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 98% confidence interval ranges from 48.60 to 51.40 hours.

Explanation:

Answer 2

The confidence interval for the population mean at a
`100*(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|<Z_(\alpha/2))=1-\alpha`, and
`\sigma_\mu=\frac\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*\frac6{√(100)},50+2.3264*\frac6{√(100)}\right)=(48.6042,51.3958)`