Elements of Statistics II

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STIIModule5.Testdocx.docx
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STIIModule5.Testdocx.docx

Module 5: Content Mastery Testing

Instructions

After studying and reviewing the concepts and practices of Module 5, answer the following:

1. Listed below are amounts of arsenic in samples of brown rice from three different states. The amount are in micrograms of arsenic and all samples have the same serving size. The data are from the Food and Drug Administration. Use a 10% significance level to test the claim that the three samples are from populations with the same mean. Use Excel to solve the exercise.

Arkansas	California	Texas
4.8	1.5	5.6
4.9	3.7	5.8
5.0	4.0	6.6
5.4	4.5	6.9
5.4	4.9	6.9
5.4	5.1	6.9
5.6	5.3	7.1
5.6	5.4	7.3
5.6	5.4	7.5
5.9	5.5	7.6
6.0	5.6	7.7
6.1	5.6	7.7

(Triola, 2018)

2. Listed below are attribute ratings of males by females who participated in speed dating events. Use a 5% significance level to test the claim that females in the different age brackets give attribute ratings with the same mean. Use Excel to solve the exercise.

Age 20-22	Age 23-26	Age 27-29
38.0	39.0	36.0
42.0	31.0	42.0
30.0	36.0	35.5
39.0	35.0	27.0
47.0	41.0	37.0
43.0	45.0	34.0
33.0	36.0	22.0
31.0	23.0	47.0
32.0	36.0	36.0
28.0	20.0	32.0

(Triola, 2018)

Be sure to review the academic expectations for your submission.

Submission Instructions:

· Submit your assignment by 11:59 PM Eastern on Sunday.

· Review the rubric to determine how your assignment will be graded.

· Your assignment will be run through Turnitin to check for plagiarism.

· Justify any assumption and show all the steps to solve the exercises.

STIIModule5Class.docx

Module 5: ANOVA

The Analysis of Variance (ANOVA) is a statistical procedure to test the hypothesis that the means of three or more populations are equal. The ANOVA requires the F-distribution.

(Triola, 2018)

If there is only one explanatory variable, the ANOVA procedure is called one-way or single factor ANOVA, in which this variable is a continuous variable with three or more categories called groups.

Hypotheses:

Null Hypothesis: H0 : μ1 = μ2 = μ3 = · · · = μk

Alternative Hypothesis: H1 : μ1 ≠ μ2 ≠ μ3 ≠ · · · ≠ μk

If the null hypothesis is rejected, we conclude that with a particular significance level (α) the population means are not all equal.

The Assumptions of the ANOVA are:

· All populations are normally distributed.

· The variances among all the populations are the same.

· The random error terms are independent and normally distributed with mean equal to zero (μ = 0), and standard deviation σ.

ANOVA Hypothesis Test:

Basics of Hypothesis Testing:

1. Given a claim, identify the Null Hypothesis H0 and the Alternative Hypothesis H1.

2. State the conclusion (reject the Null Hypothesis of fail to reject the Null Hypothesis).

Null Hypothesis: H0 : μ1 = μ2 = μ3 = · · · = μk

Alternative Hypothesis: H1 : μ1 ≠ μ2 ≠ μ3 ≠ · · · ≠ μk

Decision Rule: Reject Ho if p-value is less than the significance level

(p-value < α)

p-value is a measure of the strength of the evidence provided by the data against the null hypothesis Ho. The smaller the p-value, the stronger is the sample evidence for rejecting the null hypothesis Ho.

Example:

Autor Readability. Pages were randomly selected by the author from The Bear and the Dragon by Tom Clancy, Harry Potter and the Sorcerer’s Stone by J.K. Rowling, and War and Peace by Leo Tolstoy. The Flesch Reading Ease scores for those pages are listed below. With a significance level of 5%, do the authors appear to have the same level of readability?

Clancy	Rowling	Tolstoy
58.2	85.3	69.4
73.4	84.3	64.2
73.1	79.5	71.4
64.4	82.5	71.6
72.7	80.2	68.5
89.2	84.6	51.9
43.9	79.2	72.2
76.3	70.9	74.4
76.4	78.6	52.8
78.9	86.2	58.4
69.4	74.0	65.4
72.9	83.7	73.6

(Triola, 2018)

We will assume that the assumptions of the model are met:

· The three populations are normally distributed.

· The variances among all the populations are the same.

· The random error terms are independent and normally distributed with mean equal to zero (μ = 0), and standard deviation σ.

We are going to evaluate the ANOVA, with the support of the MS Excel program ( steps are listed in the Read Section of this module-Word and Excel documents attached).

From Excel ANOVA Single Factor:

Anova: Single Factor

ANOVA Hypothesis Testing Steps:

1) Given a claim, identify the Null Hypothesis H0 and the Alternative Hypothesis H1.

Null Hypothesis: H0 : μC = μR = μT

Alternative Hypothesis: H1 : μC ≠ μR ≠ μT

2) State the conclusion (reject the Null Hypothesis of fail to reject the Null Hypothesis).

Decision Rule: Reject Ho if p-value is less than the significance level

(p-value < α)

α = 5% or 0.05

From the Excel output: p-value = 0.000562

0.000562 < 0.05

Conclusion: Reject Ho, with a significance level of 5% there is statistical evidence that the readability population means of the three authors are different.

References:

Rajaretnam, T. (2016). Statistics for social sciences. Sage Publications, Inc. ISBN-13: 9789351506560

Triola, M. F. (2018). Elementary statistics (13th ed.). Pearson. ISBN-13: 978-0134462455 https://librarylogin-carolina.uagm.edu/login?url=https://search.ebscohost.com/login.aspx?direct=true&db=e000xww&AN=1214457&site=ehost-live&ebv=EB&ppid=pp_Cover

STIIModule5Class.docx

Module 5: ANOVA

The Analysis of Variance (ANOVA) is a statistical procedure to test the hypothesis that the means of three or more populations are equal. The ANOVA requires the F-distribution.

(Triola, 2018)

If there is only one explanatory variable, the ANOVA procedure is called one-way or single factor ANOVA, in which this variable is a continuous variable with three or more categories called groups.

Hypotheses:

Null Hypothesis: H0 : μ1 = μ2 = μ3 = · · · = μk

Alternative Hypothesis: H1 : μ1 ≠ μ2 ≠ μ3 ≠ · · · ≠ μk

If the null hypothesis is rejected, we conclude that with a particular significance level (α) the population means are not all equal.

The Assumptions of the ANOVA are:

· All populations are normally distributed.

· The variances among all the populations are the same.

· The random error terms are independent and normally distributed with mean equal to zero (μ = 0), and standard deviation σ.

ANOVA Hypothesis Test:

Basics of Hypothesis Testing:

1. Given a claim, identify the Null Hypothesis H0 and the Alternative Hypothesis H1.

2. State the conclusion (reject the Null Hypothesis of fail to reject the Null Hypothesis).

Null Hypothesis: H0 : μ1 = μ2 = μ3 = · · · = μk

Alternative Hypothesis: H1 : μ1 ≠ μ2 ≠ μ3 ≠ · · · ≠ μk

Decision Rule: Reject Ho if p-value is less than the significance level

(p-value < α)

Example:

Clancy	Rowling	Tolstoy
58.2	85.3	69.4
73.4	84.3	64.2
73.1	79.5	71.4
64.4	82.5	71.6
72.7	80.2	68.5
89.2	84.6	51.9
43.9	79.2	72.2
76.3	70.9	74.4
76.4	78.6	52.8
78.9	86.2	58.4
69.4	74.0	65.4
72.9	83.7	73.6

(Triola, 2018)

We will assume that the assumptions of the model are met:

· The three populations are normally distributed.

· The variances among all the populations are the same.

· The random error terms are independent and normally distributed with mean equal to zero (μ = 0), and standard deviation σ.

We are going to evaluate the ANOVA, with the support of the MS Excel program ( steps are listed in the Read Section of this module-Word and Excel documents attached).

From Excel ANOVA Single Factor:

Anova: Single Factor

ANOVA Hypothesis Testing Steps:

1) Given a claim, identify the Null Hypothesis H0 and the Alternative Hypothesis H1.

Null Hypothesis: H0 : μC = μR = μT

Alternative Hypothesis: H1 : μC ≠ μR ≠ μT

2) State the conclusion (reject the Null Hypothesis of fail to reject the Null Hypothesis).

Decision Rule: Reject Ho if p-value is less than the significance level

(p-value < α)

α = 5% or 0.05

From the Excel output: p-value = 0.000562

0.000562 < 0.05

Conclusion: Reject Ho, with a significance level of 5% there is statistical evidence that the readability population means of the three authors are different.

References:

Rajaretnam, T. (2016). Statistics for social sciences. Sage Publications, Inc. ISBN-13: 9789351506560

STIIModule5.Testdocx.docx

Module 5: Content Mastery Testing

Instructions

After studying and reviewing the concepts and practices of Module 5, answer the following:

Arkansas	California	Texas
4.8	1.5	5.6
4.9	3.7	5.8
5.0	4.0	6.6
5.4	4.5	6.9
5.4	4.9	6.9
5.4	5.1	6.9
5.6	5.3	7.1
5.6	5.4	7.3
5.6	5.4	7.5
5.9	5.5	7.6
6.0	5.6	7.7
6.1	5.6	7.7

(Triola, 2018)

Age 20-22	Age 23-26	Age 27-29
38.0	39.0	36.0
42.0	31.0	42.0
30.0	36.0	35.5
39.0	35.0	27.0
47.0	41.0	37.0
43.0	45.0	34.0
33.0	36.0	22.0
31.0	23.0	47.0
32.0	36.0	36.0
28.0	20.0	32.0

(Triola, 2018)

Be sure to review the academic expectations for your submission.

Submission Instructions:

· Submit your assignment by 11:59 PM Eastern on Sunday.

· Review the rubric to determine how your assignment will be graded.

· Your assignment will be run through Turnitin to check for plagiarism.

· Justify any assumption and show all the steps to solve the exercises.

STIIModule5Class.docx

Module 5: ANOVA

The Analysis of Variance (ANOVA) is a statistical procedure to test the hypothesis that the means of three or more populations are equal. The ANOVA requires the F-distribution.

(Triola, 2018)

If there is only one explanatory variable, the ANOVA procedure is called one-way or single factor ANOVA, in which this variable is a continuous variable with three or more categories called groups.

Hypotheses:

Null Hypothesis: H0 : μ1 = μ2 = μ3 = · · · = μk

Alternative Hypothesis: H1 : μ1 ≠ μ2 ≠ μ3 ≠ · · · ≠ μk

If the null hypothesis is rejected, we conclude that with a particular significance level (α) the population means are not all equal.

The Assumptions of the ANOVA are:

· All populations are normally distributed.

· The variances among all the populations are the same.

· The random error terms are independent and normally distributed with mean equal to zero (μ = 0), and standard deviation σ.

ANOVA Hypothesis Test:

Basics of Hypothesis Testing:

1. Given a claim, identify the Null Hypothesis H0 and the Alternative Hypothesis H1.

2. State the conclusion (reject the Null Hypothesis of fail to reject the Null Hypothesis).

Null Hypothesis: H0 : μ1 = μ2 = μ3 = · · · = μk

Alternative Hypothesis: H1 : μ1 ≠ μ2 ≠ μ3 ≠ · · · ≠ μk

Decision Rule: Reject Ho if p-value is less than the significance level

(p-value < α)

Example:

Clancy	Rowling	Tolstoy
58.2	85.3	69.4
73.4	84.3	64.2
73.1	79.5	71.4
64.4	82.5	71.6
72.7	80.2	68.5
89.2	84.6	51.9
43.9	79.2	72.2
76.3	70.9	74.4
76.4	78.6	52.8
78.9	86.2	58.4
69.4	74.0	65.4
72.9	83.7	73.6

(Triola, 2018)

We will assume that the assumptions of the model are met:

· The three populations are normally distributed.

· The variances among all the populations are the same.

· The random error terms are independent and normally distributed with mean equal to zero (μ = 0), and standard deviation σ.

We are going to evaluate the ANOVA, with the support of the MS Excel program ( steps are listed in the Read Section of this module-Word and Excel documents attached).

From Excel ANOVA Single Factor:

Anova: Single Factor

ANOVA Hypothesis Testing Steps:

1) Given a claim, identify the Null Hypothesis H0 and the Alternative Hypothesis H1.

Null Hypothesis: H0 : μC = μR = μT

Alternative Hypothesis: H1 : μC ≠ μR ≠ μT

2) State the conclusion (reject the Null Hypothesis of fail to reject the Null Hypothesis).

Decision Rule: Reject Ho if p-value is less than the significance level

(p-value < α)

α = 5% or 0.05

From the Excel output: p-value = 0.000562

0.000562 < 0.05

Conclusion: Reject Ho, with a significance level of 5% there is statistical evidence that the readability population means of the three authors are different.

References:

Rajaretnam, T. (2016). Statistics for social sciences. Sage Publications, Inc. ISBN-13: 9789351506560

STIIModule5.Testdocx.docx

Module 5: Content Mastery Testing

Instructions

After studying and reviewing the concepts and practices of Module 5, answer the following:

Arkansas	California	Texas
4.8	1.5	5.6
4.9	3.7	5.8
5.0	4.0	6.6
5.4	4.5	6.9
5.4	4.9	6.9
5.4	5.1	6.9
5.6	5.3	7.1
5.6	5.4	7.3
5.6	5.4	7.5
5.9	5.5	7.6
6.0	5.6	7.7
6.1	5.6	7.7

(Triola, 2018)

Age 20-22	Age 23-26	Age 27-29
38.0	39.0	36.0
42.0	31.0	42.0
30.0	36.0	35.5
39.0	35.0	27.0
47.0	41.0	37.0
43.0	45.0	34.0
33.0	36.0	22.0
31.0	23.0	47.0
32.0	36.0	36.0
28.0	20.0	32.0

(Triola, 2018)

Be sure to review the academic expectations for your submission.

Submission Instructions:

· Submit your assignment by 11:59 PM Eastern on Sunday.

· Review the rubric to determine how your assignment will be graded.

· Your assignment will be run through Turnitin to check for plagiarism.

· Justify any assumption and show all the steps to solve the exercises.

STIIModule5Class.docx

Module 5: ANOVA

The Analysis of Variance (ANOVA) is a statistical procedure to test the hypothesis that the means of three or more populations are equal. The ANOVA requires the F-distribution.

(Triola, 2018)

If there is only one explanatory variable, the ANOVA procedure is called one-way or single factor ANOVA, in which this variable is a continuous variable with three or more categories called groups.

Hypotheses:

Null Hypothesis: H0 : μ1 = μ2 = μ3 = · · · = μk

Alternative Hypothesis: H1 : μ1 ≠ μ2 ≠ μ3 ≠ · · · ≠ μk

If the null hypothesis is rejected, we conclude that with a particular significance level (α) the population means are not all equal.

The Assumptions of the ANOVA are:

· All populations are normally distributed.

· The variances among all the populations are the same.

· The random error terms are independent and normally distributed with mean equal to zero (μ = 0), and standard deviation σ.

ANOVA Hypothesis Test:

Basics of Hypothesis Testing:

1. Given a claim, identify the Null Hypothesis H0 and the Alternative Hypothesis H1.

2. State the conclusion (reject the Null Hypothesis of fail to reject the Null Hypothesis).

Null Hypothesis: H0 : μ1 = μ2 = μ3 = · · · = μk

Alternative Hypothesis: H1 : μ1 ≠ μ2 ≠ μ3 ≠ · · · ≠ μk

Decision Rule: Reject Ho if p-value is less than the significance level

(p-value < α)

Example:

Clancy	Rowling	Tolstoy
58.2	85.3	69.4
73.4	84.3	64.2
73.1	79.5	71.4
64.4	82.5	71.6
72.7	80.2	68.5
89.2	84.6	51.9
43.9	79.2	72.2
76.3	70.9	74.4
76.4	78.6	52.8
78.9	86.2	58.4
69.4	74.0	65.4
72.9	83.7	73.6

(Triola, 2018)

We will assume that the assumptions of the model are met:

· The three populations are normally distributed.

· The variances among all the populations are the same.

· The random error terms are independent and normally distributed with mean equal to zero (μ = 0), and standard deviation σ.

We are going to evaluate the ANOVA, with the support of the MS Excel program ( steps are listed in the Read Section of this module-Word and Excel documents attached).

From Excel ANOVA Single Factor:

Anova: Single Factor

ANOVA Hypothesis Testing Steps:

1) Given a claim, identify the Null Hypothesis H0 and the Alternative Hypothesis H1.

Null Hypothesis: H0 : μC = μR = μT

Alternative Hypothesis: H1 : μC ≠ μR ≠ μT

2) State the conclusion (reject the Null Hypothesis of fail to reject the Null Hypothesis).

Decision Rule: Reject Ho if p-value is less than the significance level

(p-value < α)

α = 5% or 0.05

From the Excel output: p-value = 0.000562

0.000562 < 0.05

Conclusion: Reject Ho, with a significance level of 5% there is statistical evidence that the readability population means of the three authors are different.

References:

Rajaretnam, T. (2016). Statistics for social sciences. Sage Publications, Inc. ISBN-13: 9789351506560

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