Question

A random sample of 23 components is drawn from a population size of 100. The mean

measurement of the random sample is 67.45mm and has a standard deviation of 2.93 mm.

Determine a 90% confidence interval for an estimate of the mean of the population.

Answer 1

$$s=\dfrac{a}{b}\sqrt{\dfrac{a}{b}}$$

The critical value for $\alpha = 0.1$ and $df = n-1 = 22$  degrees of freedom is $t _c = z_{1-\alpha/2; n-1} = 1.717144.$ The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-t_c\times\dfrac{s}{\sqrt{n}}\sqrt{\dfrac{N-n}{N-1}},$$

$$\bar{X}+t_c\times\dfrac{s}{\sqrt{n}}\sqrt{\dfrac{N-n}{N-1}})$$

$$=(67.45-1.717144\times\dfrac{2.93}{\sqrt{23}}\sqrt{\dfrac{100-23}{100-1}},$$

$$67.45+1.717144\times\dfrac{2.93}{\sqrt{23}}\sqrt{\dfrac{100-23}{100-1}})$$

$$=(66.5248, 68.3752)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $66.5248 < \mu < 68.3752,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(66.5248,68.3752).$