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.
Number produced | Number defective |
1-100 | 5 |
101-200 | 6 |
201-300 | 7 |
301-400 | 8 |
401-500 | 10 |
Table 11.5
A random sample was taken to determine the actual number of defects. Table 11.6 shows the results of the survey.
Number produced | Number defective |
1-100 | 5 |
101-200 | 7 |
201-300 | 8 |
301-400 | 9 |
401-500 | 11 |
Table 11.6
State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom.
To state:
The null and alternative hypothesis needed to conduct a goodness-of-fit test and the degrees of freedom.
Answer to Problem 11.1TI
Explanation of Solution
Given:
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 |
Interpretation:
The null hypothesis can be stated as:
And the alternative hypothesis can be stated as:
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;
Conclusion:
The null hypothesis can be stated as:
And the alternative hypothesis can be stated as:
And the degree of freedom = 4.
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Chapter 11 Solutions
Introductory Statistics
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