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.

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Answer 1

Textbook Question

Chapter 11, Problem 11.1TI

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.

Expert Solution & Answer

To determine

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

H0: The number of defaults fits expectations

Ha: The number of defaults does not fit expectations

df=4.

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:

H0: The number of defaults fits expectations

And the alternative hypothesis can be stated as:

Ha: 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

Conclusion:

The null hypothesis can be stated as:

H0: The number of defaults fits expectations

And the alternative hypothesis can be stated as:

Ha: The number of defaults does not fit expectations

And the degree of freedom = 4.

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Answer 2

Textbook Question

Chapter 11, Problem 11.1TI

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.

Expert Solution & Answer

To determine

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

H0: The number of defaults fits expectations

Ha: The number of defaults does not fit expectations

df=4.

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:

H0: The number of defaults fits expectations

And the alternative hypothesis can be stated as:

Ha: 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

Conclusion:

The null hypothesis can be stated as:

H0: The number of defaults fits expectations

And the alternative hypothesis can be stated as:

Ha: The number of defaults does not fit expectations

And the degree of freedom = 4.

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Answer 3

Textbook Question

Chapter 11, Problem 11.1TI

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.

Expert Solution & Answer

To determine

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

H0: The number of defaults fits expectations

Ha: The number of defaults does not fit expectations

df=4.

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:

H0: The number of defaults fits expectations

And the alternative hypothesis can be stated as:

Ha: 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

Conclusion:

The null hypothesis can be stated as:

H0: The number of defaults fits expectations

And the alternative hypothesis can be stated as:

Ha: The number of defaults does not fit expectations

And the degree of freedom = 4.

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Answer 4

Textbook Question

Chapter 11, Problem 11.1TI

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.

Expert Solution & Answer

To determine

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

H0: The number of defaults fits expectations

Ha: The number of defaults does not fit expectations

df=4.

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:

H0: The number of defaults fits expectations

And the alternative hypothesis can be stated as:

Ha: 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

Conclusion:

The null hypothesis can be stated as:

H0: The number of defaults fits expectations

And the alternative hypothesis can be stated as:

Ha: The number of defaults does not fit expectations

And the degree of freedom = 4.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

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

H0: The number of defaults fits expectations

Ha: The number of defaults does not fit expectations

df=4.

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:

H0: The number of defaults fits expectations

And the alternative hypothesis can be stated as:

Ha: 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

Conclusion:

The null hypothesis can be stated as:

H0: The number of defaults fits expectations

And the alternative hypothesis can be stated as:

Ha: The number of defaults does not fit expectations

And the degree of freedom = 4.

Answer 8

Expert Solution & Answer

To determine

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

H0: The number of defaults fits expectations

Ha: The number of defaults does not fit expectations

df=4.

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:

H0: The number of defaults fits expectations

And the alternative hypothesis can be stated as:

Ha: 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

Conclusion:

The null hypothesis can be stated as:

H0: The number of defaults fits expectations

And the alternative hypothesis can be stated as:

Ha: The number of defaults does not fit expectations

And the degree of freedom = 4.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

To state:

The null and alternative hypothesis needed to conduct a goodness-of-fit test and the degrees of freedom.

Answer 12

Answer to Problem 11.1TI

H0: The number of defaults fits expectations

Ha: The number of defaults does not fit expectations

df=4.

Answer 13

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: