Question

2. Suppose that 100 people are sampled at random and their average web browsing time per week is 3.2 hours with a standard deviation of 2.5 hours. Construct the 99% confidence interval for the population mean.

Answer 1

Conditions

2. Suppose that 100 people are sampled at random and their average web browsing time per week is 3.2 hours with a standard deviation of 2.5 hours. Construct the 99% confidence interval for the population mean.

Solution

$$\mathbb{P} \left( \bar{X} - z_{1-\alpha} \frac{\sigma}{\sqrt{n}} \leq \mu \leq \bar{X} + z_{1-\alpha} \frac{\sigma}{\sqrt{n}} \right) = \alpha$$ $$z_{0.99} = 2.326$$

Lower point of the confidence interval:

$$3.2 - 2.326 \cdot \frac{2.5}{10} = 2.6185$$

Upper point of the confidence interval:

$$3.2 + 2.326 \cdot \frac{2.5}{10} = 3.7815$$

Answer: $2.6185 \leq \mu \leq 3.7815$