An educational organization in California is interested in estimating the mean number of minutes per day that children between the age of 6 and 18 spend watching television per day. A previous study showed that the population standard deviation was 21.5 minutes. The organization selected a random sample of n = 200 children between the age of 6 and 18 and recorded the number of minutes of TV that each person watched on a particular day. The mean time was 191.3 minutes. If the leaders of the organization wish to develop an interval estimate with 98 percent confidence, what would be the upper and lower limits of the interval estimate?
A) Approximately 187.76 minutes - 194.84 minutes
B) About 141.21 minutes - 241.40 minutes
C) Approximately 188.3 minutes - 194.3 minutes
D) None of the above

Question

Quizplus · Accepted Answer

To calculate the confidence interval, we use the formula for the margin of error: $ME = Z * (\sigma / \sqrt{n})$, where $Z$ is the Z-score corresponding to the confidence level (for 98% confidence, $Z \approx 2.33$), $\sigma$ is the population standard deviation (21.5 minutes), and $n$ is the sample size (200). Calculating the margin of error gives us $ME = 2.33 * (21.5 / \sqrt{200}) \approx 3.54$. Adding and subtracting this margin of error from the sample mean (191.3 minutes) gives us the confidence interval: $191.3 \pm 3.54$, which is approximately 187.76 minutes to 194.84 minutes.

An Educational Organization in California Is Interested in Estimating the Mean