Need Help?

Subscribe to Probability

Subscribe
  • Notes
  • Comments & Questions

The life in hours of a 75 -watt light bulb is known to be
normally distributed with $$\sigma=25$$ hours. A random sample of
20 bulbs has a mean life of $$\overline{x}=1014$$ hours.

(a) Construct a $$95\%$$ two-sided confidence interval on the
mean life.
(b) Construct a $$95\%$$ lower-confidence bound on the mean
life.

$$\sigma=25, n=20, \overline{x}=1014$$

(a) $$95\%$$ CI Two sided

(b) $$95\%$$ CI lower-confidence bound

(a) $$\alpha=1-0.95=0.05$$

$$\overline{x}-Z_\frac{\alpha}{2}\left(\frac{\sigma}{\sqrt{n}}\right)\leq\mu \leq \overline{x}+Z_\frac{\alpha}{2}\left(\frac{\sigma}{\sqrt{n}}\right)$$

$$\rightarrow \frac{\alpha}{2}=0.025$$

$$Z_{0.025}=1.96$$

$$1014-1.96 \times \frac{25}{\sqrt{20}} \leq \mu \leq 1014+1.96\left(\frac{25}{\sqrt{20}}\right)$$

$$1003 \leq \mu \leq 1025$$

$$u=\infty$$

(b) $$\overline{x}-Z_{\alpha} \frac{\sigma}{\sqrt{n}} \leq \mu$$

$$Z_{\alpha}=Z_{0.05}$$

$$Z=1.65$$

$$1014-1.65 \times \frac{25}{\sqrt{20}} \leq \mu$$

$$1005 \leq \mu$$

Suppose that in first Exercise we wanted to be $$95\%$$
confident that the error in estimating the mean life is less than
five hours. What sample size should be used?

$$95 \mathrm{CI} \rightarrow E=5$$

$$\alpha=0.05$$

$$\sigma=25$$

$$n ?!$$

$$E=5$$

$$n=\left(\frac{Z_{\frac{\alpha}{2}} \sigma}{E}\right)^{2}$$

$$\frac{\alpha}{2}=0.025 \rightarrow Z_{0.025}=1.96$$

$$n=\left(\frac{1.96 \times 25}{5}\right)^{2}$$

$$=96.04$$

$$n=97$$

No comments yet

Join the conversation

Join Notatee Today!