Chapter 11, Problem 8RE

(a)

To determine

To find:

The null and alternate hypotheses for the given data.

Answer to Problem 8RE

Null hypothesis: Machine $A$ produces greater than or equal proportion of rods than machine $B$ produces.

  $H_0:{\widehat{p}}_1\ge {\widehat{p}}_2$

Alternate hypothesis: Machine $A$ produces less than the proportion of rods of machine $B$ produces.

  $H_0:{\widehat{p}}_1<{\widehat{p}}_2$

Explanation of Solution

Given Information:

The number of rods taken from machine $\text{A is }{n}_1=1000$

The number of rods met specifications regarding length and diameter is $x_1=960$

The number of rods taken from machine $\text{B is }{n}_2=600$

The number of rods met specifications regarding length and diameter is $x_2=582$

Formula used:

  ${\widehat{p}}_1=\frac{x_1}{n_1}$

Calculation:

The number of rods taken from machine $\text{A is }{n}_1=1000$

The number of rods met specifications regarding length and diameter is $x_1=960$

The proportion of rods met specifications regarding length and diameter of machine $A$ is as follows:

  $\begin{array}{c}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{960}{1000}\\ =0.96\end{array}$

The number of rods taken from machine $\text{B is }{n}_2=600$

The number of rods met specifications regarding length and diameter is $x_2=582$

The proportion of rods met specifications regarding length and diameter of machine $B$ is as follows:

  $\begin{array}{c}{\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{582}{600}\\ =0.97\end{array}$

Null hypothesis: Machine $A$ produces greater than or equal proportion of rods than machine $B$ produces.

  $H_0:{\widehat{p}}_1\ge {\widehat{p}}_2$

Alternate hypothesis: Machine $A$ produces less than the proportion of rods of machine $B$ produces.

  $H_0:{\widehat{p}}_1<{\widehat{p}}_2$

(b)

To determine

To find:

The machine which is to be used.

Answer to Problem 8RE

Machine $A$ produces greater than or equal proportion of rods than machine $B$ produces.

Explanation of Solution

Given Information:

The number of rods taken from machine $\text{A is }{n}_1=1000$

The number of rods met specifications regarding length and diameter is $x_1=960$

Calculation:

The proportion of rods met specifications regarding length and diameter of machine $A$ is as follows:

  $\begin{array}{c}z=\frac{{\widehat{p}}_1-{\widehat{p}}_2}{\sqrt{\frac{p_1{q}_1}{n_1}+\frac{p_2{q}_2}{n_2}}}\\ =\frac{0.96-0.97}{\sqrt{\frac{0.96\left(1-0.96\right)}{1000}+\frac{0.97\left(1-0.97\right)}{600}}}\\ =\frac{-0.01}{0.00932}\\ =-1.0727\end{array}$

  $\begin{array}{c}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{960}{1000}\\ =0.96\end{array}$

The number of rods taken from machine $\text{B is }{n}_2=600$

The number of rods met specifications regarding length and diameter is $x_2=582$

The proportion of rods met specifications regarding length and diameter of machine $B$ is as follows:

  $\begin{array}{c}{\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{582}{600}\\ =0.97\end{array}$

  $$

The critical value at $5\%$ level of significance is $-1.645$ . Compare test statistic and the critical value. The test statistic value is less than the critical value i.e.

  $\left|z_{Cal}\right|=1.0727,\left|z_{\text{Critical}}\right|=1.645\Rightarrow \left|z_{Cal}\right|<\left|z_{\text{Critical}}\right|$ . The researcher can fail to reject the null hypothesis.

Thus, the machine $A$ produces greater than or equal proportion of rods than machine $B$ produces.