Chapter 9, Problem T9.5SPT

To determine

To find: The standardized test statistic using the provided information.

Answer to Problem T9.5SPT

The test statistic is $\frac{0.59-0.50}{\sqrt{\frac{0.50\left(1-0.50\right)}{100}}}$ .

Explanation of Solution

Given:

  $\begin{array}{l}\text{Number of trials}\left(n\right)=100\\ \text{Number of events}\left(x\right)=59\\ \text{Level of significance}\left(\alpha \right)=0.05\end{array}$

The null and alternative hypotheses are:

  $\begin{array}{l}{H}_0:p=0.5\\ H_a:p>0.5\end{array}$

Formula used:

The formula to compute the test statistic is:

  $z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}$

Where,

  $\begin{array}{l}{p}_0\text{ is the population proportion}\\ \widehat{p}\text{ is the sample proportion}\end{array}$

Calculation:

The sample proportion can be computed as:

  $\begin{array}{c}\text{Sample proportion}\left(\widehat{p}\right)=\frac{x}{n}\\ =\frac{59}{100}\\ =0.59\end{array}$

Now, test statistic can be calculated as:

  $\begin{array}{c}z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\\ =\frac{0.59-0.50}{\sqrt{\frac{0.50\left(1-0.50\right)}{100}}}\end{array}$

Thus, the test statistic is $\frac{0.59-0.50}{\sqrt{\frac{0.50\left(1-0.50\right)}{100}}}$ .

Hence, the correct option is (d).