0
1.4kviews
Given the following data for utilisation and time spent in system for the Able-Baker carhop problem. Calculate the overall point estimators, standard error and 95% confidence interval for the same.

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
1 Answer
0
1views

Part - I

Given that the four estimates, $\overline{\rho_r}$ of Able's utilization

  1. Compute overall point estimate

    $\begin{aligned} \hat{\rho_r} &= \frac{1}{R} \sum_{r=1}^R \space \hat{P_r} \\ &= \frac{1}{4} (0.808 + 0.875 + 0.708 + 0.842) \\ &= 0.808 \end{aligned}$

  2. Compute the estimate of variable of $\hat{\rho}$

    $\begin{aligned} \hat{\sigma}^2 (\hat{\rho}) &= \frac{1}{(R-1)} \sum^R_{R_{r-1}} (\hat{\rho_r} - \hat{\rho})^2 \\ &= \frac{1}{(3)(4)} \bigg\{ (0.808 - 0.808)^2 \\ &+ (0.875 - 0.808)^2 + (0.708 - 0.808)^2 \\ &+ (0.842 - 0.808)^2 \bigg\} \\ &= \frac{0.015645}{12} \\ &= 0.0013 \end{aligned}$

  3. Compute the standard errors of $\hat{\rho}$

    $S.E. (\hat{\rho}) = \hat{\sigma} (\hat{\rho}) = \sqrt{0.0013} = 0.036 $

  4. 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 $3.46 \le w \le 4.58$

Please log in to add an answer.

Continue reading...

The best way to discover useful content is by searching it.

Search