Answer to Question #113809 in Complex Analysis for Windia paklungang

Question

We have a sample of 50 micro-drills for drilling holes in low-carbon alloy steel. The average lifetime, expressed as the number of holes drilled before failure, is 12.68. The standard deviation of the sample of 50 is 6.83 holes. Find the 95% confidence interval for this sample.

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.

We need to construct the 95% confidence interval for the population mean $\mu.$ The following information is provided:

$\bar{X}=12.68, \sigma=6.83, n=50,$

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:

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

$$=\big(12.68-1.96\times{6.83\over\sqrt{50}},12.68+1.96\times{6.83\over\sqrt{50}}\big)\approx$$

$$\approx(10.787, 14.573)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $10.787<\mu<14.573,$ which indicates that we are 95% confident that the true population mean $\mu$

is contained by the interval $(10.787, 14.573).$