A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in Table 11.5.

$\begin{array}{|ll|}\hline \text{Number produced}& \text{Number defective}\\ 0-100& 5\\ 101-200& 6\\ 201-300& 7\\ 301-400& 8\\ 401-500& 10\\ \hline\end{array}$

A random sample was taken to determine the actual number of defects. Table 11.6 shows the results of the survey.

$\begin{array}{|ll|}\hline \text{Number produced}& \text{Number defective}\\ 0-100& 5\\ 101-200& 7\\ 201-300& 8\\ 301-400& 9\\ 401-500& 11\\ \hline\end{array}$

State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom.

Given data:

A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in Table 11.5.

$\begin{array}{|ll|}\hline \text{Number produced}& \text{Number defective}\\ 0-100& 5\\ 101-200& 6\\ 201-300& 7\\ 301-400& 8\\ 401-500& 10\\ \hline\end{array}$

A random sample was taken to determine the actual number of defects. Table 11.6 shows the results of the survey.

$\begin{array}{|ll|}\hline \text{Number produced}& \text{Number defective}\\ 0-100& 5\\ 101-200& 7\\ 201-300& 8\\ 301-400& 11\\ 401-500& 9\\ \hline\end{array}$

The null hypothesis is expressed as follows:

$H_0$: The number of defaults is consistent with expectations.

The alternate hypothesis is as follows:

$H_a$: The number of defaults is higher than expected.

Because all of the entries are larger than or equal to five, it is best to utilize the goodness-of-fit test in this circumstance.

The degree of freedom can be computed.

Degree of freedom $df=numberofcells-1$

$=5-1$

$=4$