Chapter 13.5, Problem 133E

a:

To determine

Testing hypothesis.

Explanation of Solution

Null hypothesis (H₀) is $\left({\text{p}}_1-{\text{p}}_2\right)=0$ and alternate hypothesis is (H₁) $\left({\text{p}}_1-{\text{p}}_2\right)>0$.

The Z table value can be calculated as follows:

$\begin{array}{c}\text{z}>{\text{Z}}_{\alpha }\\ ={\text{Z}}_{0.05}\\ =1.645\end{array}$

The Z table value is 1.645.

The Z value (z) can be calculated as follows:

$\begin{array}{c}\text{z}=\frac{{\text{(}\widehat{\text{p}}}_{\text{1}}-{\widehat{\text{p}}}_{\text{2}}\text{)}}{\sqrt{\widehat{\text{p}}\text{(1}-\widehat{\text{p}}\text{)}\left(\frac{\text{1}}{{\text{n}}_{\text{1}}}+\frac{\text{1}}{{\text{n}}_{\text{2}}}\right)}}\\ =\frac{(0.56-0.46)}{\sqrt{0.518(1-0.518)\left(\frac{1}{1,100}+\frac{1}{800}\right)}}\\ =4.31\end{array}$

The Z value is 4.31. The p value is 0. Since the p value is equal to 0, the alternate hypothesis is accepted.

b:

To determine

Testing a new hypothesis.

Explanation of Solution

Null hypothesis (H₀) is $\left({\text{p}}_1-{\text{p}}_2\right)=0$ and alternate hypothesis is (H₁) $\left({\text{p}}_1-{\text{p}}_2\right)>0$.

The Z table value can be calculated as follows:

$\begin{array}{c}\text{z}>{\text{Z}}_{\alpha }\\ ={\text{Z}}_{0.05}\\ =1.645\end{array}$

The Z table value is 1.645.

The Z value (z) can be calculated as follows:

$\begin{array}{c}\text{z}=\frac{{\text{(}\widehat{\text{p}}}_{\text{1}}-{\widehat{\text{p}}}_{\text{2}}\text{)}-{\text{(p}}_{\text{1}}-{\text{p}}_{\text{2}}\text{)}}{\sqrt{\frac{{\widehat{\text{p}}}_{\text{1}}\text{(1}-{\widehat{\text{p}}}_{\text{1}}\text{)}}{{\text{n}}_{\text{1}}}+\frac{{\widehat{\text{p}}}_{\text{2}}\text{(1}-{\widehat{\text{p}}}_{\text{2}}\text{)}}{{\text{n}}_{\text{2}}}}}\\ =\frac{(0.56-0.46)-0.05}{\sqrt{\frac{0.56(1-0.56)}{1,100}+\frac{0.46(1-0.46)}{800}}}\\ =2.16\end{array}$

The Z value is 2.16.

The p value (p) can be calculated as follows. Since the p value is equal to 0, the alternate hypothesis is accepted.

$\begin{array}{c}\text{p}=\left(\text{1}-\text{P}\left(Z<2.16\right)\right)\\ =\left(1-0.9846\right)\\ =0.0154\end{array}$

The p value is 0.0154.

c:

To determine

Calculate the upper and lower limits.

Explanation of Solution

The values of upper limit (UL) and lower limit (LL) can be calculated as follows:

$\begin{array}{c}\text{UL}={(\widehat{\text{p}}}_{\text{1}}-{\widehat{\text{p}}}_{\text{2}}\text{)}\pm {\text{z}}_{\frac{\text{α}}{2}}\sqrt{\frac{{\widehat{\text{p}}}_{\text{1}}\text{(1}-{\widehat{\text{p}}}_{\text{1}}\text{)}}{{\text{n}}_{\text{1}}}+\frac{{\widehat{\text{p}}}_{\text{2}}\text{(1}-{\widehat{\text{p}}}_{\text{2}}\text{)}}{{\text{n}}_{\text{2}}}}\\ ={(\widehat{\text{p}}}_{\text{1}}-{\widehat{\text{p}}}_{\text{2}}\text{)}\pm {\text{z}}_{\frac{\text{95\%}}{2}}\sqrt{\frac{{\widehat{\text{p}}}_{\text{1}}\text{(1}-{\widehat{\text{p}}}_{\text{1}}\text{)}}{{\text{n}}_{\text{1}}}+\frac{{\widehat{\text{p}}}_{\text{2}}\text{(1}-{\widehat{\text{p}}}_{\text{2}}\text{)}}{{\text{n}}_{\text{2}}}}\\ =(0.56-0.46)\pm 1.96\sqrt{\frac{0.56(1-0.56)}{1,100}+\frac{0.46(1-0.46)}{800}}\\ =0.055\text{ and 0}\text{.145}\end{array}$

The lower limit is 0.055 and the upper limit is 0.145.