Question

The yield of a chemical process is being studied. From a previous experience yield is known to be normally distributed and σ = 3. The past 5 days if plant operation have to resulted in the following percent yields: 91.6,88.75,90.8,89.95, and 91.3. Find a 95% two-sided confidence interval on the true mean yield.

Answer 1

Sample mean

$$\bar{x}=\dfrac{1}{5}(91.6+88.75+90.8+89.95+91.3)=90.48$$

 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:

$$CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$

$$=(90.48-1.96\times\dfrac{3}{\sqrt{5}}, 90.48+1.96\times\dfrac{3}{\sqrt{5}})$$

$$=(87.85, 93.11)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $87.85 < \mu < 93.11,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(87.85, 93.11).$