A newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the results of a poll, showed that 46% would like more discussion about the family's financial situation, 37% would like to talk about school, and 30% would like to talk about religion.
These and other percentages were based on a national sampling of 508 teenagers. Estimate the proportion of all teenagers who want more family discussions about school. Use a 99% confidence
Level.

A) $.63 \pm .055$
B) $.63 \pm .002$
C). $37 \pm .002$
D). $37 \pm .055$

Question

A newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the results of a poll, showed that 46% would like more discussion about the family's financial situation, 37% would like to talk about school, and 30% would like to talk about religion.
These and other percentages were based on a national sampling of 508 teenagers. Estimate the proportion of all teenagers who want more family discussions about school. Use a 99% confidence
Level.

A) $.63 \pm .055$
B) $.63 \pm .002$
C). $37 \pm .002$
D). $37 \pm .055$

Quizplus · Accepted Answer

The sample proportion of teenagers who want more family discussions about school is 0.37. To estimate the population proportion with 99% confidence, we can use the formula:sample proportion +/- margin of errorwhere the margin of error is:z * sqrt((sample proportion * (1 - sample proportion)) / sample size)where z is the z-score corresponding to the desired confidence level. For a 99% confidence level, z = 2.576.Plugging in the values, we get:0.37 +/- 2.576 * sqrt((0.37 * (1 - 0.37)) / 508)= 0.37 +/- 0.055Therefore, the 99% confidence interval for the population proportion of teenagers who want more family discussions about school is (0.37 - 0.055, 0.37 + 0.055) = (0.315, 0.425).

A Newspaper Reported on the Topics That Teenagers Most Want .63±.055.63 \pm .055.63±.055

A Newspaper Reported on the Topics That Teenagers Most Want $.63 \pm .055$