Question

From a random sample of 41 teens, it is found that on average they spend 43.1 hours each week online with a population standard deviation of 5.91 hours. What is the 90% confidence interval for the amount of time they spend online each week?

Answer 1

Explanation:

We want to determine a 90% confidence interval for the mean amount of time that teens spend online each week.

Number of sample, n = 41

Mean, u = 43.1 hours

Standard deviation, s = 5.91 hours

For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean +/- z ×standard deviation/√n

It becomes

43.1 ± 1.645 × 5.91/√41

= 43.1 ± 1.645 × 0.923

= 43.1 ± 1.52

The lower end of the confidence interval is 43.1 - 1.52 =41.58

The upper end of the confidence interval is 43.1 + 1.52 =44.62

Therefore, with 90% confidence interval, the mean amount of time that teens spend online each week is between 41.58 and 44.62