Question

One step in producing a spare part, involves drilling 4 holes. From 150 samples, the average time needed for this process is 72 seconds and SD = 10 seconds.
(a) calculate the CIs each with 95% CIs and 99.5% for the average time needed for the process.
(b) How many samples must be taken to get a 99.5% CI for averages in Ŧ 1.5 seconds from the actual price?

Answer 1

The Central Limit Theorem

Let $X_1,X_2,...,X_n$ be a random sample from a distribution with mean $\mu$ and variance $\sigma^2.$ Then if $n$

is sufficiently large, $\bar{X}$ has approximately a normal distribution with $\mu_{\bar{X}}=\mu$ and $\sigma_{\bar{X}}^2=\sigma^2/n.$

If $n>30$ the Central Limit Theorem can be used.

The following information is provided: $\bar{X}=72, \sigma=10, n=150.$

We need to construct the 95% confidence interval for the population mean $\mu.$

The critical value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$ The corresponding confidence interval is computed as shown below:

$$CI=\big(\bar{X}-z_c{\sigma\over\sqrt{n}},\bar{X}+z_c{\sigma\over\sqrt{n}}\big)=$$

$$=\big(72-1.96\times{10\over\sqrt{150}},72+1.96\times{10\over\sqrt{150}}\big)\approx$$

$$\approx(70.400, 73.600)$$

We need to construct the 99.5% confidence interval for the population mean $\mu.$

The critical value for $\alpha=0.005$ is $z_c=z_{1-\alpha/2}=2.807.$

The corresponding confidence interval is computed as shown below:

$$CI=\big(\bar{X}-z_c{\sigma\over\sqrt{n}},\bar{X}+z_c{\sigma\over\sqrt{n}}\big)=$$

$$=\big(72-2.807\times{10\over\sqrt{150}},72+2.807\times{10\over\sqrt{150}}\big)\approx$$

$$\approx(69.708, 74.292)$$

(b) How many samples must be taken to get a 99.5% CI for averages in $\pm1.5$ seconds from the actual price?

$$z_c{\sigma\over\sqrt{n}}\leq1.5$$

$$n\geq\bigg({z_c\sigma\over1.5}\bigg)^2$$

$$n\geq\bigg({2.807\cdot10\over1.5}\bigg)^2$$

$$n\geq351$$