Question

# The waiting time (in minutes) at a bus stop has exponential distribution with mean >0. The...

The waiting time (in minutes) at a bus stop has exponential distribution with mean >0. The waiting times on ten occasions were recorded as follows:

6.2, 5.8, 4.5, 6.1, 4.6, 4.8, 5.3, 5.0, 3.8, 4.0

a. Construct a 95% two-sided confidence interval for the true average waiting time.

b. Construct a 95% two-sided confidence interval for the true variance of the waiting time.

 Values ( X ) $\Sigma (X_{i} - \bar{X})^{2}$ 6.2 1.4161 5.8 0.6241 4.5 0.2601 6.1 1.1881 4.6 0.1681 4.8 0.0441 5.3 0.0841 5 0.0001 3.8 1.4641 4 1.0201 Total 50.1 6.269

Mean $\bar{X} = \Sigma X_{i} / n$
$\bar{X} = 50.1 / 10 = 5.01$
Standard deviation $S_{X} = \sqrt{\Sigma (X_{i} - \bar{X})^{2}/n-1}$
$S_{X} = \sqrt{ 6.269 / 10 -1} = 0.8346$

Part a)

Confidence Interval
$\bar{X} \pm t_{\alpha /2, n-1} S/\sqrt{n}$
$t_{\alpha /2, n-1} = t_{ 0.05 /2, 10- 1 } = 2.262$
$5.01 \pm t_{\ 0.05/2, 10 -1} * 0.8346/\sqrt{ 10}$
Lower Limit = $5.01 - t_{\ 0.05/2, 10 -1}0.8346/\sqrt{ 10}$
Lower Limit = 4.413
Upper Limit = $5.01 + t_{\ 0.05/2, 10 -1}0.8346/\sqrt{ 10}$
Upper Limit = 5.607
95% Confidence interval is ( 4.413 , 5.607 )

part b)

$S^2 = 0.6966$
\alpha = 0.05
n = 10
$((n-1)S^2 / \chi_{\alpha/2}^{2} ) < \sigma^2 < ((n-1)S^2 / \chi_{1 - \alpha/2}^{2} )$
$(( 10-1 ) 0.6966^2 / \chi_{0.05/2}^{2} ) < \sigma^2 < ((10-1)0.6966^2 / \chi_{1 - 0.05/2}^{2} )$
$\chi_{0.05/2}^{2} = 19.0228$
$\chi_{1 - 0.05/2}^{2} ) = 2.7004$
Lower Limit = $(( 10-1 ) 0.6966^2 / \chi_{0.05/2}^{2} ) = 0.3296$
Upper Limit = $(( 10-1 ) 0.6966^2 / \chi_{0.05/2}^{2} ) = 2.3217$
95% Confidence interval is ( 0.3296 , 2.3217 )
( 0.3296 < $\sigma^2$ < 2.3217 )

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