Refer to the following partial anova results from excel (some information is missing).

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Question 1 Refer to the following partial ANOVA results from Excel (some information is missing). Picture  The number of observations in the entire sample is Answer 20 19 22 1 points  Question 2 Here is an Excel ANOVA table that summarizes the results of an experiment to assess the effects of ambient noise level and plant location on worker productivity. The test used a= 0.05. Picture  Is the effect of plant location significant at a= .05? Answer Yes No Need more information to say 1 points  Question 3 Using one-factor ANOVA with 30 observations we find at a= .05 that we cannot reject the null hypothesis of equal means. We increase the sample size from 30 observations to 60 observations and obtain the same value for the sample F test statistic. Which is correct? Answer We might now be able to reject the null hypothesis We surely must reject H0 for 60 observations C.We cannot reject H0 since we obtained the same F-value It is impossible to get the same F-value for n = 60 as for n = 30 1 points  Question 4 What are the degrees of freedom for Tukey's test statistic for a one-factor ANOVA with with n1= 6, n2 = 6, n3 = 6? Answer 3, 6 6, 3 6, 15 3, 15 1 points  Question 5 The Internal Revenue Service wishes to study the time required to process tax returns in three regional centers. A random sample of three tax returns is chosen from each of three centers. The time (in days) required to process each return is recorded as shown below Picture  The test to use to compare the means would require Answer Three-factor ANOVA One-factor ANOVA Two-factor ANOVA without replication Two-factor ANOVA with replication 1 points  Question 6 Here is an Excel ANOVA table that summarizes the results of an experiment to assess the effects of ambient noise level and plant location on worker productivity. The test used a= 0.05. Picture  Is the effect of noise level significant at a= .01? Answer Yes No Need more information to say 1 points  Question 7 Refer to the following MegaStat output (some information is missing). The sample size was n = 65 in a one-factor ANOVA. Picture  At a = .05, which is the critical value of the test statistic for a two-tailed test for a significant difference in means that are to be compared simultaneously? Note: this question requires a Tukey table. Answer 2.81 2.54 2.33 1.96 1 points  Question 8 Refer to the following partial ANOVA results from Excel (some information is missing). The response variable was Y = maximum amount of water pumped from wells (gallons per minute). Picture  The MS for interaction is Answer 7.25 8.17 8.37 9.28 1 points  Question 9 Refer to the following partial ANOVA results from Excel (some information is missing).  ANOVA Table Picture  The number of observations in the original sample was Answer 59 60 58 54 1 points  Question 10 To compare the cost of three shipping methods, a random sample of four shipments is taken for each of three firms. The average cost per shipment is shown below. Picture  In a one-factor ANOVA, degrees of freedom for the within-groups sum of squares will be Answer 11 3 9 2 1 points  Question 11 Which is the Excel function to find the right-tail p-value for Fcalc = 4.52 with n1 = 15, n2 = 12? Answer =FDIST(4.52, 15, 12) =FINV(4.52, 14, 11) =FDIST(4.52, 14, 11)

=FINV(4.52, 15, 12)

Refer to the following partial ANOVA results from Excel (some information is missing).

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Refer to the following partial anova results from excel (some information is missing).

Question 1

Refer to the following partial ANOVA results from Excel (some information is missing).

Picture The number of observations in the entire sample is

Answer

20

19

221 points 

Question 2

Here is an Excel ANOVA table that summarizes the results of an experiment to assess the effects of ambient noise level and plant location on worker productivity. The test used a= 0.05.

Picture Is the effect of plant location significant at a= .05?

Answer

Yes

No

Need more information to say1 points 

Question 3

Using one-factor ANOVA with 30 observations we find at a= .05 that we cannot reject the null hypothesis of equal means. We increase the sample size from 30 observations to 60 observations and obtain the same value for the sample F test statistic. Which is correct?
Answer

We might now be able to reject the null hypothesis

We surely must reject H0 for 60 observations

C.We cannot reject H0 since we obtained the same F-value

It is impossible to get the same F-value for n = 60 as for n = 301 points 

Question 4

What are the degrees of freedom for Tukey’s test statistic for a one-factor ANOVA with with n1= 6, n2 = 6, n3 = 6?
Answer

3, 6

6, 3

6, 15

3, 151 points 

Question 5

The Internal Revenue Service wishes to study the time required to process tax returns in three regional centers. A random sample of three tax returns is chosen from each of three centers. The time (in days) required to process each return is recorded as shown below

Picture The test to use to compare the means would require

Answer

Three-factor ANOVA

One-factor ANOVA

Two-factor ANOVA without replication

Two-factor ANOVA with replication1 points 

Question 6

Here is an Excel ANOVA table that summarizes the results of an experiment to assess the effects of ambient noise level and plant location on worker productivity. The test used a= 0.05.

Picture Is the effect of noise level significant at a= .01?

Answer

Yes

No

Need more information to say1 points 

Question 7

Refer to the following MegaStat output (some information is missing). The sample size was n = 65 in a one-factor ANOVA.

Picture At a = .05, which is the critical value of the test statistic for a two-tailed test for a significant difference in means that are to be compared simultaneously? Note: this question requires a Tukey table.

Answer

2.81

2.54

2.33

1.961 points 

Question 8

Refer to the following partial ANOVA results from Excel (some information is missing). The response variable was Y = maximum amount of water pumped from wells (gallons per minute).

Picture The MS for interaction is

Answer

7.25

8.17

8.37

9.281 points 

Question 9

Refer to the following partial ANOVA results from Excel (some information is missing). 
ANOVA Table

Picture The number of observations in the original sample was

Answer

59

60

58

541 points 

Question 10

To compare the cost of three shipping methods, a random sample of four shipments is taken for each of three firms. The average cost per shipment is shown below.

Picture In a one-factor ANOVA, degrees of freedom for the within-groups sum of squares will be

Answer

11

3

9

21 points 

Question 11

Which is the Excel function to find the right-tail p-value for Fcalc = 4.52 with n1 = 15, n2 = 12?
Answer

=FDIST(4.52, 15, 12)

=FINV(4.52, 14, 11)

=FDIST(4.52, 14, 11)

=FINV(4.52, 15, 12)

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Question “Refer to the following partial one-factor ANOVA results from excel (some information is missing. And you…”

  1. Refer to the following partial one-factor ANOVA results from excel (some information is missing. And you need to work it out in

    the questions below.) Now, the F statistic is equal to:

Source of Variation

Sum of Squares

Degrees of Freedom

Mean Square

F statistic

Between Groups

210.2788

Within Groups

1483

74.15

Total

2113.833

  1. Referring to the table in question 1. The sum of squares for
    between groups variation is:
  1. 129.99
  2. 630.83
  3. 1233.4
  4. We cannot tell from given information
  1. Referring to the table in question 1, the Degree of Freedom for between groups variation is (hint: use the results from Q2 and the

    mean square of between groups):

Using above ANOVA result we get,

i) The F statistic is.

F = MS Between and MS Within = 209.2788 / 2.84

Answer: d) 2.84

ii) The sum squares of between group variation (SS Between) is,

SS Between = SS SS Within = 2113.833 – 1483 = 630.833

Answer: b) 630.83

iii). Degrees of freedom for variation between groups ( DF Between)

DF Between = MS Between/ MS between = 603.83 / 209.2788 = 2.999

DF Between = “strongTag4$”

Answer: a) 3

DF Within = MS Within = 1483 / 20

DF Total = Df Between & DF Within = 2 + 20 = 23

Conclusion

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