Question

in a survey of 658 lesbian, gay, bisexual, or transgender adults in a certain country, 325 said that they consider themselves bisexual. construct a 99% confidence interval for the population proportion. interpret the results. a 99% confidence interval for the population proportion is (□,□). (round to three decimal places as needed.)

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 325$ (number of successes) and $n=658$ (sample size).
$\hat{p}=\frac{325}{658}\approx0.494$

Step2: Find $z$-value

For a $99\%$ confidence interval, the significance level $\alpha=1 - 0.99=0.01$. Then $\frac{\alpha}{2}=0.005$. The $z$-value $z_{\frac{\alpha}{2}}$ is the $z$-score such that the area to the right of it is $0.005$. From the standard normal table, $z_{\frac{\alpha}{2}} = 2.576$

Step3: Calculate margin of error

The margin of error $E=z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.494$, $n = 658$, and $z_{\frac{\alpha}{2}}=2.576$
$E=2.576\sqrt{\frac{0.494\times(1 - 0.494)}{658}}$
First, calculate $0.494\times(1 - 0.494)=0.494\times0.506 = 0.249$
Then $\sqrt{\frac{0.249}{658}}\approx\sqrt{0.000378}\approx0.0194$
$E=2.576\times0.0194\approx0.050$

Step4: Calculate confidence interval

The confidence interval is $\hat{p}-E$0.494-0.050 < p<0.494 + 0.050$
$0.444

Answer:

$(0.444,0.544)$

Interpretation: We are $99\%$ confident that the true proportion of lesbian, gay, bisexual, or transgender adults in the country who consider themselves bisexual lies between $0.444$ and $0.544$.