Answer to Question #343929 in Statistics and Probability for yow

Question

The average weight of 25 chocolate bars selected from a normally distributed population is 200 g with a standard deviation of 10 g. Find the interval estimate using 98% confidence level.

Answer 1

The critical value for $\alpha = 0.02$ is $z_c = z_{1-\alpha/2} = 2.3263.$

The corresponding confidence interval is computed as shown below:

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

$$=(200-2.3263\times\dfrac{10}{\sqrt{25}}, 200+2.3263\times\dfrac{10}{\sqrt{25}})$$

$$=(195.3474, 204.6526)$$

Therefore, based on the data provided, the 98% confidence interval for the population mean is $195.3474 < \mu < 204.6526,$ which indicates that we are 98% confident that the true population mean $\mu$ is contained by the interval $(195.3474, 204.6526).$