A newspaper reported the results of a poll concerning topics that teenagers most want to discuss with their parents. In the the poll 37% of teenagers said they would like to talk with their parents about school. These and other percentages were based on a national sampling of 536 teenagers. Estimate the proportion of all teenagers that want more discussions with their parents about school. Use a 99% confidence level. Right-click this Excel file to open a worksheet template.

Step-by-step explanation:

Confidence interval is written as

Sample proportion ± margin of error

Margin of error = z × √pq/n

Where

z represents the z score corresponding to the confidence level

p = sample proportion. It also means probability of success

q = probability of failure

q = 1 - p

p = x/n

Where

n represents the number of samples

x represents the number of success

From the information given,

n = 536

p = 37/100 = 0.37

q = 1 - 0.37 = 0.63

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.005 = 0.995

The z score corresponding to the area on the z table is 2.53. Thus, confidence level of 99% is 2.58

Therefore, the 99% confidence interval is

0.37 ± 2.58 × √(0.37)(0.63)/536

= 0.37 ± 0.054

The lower limit of the confidence interval is

0.37 - 0.054 = 0.316

The upper limit of the confidence interval is

0.37 + 0.054 = 0.424

Therefore, with 99% confidence interval, the proportion of all teenagers that want more discussions with their parents about school is between 0.316 and 0.424

150

step-by-step explanation:

(sorry i'm not typing it again just read the other two explanation sections i just typed it'll be the same.)