Question

an article in a florida newspaper reported on the topics that teenagers most want to discuss with their parents. the findings, the results of a poll, showed that 253 would like more discussion about the family's financial situation, 203 would like to talk about school, and 93 would like to talk about religion. these were based on a national sampling of 549 teenagers. estimate the proportion of all teenagers who want more discussions about school. use a 99% confidence level.

Answer 1

Based on a national sample of 549 teenagers, 203 expressed a desire for more discussions about school. With a 99% confidence level, the estimated proportion of all teenagers wanting more school discussions is between 33.5% and 40.2%.

To estimate the proportion of all teenagers who want more discussions about school, we can use the formula for calculating the confidence interval for a proportion. The formula is given by:

`\[ \text{Confidence Interval} = \hat{p} \pm Z * \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]`

Where:

-
`\(\hat{p}\)` is the sample proportion,

- Z is the Z-score corresponding to the desired confidence level,

- n is the sample size.

In this case, the sample proportion
`(\(\hat{p}\))` is the proportion of teenagers who want to talk about school, which is 203/549. The Z-score for a 99% confidence level is approximately 2.576. The sample size n is 549.

Let's plug these values into the formula:

`\[ \text{Confidence Interval} = (203)/(549) \pm 2.576 * \sqrt{((203)/(549) * \left(1 - (203)/(549)\right))/(549)} \]`

Now, calculate the values:

`\[ \text{Confidence Interval} = 0.369 \pm 2.576 * \sqrt{(0.369 * 0.631)/(549)} \]`

Calculate the standard error:

`\[ \text{Standard Error} = \sqrt{(0.369 * 0.631)/(549)} \]\[ \text{Standard Error} \approx 0.026 \]`

Now, substitute the standard error into the confidence interval formula:

`\[ \text{Confidence Interval} = 0.369 \pm 2.576 * 0.026 \]`

Now, calculate the upper and lower bounds of the confidence interval:

`\[ \text{Lower Bound} = 0.369 - (2.576 * 0.026) \]\[ \text{Upper Bound} = 0.369 + (2.576 * 0.026) \]`

With a 99% confidence level, the estimated proportion of all teenagers wanting more school discussions is between 33.5% and 40.2%. This gives you the 99% confidence interval for the proportion of teenagers who want more discussions about school.

Keep in mind that the confidence interval is an estimate, and it indicates a range within which we are reasonably confident the true proportion lies. In this case, it represents the proportion of all teenagers who want more discussions about school.