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 $p_1-p_2$. Interpret this interval. Can we be $95$ percent confident that $p_1-p_2$ is less than $0$? That is, can we be $95$ percent confident that $p_1$ is less than $p_2$? Explain.

Which proportions are equivalent to $\dfrac{x}{12}=\dfrac{3}{4}$ ?

$$\dfrac{x}{4}=\dfrac{12}{3}$$

Use each set of numbers to form two proportions. 30, 6, 5, 1

Show that the given proportions are eq Find the value of $x$.

$$\dfrac{x-1}{x+2}=\dfrac{10}{3 x-2}$$

Consider the proportions given below:

$$\frac{m}{n}=\frac{1}{2}\text{~~~and~~~~}\frac{n}{k}=\frac{2}{5}$$

What is $\dfrac{m}{k}$ ? Explain your reasoning.

Why is it not possible in the earlier example to have 100% confidence? Explain.

Given that

$$\frac{A}{B}=\frac{C}{D^{\prime}}$$

write three other proportions using A,B,C, and D.

Are the proportions $\frac{e}{f}=\frac{g}{5}$ and $\frac{5}{f}=\frac{g}{e}$ equivalent? Why?

Determine the proportions $\hat{p}$ and $\hat{q}$ for each.

a. $n=48, X=34$ b. $n=75, X=28$ c. $n=100, X=50$ d. $n=24, X=6$ e. $n=144, X=12$

Solve these proportions.

$$\dfrac{18}{c} = \dfrac{2}{7}$$

Given specific sample data, such as the data given in the Exercise, which confidence interval is wider: the $95 \%$ confidence interval or the $80 \%$ confidence interval? Why is it wider?

Find the proportions $\hat{p}$ and $\hat{q}$ for each.

a. $n=48, X=34$ b. $n=75, X=28$ c. $n=100, X=50$ d. $n=24, X=6$ e. $n=144, X=12$

Solve the following proportions. $\frac{4}{w}=\frac{5}{8}$

Show that the given proportions are eq Find the value of $x$.

$$\dfrac{x(x+5)}{4 x+4}=\dfrac{9}{5}$$