Question

A television news program conducts a call in poll about a proposed city ban on handgun ownership. Of the 2372 calls,1921 oppose the ban. The station, following recommended practice, makes a confidence statement: "81% of the Channel 13 Pulse Poll sample opposed the ban. We can be 95% confident that the true proportion of citizens opposing a handgun ban is within 1.6% of the sample result." Is this conclusion justified?

Answer 1

We can be 95% confident that the true proportion of citizens opposing a handgun ban is within 1.6% of the sample result.

is true.

Explanation:

Given that a television news program conducts a call in poll about a proposed city ban on handgun ownership. Of the 2372 calls,1921 oppose the ban. The station, following recommended practice, makes a confidence statement:

Sample proportion p =
`(1921)/(2372) \\=0.8099`

q=1-p =
`0.1901\\`

Std error of p =
`\sqrt{(pq)/(n) } \\=0.0081`

Margin of error for 95%=1.96*std error = 0.0158

Since margin of error is 0.0158 i.e. nearly 15.8% and proportion is 0.8099 = 81%

the statement is correct.

We can be 95% confident that the true proportion of citizens opposing a handgun ban is within 1.6% of the sample result.

is true.