Question

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.

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

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

Short answer

\(H_{0}:\)The number of defaults fits expectations

\(H_{a}:\) The number of defaults does not fit expectations

\(df=4\)

Step-by-step solution

Step 1. Given information

A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in the below given table:

Number produced	Number defective
0 – 100	5
101 – 200	6
201 – 300	7
301 – 400	8
401 – 500	10

A random sample was taken to determine the actual number of defects. The given table shows the results of the survey;

Number produced	Number defective
0 – 100	5
101 – 200	7
201 – 300	8
301 – 400	9
401 – 500	11

Step 2. Calculation

The null hypothesis can be stated as:

\(H_{0}:\)The number of defaults fits expectations

And the alternative hypothesis can be stated as:

\(H_{a}:\) The number of defaults does not fit expectations

In the given case, it is best to apply goodness-of-fit test because all entries are greater than or equal to five.

Therefore, the degree of freedom can be calculated as shown below;

degree of freedom\(=df=number of cells -1\)

\(=5-1=4\)

The null hypothesis can be stated as:

\(H_{0}:\) The number of defaults fits expectations

And the alternative hypothesis can be stated as:

\(H_{a}:\) The number of defaults does not fit expectations

And the degree of freedom \(= 4\).