Question

in a survey of 668 lesbian, gay, bisexual, or transgender adults in a certain country, 385 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 $\hat{p}$

Sample proportion $\hat{p}=\frac{x}{n}$, where $x = 385$ (number of successes) and $n=668$ (sample size).
$\hat{p}=\frac{385}{668}\approx0.576$

Step2: Calculate $q = 1-\hat{p}$

$q=1 - 0.576=0.424$

Step3: Find $z$-value for 99% confidence interval

For 99% confidence interval, $z_{\alpha/2}=2.576$ (from standard normal distribution table)

Step4: Calculate margin of error $E$

Margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}\hat{q}}{n}}$
$E=2.576\sqrt{\frac{0.576\times0.424}{668}}$
First calculate $\frac{0.576\times0.424}{668}=\frac{0.244224}{668}\approx0.0003656$
$\sqrt{0.0003656}\approx0.01912$
$E=2.576\times0.01912\approx0.0493$

Step5: Calculate confidence interval

Lower limit $=\hat{p}-E=0.576 - 0.0493=0.527$
Upper limit $=\hat{p}+E=0.576+ 0.0493=0.625$

Answer:

$(0.527,0.625)$

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 $0.527$ and $0.625$.