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 95 percent confidence, what will the margin of error be?
a) about ±2.98 minutes
b) approximately ±1.52 minute
c) approximately ±42.14 minutes
d) z = 1.96

Option A) About ±2.98 minutes

Step-by-step explanation:

We are given the following information in the question:

Sample mean, = 191.3 minutes

Sample size, n = 200

Population standard deviation, σ = 21.5 minutes

Alpha, α = 0.05

The leaders of the organization wish to develop an interval estimate with 95 percent confidence.

Margin of error =

Putting the values, we get:

Option A) About ±2.98 minutes

in the math equation 8 x 7 = 56, the 8 and the 7 are both factors of 56, since in this lesson we will use square tiles to figure out the factors of a whole find as many different arrays as you can, but follow one important rule: all the rows must have the i can make two rows of nine, so 2 and 9 are both factors of 18.

step-by-step explanation:

3

step-by-step explanation:

divide 13.5 by 9 to get the scale factor. it equals 1.5

2 * 1.5 = 3