Question

A researcher wants to estimate the numbers of hours that 5 years old children spend watching television. A sample of 50 five-year old children was observed to have a mean viewing time of 5 hours. The population is normally distributed with a population standard deviation of 0.5 hours. Find the 95% confidence interval of the population mean.

Answer 1

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:

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

$$=(5-1.96\dfrac{0.5}{\sqrt{50}}, 5+1.96\dfrac{0.5}{\sqrt{50}})$$

$$=(4.861, 5.139)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $4.861 < \mu < 5.139,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(4.861, 5.139).$