Question

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 ages 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 critical value should be used? Question 31 options: t = 2.38 z = 1.645 Approximately z = 2.33 Can't be determined without knowing the margin of error.

Answer 1

Answer:

The critical value for a 98% CI is z=2.33.

The 98% confidence interval for the mean is (187.76, 194.84).

Step-by-step explanation:

We have to develop a 98% confidence interval for the mean number of minutes per day that children between the age of 6 and 18 spend watching television per day.

We know the standard deveiation of the population (σ=21.5 min.).

The sample mean is 191.3 minutes, with a sample size n=200.

The z-value for a 98% CI is z=2.33, from the table of the standard normal distribution.

The margin of error is:

$E=z\cdot \sigma/\sqrt{n}=2.33*21.5/\sqrt{200}=50.095/14.142=3.54$

With this margin of error, we can calculate the lower and upper bounds of the CI:

$LL=\bar x-z\cdot\sigma/\sqrt{n}=191.3-3.54=187.76\\\\\\UL=\bar x+z\cdot\sigma/\sqrt{n}=191.3+3.54=194.84$

The 98% confidence interval for the mean is (187.76, 194.84).