AP Statistics - How to find confidence intervals for a difference between two proportions

In a simple random sample of people, are left-handed. Find a confidence interval for the true proportion of left-handed people in the entire population.

$0.261\pm1.96 \sqrt{\frac{(0.261)(0.739)}{1000}}$

$0.261\pm 1.64\sqrt{\frac{(0.261)(0.739)}{1000}}$

$0.261\pm 1.64\sqrt{\frac{(0.25)(0.75)}{1000}}$

$0.261\pm \sqrt{\frac{(0.261)(0.739)}{1000}}$

The sample is not sufficiently large. A confidence interval cannot be used.

Explanation

Step 1: Determine the approporiate formula. In this case, the statistic is a proportion because the question says 261 people out of 1000 are left-handed. We want to use a confidence interval for proportions. The equation is

$\hat{p}\pm z^{*} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Step 2: $\hat{p}$ , or the sample proportion is equal to $\frac{261}{1000} = 0.261$

Step 3: The questions asks for a 95% confidence interval. You can assume a normal distribution because of the large sample size. Therefore, in a normal distribution, 95% of the data is contained within $z=\pm 1.96$ .

$z^{*}=1.96$

Step 4: Substitue all the values into the equation to get the answer.

$0.261\pm1.96 \sqrt{\frac{(0.261)(0.739)}{1000}}$

How to find confidence intervals for a difference between two proportions

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