written 5.6 years ago by | • modified 4.5 years ago |
Given t$_{0.025,3}$=3.18
Run r | 1 | 2 | 3 | 4 |
---|---|---|---|---|
Able's Utilization $\rho _r$ | 0.808 | 0.875 | 0.708 | 0.842 |
Average system time w$_r$ (mins) | 3.74 | 4.53 | 3.84 | 3.98 |
written 5.6 years ago by | • modified 4.5 years ago |
Given t$_{0.025,3}$=3.18
Run r | 1 | 2 | 3 | 4 |
---|---|---|---|---|
Able's Utilization $\rho _r$ | 0.808 | 0.875 | 0.708 | 0.842 |
Average system time w$_r$ (mins) | 3.74 | 4.53 | 3.84 | 3.98 |
written 5.6 years ago by | • modified 5.6 years ago |
Part - I
Given that the four estimates, $\overline{\rho_r}$ of Able's utilization
Compute overall point estimate
$\begin{aligned} \hat{\rho_r} &= \frac{1}{R} \sum_{r=1}^R \space \hat{P_r} \\ &= \frac{1}{4} (0.808 + 0.875 + 0.708 + 0.842) \\ &= 0.808 \end{aligned}$
Compute the estimate of variable of $\hat{\rho}$
$\begin{aligned} \hat{\sigma}^2 (\hat{\rho}) &= \frac{1}{(R-1)} \sum^R_{R_{r-1}} (\hat{\rho_r} - \hat{\rho})^2 \\ &= \frac{1}{(3)(4)} \bigg\{ (0.808 - 0.808)^2 \\ &+ (0.875 - 0.808)^2 + (0.708 - 0.808)^2 \\ &+ (0.842 - 0.808)^2 \bigg\} \\ &= \frac{0.015645}{12} \\ &= 0.0013 \end{aligned}$
Compute the standard errors of $\hat{\rho}$
$S.E. (\hat{\rho}) = \hat{\sigma} (\hat{\rho}) = \sqrt{0.0013} = 0.036 $
Compute the 95% confidence interval using t-distribution
As 95 % confidence interval is expected
$\Rightarrow \quad 100(1-\alpha)=95$
$1-\alpha=\frac{95}{100}$
$\alpha=1-\frac{95}{100}=5 \%=0.05$
Also, degree of freedom $f = R - 1 = 4 - 1 = 3$
$\therefore t_{\alpha / 2, f}=t_{0.025,3}=3 \cdot 18 \text{ (given) }$
$\therefore$ 95% of confidence interval is $\hat{\rho} \pm t_{0.02, 3} \hat{\sigma} (\hat{\rho})$
$ = 0.808 \pm 3.18 (0.036)$
$= 0.808 \pm 0.114$
$\Rightarrow 0.694 \le \hat{\rho} \le 0.922$
Part - II:
Given that four estimates, $\hat{\omega_r}$ of time spent in the system,
(1) Compute overall point estimate
$\begin{aligned} \hat{w} &= \frac{1}{R} \sum_{r=1}^R \hat{w_r} \\ &=\frac{3-74+4 \cdot 53+3-84+3 \cdot 98}{4}=4.02 \\ \end{aligned}$
(2) Compute the estimate of the variance of $\hat{w}$
$\begin{aligned} \hat{\sigma}^{2}(\hat{w}) &=\frac{1}{(R-1)(R)} \sum_{r=1}^{R}\left(\hat{w_{r}}-\hat{w}\right)^2 \\ &= \frac{1}{(3)(4)} \bigg\{ (3.74 - 4.02)^2 \\ &+ (4.53 - 4.02)^2 \\ &+ (3.84 - 4.02)^2 \\ &+ (3.98 - 4.02)^2 \bigg\} \\ &= \frac{0.3725}{12} \\ &= 0.031 \end{aligned}$
(3) Compute the standard error of $\hat{w}$
$S.E. (\hat{w}) = \hat{\sigma}(\hat{\omega})=\sqrt{0.031}=0.176$
(4) Compute the 95% confidence interval for time spent in system, w
$\hat{w} \pm t_{0.025,3} \hat{\sigma}(\hat{w})$
$=4.02 \pm 3.18(0.176)$
$=\lambda 4.02+0.56$
OR $3.46 \le w \le 4.58$