because of the independence, the variance of the estimator is

\(\begin{aligned}{*{20}{c}}{V(\hat \theta ) = V\left( {{{\bar X}_{i..}} - {{\bar X}_{i'..}}} \right) = V\left( {{{\bar X}_{i..}}} \right) + V\left( {{{\bar X}_{i'..}}} \right)}\\{ = \frac{{{\sigma ^2}}}{{JK}} + \frac{{{\sigma ^2}}}{{JK}} = \frac{{2{\sigma ^2}}}{{JK}}}\end{aligned}\)

Remember that

and the variance of the error is

which indicates that

and the fact that

The estimator of the variance is the mean square error (MSE); thus, the estimator of the standard deviation required to compute the confidence interval, is

\({\hat \sigma _{\hat \theta }} = \sqrt {\frac{{2 \cdot MSE}}{{JK}}} \)

The random variable has student distribution with I J(K-1) degrees of freedom. The confidence interval now becomes

\({\bar x_{{i_{..}}}} - {\bar x_{i' \ldots \pm }} \pm {t_{\alpha /2,IJ(K - 1)}} \cdot \sqrt {\frac{{2 \cdot MSE}}{{JK}}} \)

There is not enough data in the exercise 19 to compute the confidence interval. However, the data obtained in the mentioned exercise is

\(\begin{aligned}{*{20}{c}}{I = 3;J = 6;K = 2}\\{MSE = 18.99}\end{aligned}\)

The estimates of \(\alpha _i^{}\)and \(\alpha _i^\prime \) usually can be easily computed from the data. The values of \({t_{\alpha /2,I.J}}(K - 1)\) can be found in the appendix of the table.