Question

In a sample of 100 households, the mean number of hours spent on social networking sites during the month of January was 50 hours. In a much larger study, the standard deviation was determined to be 6 hours. Assume the population standard deviation is the same. What is the 98% confidence interval for the mean hours devoted to social networking in January?

Answer 1

The confidence interval for the population mean at a $100\times(1-\alpha)\%$ confidence level is $\mu\pm Z_{\alpha/2}\sigma_\mu$. Here $\mu=50$ is the sample mean, $Z_{\alpha/2}$ is the critical value such that $\mathbb P(|Z|$, and $\sigma_\mu=\dfrac\sigma{\sqrt n}$, where in turn $\sigma=6$ is the assumed population standard deviation and $n=100$ is the sample size.

First find the critical value. By the empirical rule, you might already know that about 95% of a normal distribution lies within two standard deviations of the mean, and that about 99.7% lies within three standard deviations, so you can expect $Z_{\alpha/2}$ to lie somewhere between 2 and 3. To be more precise, you can consult a z-score table to find that $Z_{\alpha/2}\approx2.3264$.

Plug in everything you know and you find that the 98% confidence interval is

$\left(50-2.3264\times\dfrac6{\sqrt{100}},50+2.3264\times\dfrac6{\sqrt{100}}\right)=(48.6042,51.3958)$