Question

From 50 life samples of mini drill bits made of low-carbon alloy steel, (expressed in the number of holes that can be drilled before it breaks), an average of 12.68 can be calculated, with a standard deviation of 6.83. Calculate a 95% confidence interval for the average mini drill in this condition.

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 $(10.787, 14.573).$