Chapter 9.2, Problem 59E

To determine

The correct option which shows the appropriate standardized test statistic is:

  $z=\frac{0.64-0.5}{\sqrt{\frac{0.64\left(0.36\right)}{100}}}$

  $t=\frac{0.64-0.5}{\sqrt{\frac{0.64\left(0.36\right)}{100}}}$

  $z=\frac{0.64-0.5}{\sqrt{\frac{0.5\left(0.5\right)}{100}}}$

  $z=\frac{0.64-0.5}{\sqrt{\frac{0.64\left(0.36\right)}{64}}}$

  $z=\frac{0.5-0.64}{\sqrt{\frac{0.5\left(0.5\right)}{100}}}$

Answer to Problem 59E

The correct option is C. The appropriate standardized test statistic is $z=\frac{0.64-0.5}{\sqrt{\frac{0.5\left(0.5\right)}{100}}}$

Explanation of Solution

Given:

  $\begin{array}{l}n=100\\ x=64\\ p=50\%=0.50\end{array}$

Concept Used:

The test statistic is calculated by using the following expression:

  $z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$

Where,

Sample proportion: $\widehat{p}$

Sample size: $n$

Calculation:

Firstly need to calculate the sample proportion which is the ratio of the number of successes and the sample size:

  $\begin{array}{l}\widehat{p}=\frac{x}{n}\\ \widehat{p}=\frac{64}{100}\\ \widehat{p}=0.64\end{array}$

The test statistic is calculated by using the following expression:

  $z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$

Put the values from given data in the above eq.

  $\begin{array}{l}z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}\\ z=\frac{0.64-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{100}}}\\ z=\frac{0.64-0.5}{\sqrt{\frac{0.5\left(0.5\right)}{100}}}\end{array}$

Hence, the correct option is C.

Conclusion:

Hence, the correct option is C.