7. Randomized block design A company would like to choose among three leading Internet service providers (ISPs)—ISP A, ISP B, and ISP C—for the broadband connection in its downtown office. As an analyst at the company, you have been asked to test whether there is a difference in download speeds among the three ISPs. You perform an experiment in which each measurement consists of downloading a 100-megabyte file from a particular website and measuring the average download speed in megabits per second (Mbps). Because download times vary by time of day, you conduct the experiment at five different times during a particular day: 10:00 AM, 1:00 PM, 4:00 PM, 7:00 PM, and 10:00 PM. For each time of day, three measurements are made, one for each of the three ISPs. In the experiment above, the response variable is_______ , the factor of interest is theISP chosen________ , and the blocks are the _____ . The experiment hasa complete block design. The results of the study are presented in the data table below. All figures are in units of Mbps. Complete the table by using the dropdowns to fill in the sample means for the first treatment (ISP A), the first block (10:00 AM), and the overall set of data. Treatments Row or Block Totals Block Sample Means ISP A ISP B ISP C 10:00 AM 3.41 3.42 3.49 10.32 3.44 1:00 PM 3.29 3.49 3.25 10.03 3.34 Blocks 4:00 PM 3.70 3.74 3.63 11.07 3.69 7:00 PM 3.68 4.08 3.88 11.64 3.88 10:00 PM 3.93 4.19 3.74 11.86 3.95 Column or Treatment Totals 18.01 18.92 17.99 54.92 Treatment Sample Means 3.60 3.78 3.60 Overall Sample Mean 3.66 The three treatments in the experiment define three populations of interest. You will use analysis of variance (ANOVA) to test the hypothesis that the three population means are equal. The results of your analysis will be presented in the following ANOVA table. Go through the steps listed below to fill in the rest of the table. ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments Blocks 0.2132 Error Total 14 1. The contribution to the total sum of squares (SST) from the observation assigned to the first block and the first treatment is ______ . Given that the contributions due to the other observations sum to 1.0147, select the appropriate value for the total sum of squares in the ANOVA table. 2. Select the appropriate value for the sum of squares due to treatments (SSTR) in the ANOVA table. 3. Select the appropriate value for the sum of squares due to blocks (SSBL) in the ANOVA table. 4. Select the appropriate value for the sum of squares due to error (SSE) in the ANOVA table. 5. Select the appropriate degrees of freedom for the SSTR, the SSBL, and the SSE in the ANOVA table. 6. Select the appropriate values for the mean square due to treatments, the mean square due to error, and the F test statistic in the ANOVA table. 7. Use the Distributions tool above to find the p-value of the F test statistic. Select this p-value in the ANOVA table. At a significance level of α = 0.05, test the null hypothesis that the population means for all treatments are equal. The null hypothesis is _______ . You _____ conclude that choosing a different ISP would affect one’s download speed.

7. Randomized block design A company would like to choose among three leading Internet service providers (ISPs)—ISP A, ISP B, and ISP C—for the broadband connection in its downtown office. As an analyst at the company, you have been asked to test whether there is a difference in download speeds among the three ISPs. You perform an experiment in which each measurement consists of downloading a 100-megabyte file from a particular website and measuring the average download speed in megabits per second (Mbps). Because download times vary by time of day, you conduct the experiment at five different times during a particular day: 10:00 AM, 1:00 PM, 4:00 PM, 7:00 PM, and 10:00 PM. For each time of day, three measurements are made, one for each of the three ISPs. In the experiment above, the response variable is_ , the factor of interest is theISP chosen , and the blocks are the _ . The experiment hasa complete block design. The results of the study are presented in the data table below. All figures are in units of Mbps. Complete the table by using the dropdowns to fill in the sample means for the first treatment (ISP A), the first block (10:00 AM), and the overall set of data. Treatments Row or Block Totals Block Sample Means ISP A ISP B ISP C 10:00 AM 3.41 3.42 3.49 10.32 3.44 1:00 PM 3.29 3.49 3.25 10.03 3.34 Blocks 4:00 PM 3.70 3.74 3.63 11.07 3.69 7:00 PM 3.68 4.08 3.88 11.64 3.88 10:00 PM 3.93 4.19 3.74 11.86 3.95 Column or Treatment Totals 18.01 18.92 17.99 54.92 Treatment Sample Means 3.60 3.78 3.60 Overall Sample Mean 3.66 The three treatments in the experiment define three populations of interest. You will use analysis of variance (ANOVA) to test the hypothesis that the three population means are equal. The results of your analysis will be presented in the following ANOVA table. Go through the steps listed below to fill in the rest of the table. ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments Blocks 0.2132 Error Total 14 1. The contribution to the total sum of squares (SST) from the observation assigned to the first block and the first treatment is . Given that the contributions due to the other observations sum to 1.0147, select the appropriate value for the total sum of squares in the ANOVA table. 2. Select the appropriate value for the sum of squares due to treatments (SSTR) in the ANOVA table. 3. Select the appropriate value for the sum of squares due to blocks (SSBL) in the ANOVA table. 4. Select the appropriate value for the sum of squares due to error (SSE) in the ANOVA table. 5. Select the appropriate degrees of freedom for the SSTR, the SSBL, and the SSE in the ANOVA table. 6. Select the appropriate values for the mean square due to treatments, the mean square due to error, and the F test statistic in the ANOVA table. 7. Use the Distributions tool above to find the p-value of the F test statistic. Select this p-value in the ANOVA table. At a significance level of α = 0.05, test the null hypothesis that the population means for all treatments are equal. The null hypothesis is _ . You ___ conclude that choosing a different ISP would affect one’s download speed.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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7. Randomized block design

A company would like to choose among three leading Internet service providers (ISPs)—ISP A, ISP B, and ISP C—for the broadband connection in its downtown office. As an analyst at the company, you have been asked to test whether there is a difference in download speeds among the three ISPs.

You perform an experiment in which each measurement consists of downloading a 100-megabyte file from a particular website and measuring the average download speed in megabits per second (Mbps). Because download times vary by time of day, you conduct the experiment at five different times during a particular day: 10:00 AM, 1:00 PM, 4:00 PM, 7:00 PM, and 10:00 PM. For each time of day, three measurements are made, one for each of the three ISPs.

In the experiment above, the response variable is_______ , the factor of interest is theISP chosen________ , and the blocks are the _____ .

The experiment hasa complete block design.

The results of the study are presented in the data table below. All figures are in units of Mbps. Complete the table by using the dropdowns to fill in the sample means for the first treatment (ISP A), the first block (10:00 AM), and the overall set of data.

		Treatments	Row or Block Totals	Block Sample Means
ISP A	ISP B	ISP C
	10:00 AM	3.41	3.42	3.49	10.32	3.44
	1:00 PM	3.29	3.49	3.25	10.03	3.34
Blocks	4:00 PM	3.70	3.74	3.63	11.07	3.69
	7:00 PM	3.68	4.08	3.88	11.64	3.88
	10:00 PM	3.93	4.19	3.74	11.86	3.95
	Column or Treatment Totals	18.01	18.92	17.99	54.92
	Treatment Sample Means	3.60	3.78	3.60	Overall Sample Mean	3.66

The three treatments in the experiment define three populations of interest. You will use analysis of variance (ANOVA) to test the hypothesis that the three population means are equal. The results of your analysis will be presented in the following ANOVA table. Go through the steps listed below to fill in the rest of the table.

ANOVA Table
Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p-value
Treatments
Blocks			0.2132
Error
Total		14

1. The contribution to the total sum of squares (SST) from the observation assigned to the first block and the first treatment is ______ . Given that the contributions due to the other observations sum to 1.0147, select the appropriate value for the total sum of squares in the ANOVA table.

2. Select the appropriate value for the sum of squares due to treatments (SSTR) in the ANOVA table.

3. Select the appropriate value for the sum of squares due to blocks (SSBL) in the ANOVA table.

4. Select the appropriate value for the sum of squares due to error (SSE) in the ANOVA table.

5. Select the appropriate degrees of freedom for the SSTR, the SSBL, and the SSE in the ANOVA table.

6. Select the appropriate values for the mean square due to treatments, the mean square due to error, and the F test statistic in the ANOVA table.

7. Use the Distributions tool above to find the p-value of the F test statistic. Select this p-value in the ANOVA table.

At a significance level of α = 0.05, test the null hypothesis that the population means for all treatments are equal. The null hypothesis is _______ . You _____ conclude that choosing a different ISP would affect one’s download speed.

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