Question

An Article 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 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 sampling were based on 522 teenagers. Estimate the proportion of all teenagers who want more family discussions about school. Use a 90% confidence level. Express the answer in the form P hat+- E

Answer 1

The estimate is
`P__(hat)} \pm E = 0.37 \pm 0.0348`

Explanation:

From the question we are told that

The sample size is n = 522

The sample proportion of students would like to talk about school is
`\r p__(hat)} = 0.37`

Given that the confidence level is 90 % then the level of significance can be mathematically evaluated as

`\alpha = 100 - 90`

`\alpha = 10\%`

`\alpha = 0.10`

Next we obtain the critical value of
`(\alpha )/(2)` from the normal distribution table, the value is

`Z_{(\alpha )/(2) } =Z_{(0.10)/(2) } = 1.645`

Generally the margin of error can be mathematically represented as

`E = Z_{(\alpha )/(2) } * \sqrt{(\r P_(hat)(1- \r P_(hat) ))/(n ) }`

=>
`E = 1.645 * \sqrt{(0.37 (1- 0.37 ))/(522 ) }`

=>
`E = 0.0348`

Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is

`P__(hat)} \pm E`

substituting values

`0.37 \pm 0.0348`