91影视

It is given that$n=18,\stackrel{炉}{x}=334.8,s=6.3$

a.

The 95% confidence interval can be calculated using the formula,

$\sqrt{\frac{\left(n-1\right)s^2}{{蠂}_{\frac{伪}{2}}^2}}猢�蟽猢�\sqrt{\frac{\left(n-1\right)s^2}{{蠂}_{\left(1-\frac{伪}{2}\right)}^2}}$

The degrees of freedom is 17, at 0.05 level of significance, from the table value we have

${蠂}_{0.025}^2=30.191$and${蠂}_{0.975}^2=7.564$

Substitute the values to get the required confidence interval.

$$\begin{gathered} \sqrt{\frac{\left(18-1\right){6.3}^2}{30.191}}猢�蟽猢�\sqrt{\frac{\left(18-1\right){6.3}^2}{7.564}} \\ 4.7274猢�蟽猢�9.4447 \end{gathered}$$

Therefore, the 95% confidence interval for$蟽$ is (4.7274,9.4447).

b.

From the 95% confidence interval we can say that, we are 95% confidence that the population standard deviation of conduction times of the prototype system will lies between 4.7274 nanoseconds and 9.4447 nanoseconds.

c.

It is given that, a system is considered to have low throughout jitter if the true conduction time standard deviation is less than 7 thousands. This condition is satisfies by the given prototype system because then confidence interval is obtained for the population standard deviation is contains the given true population standard deviation.