Related questions with answers
We assume that we have selected two independent random samples from populations having proportions $p_1$ and $p_2$ and that $\hat{p}_1=800 / 1,000=.8$ and $\hat{p}_2=950 / 1,000=.95$.
Calculate a $95$ percent confidence interval for $p_1p_2$. Interpret this interval. Can we be $95$ percent confident that $p_1p_2$ is less than $0$? That is, can we be $95$ percent confident that $p_1$ is less than $p_2$? Explain.
Solutions
VerifiedThe goal of this task is to compute a confidence interval for $p_1p_2$ with a $95\%$ confidence interval.
When we have to compute a $100(1\alpha)\%$ confidence interval for $p_1p_2$, the formula we have to use is
$\left[ (\^p_1\^p_2) \pm z_{\alpha/2} \cdot \sqrt{\frac{\^p_1(1\^p_1)}{n_1}+\frac{\^p_2(1\^p_2)}{n_2}} \right].$
Let $p_1$ and $p_2$ be the proportions of two independent random samples of $1000$ population $n$, where:

$\hat{p}_1=\dfrac{800}{1000}=0.8$, $n_1=1000$

$\hat{p}_2=\dfrac{950}{1000}=0.9$, $n_2=1000$
Create an account to view solutions
Create an account to view solutions
More related questions
1/4
1/7