Question

in a survey of 711 lesbian, gay, bisexual, or transgender adults in a certain country, 427 said that they consider themselves bisexual. construct a 90% confidence interval for the population proportion. interpret the results. a 90% 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 $x = 427$ (number of successes) and $n=711$ (sample size).
$\hat{p}=\frac{427}{711}\approx0.6006$

Step2: Calculate $z -$ value for 90% confidence interval

For a 90% confidence interval, the significance level $\alpha=1 - 0.90=0.10$, and $\alpha/2=0.05$. The $z -$ value $z_{\alpha/2}=z_{0.05}\approx1.645$ (from standard normal distribution table)

Step3: Calculate the margin of error $E$

The formula for margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.6006$, $n = 711$, and $z_{\alpha/2}=1.645$

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Step4: Calculate the confidence interval

The confidence interval is $\hat{p}-ESubstitute $\hat{p}=0.6006$ and $E = 0.0303$
$0.6006-0.0303$0.570

Answer:

$(0.570,0.631)$

Interpretation: We are 90% confident that the true proportion of lesbian, gay, bisexual, or transgender adults in the country who consider themselves bisexual lies between $0.570$ and $0.631$.