Question

This question uses the "Morphine" data set, available on Canvas. As usual, you have been randomly assigned one of 10 versions of this data. Your version is 10. On Canvas there is a .pdf document that describes the study this data is based on. You can find this under Additional Class Stuff / Data. Please read this before you start analyzing data. You will be asked some questions that require you to interpret results in light of the study's hypotheses, and these questions will not make sense if you haven't read the study description. 1. Use Analyze / Fit Y by X to tell JMP what your predictor and response variables are. Then select Means / ANOVA from the drop down menu. Report the values of the following statistics: a. SSG (JMP calls this "Treatment sum of squares"): b. SSE: c. MSG: d. MSE: e. The test statistic that tests H0:μM_S = μM_M = μS_S = μS_M = μM_M new: 2. Based on the result of this analysis, we can say (select all that apply): (you have five attempts) We reject H0 and conclude that all group means are equal We reject H0 and conclude that all group means differ from one another We reject H0 and conclude that at least one group mean differs from the rest The test statistic is bigger than we'd expect it to be if all group means were equal The test statistic is smaller than we'd expect it to be if all group means were equal The test statistic is about what we'd expect it to be if all group means were equal The average between group variation is large relative to the average within group variation The average between group variation is small relative to the average within group variation The average between group variation is not very different from the average within group variation 3. From the "Oneway Analysis" drop down menu, select Compare Means / Each Pair, Student's t. a. Select the pairwise comparisons that are significant, using a comparison-wise error rate of α=0.05 (you have three attempts) μM_S - μM_M μM_S - μS_S μM_S - μS_M μM_S - μM_M(new) μM_M - μS_S μM_M - μS_M μM_M - μM_M(new) μS_S - μS_M μS_S - μM_M(new) μS_M - μM_M(new) b. If we want to use a Bonferroni correction to control the family-wise error rate at α=0.05, what comparison-wise error rate should we use? c. Select the pairwise comparisons that are significant, using a Bonferroni correction to control the family-wise error rate atα=0.05 (you have three attempts) μM_S - μM_M μM_S - μS_S μM_S - μS_M μM_S - μM_M(new) μM_M - μS_S μM_M - μS_M μM_M - μM_M(new) μS_S - μS_M μS_S - μM_M(new) μS_M - μM_M(new) There is another popular multiple testing adjustment specially designed for cases when you want to compare each group mean to each other group means; i.e. when you want to run "all pairwise comparisons". This adjustment method is called Tukey's HSD (or Tukey-Kramer HSD). It is the option in JMP just underneath the "Each Pair, Student's t" when you select "Compare Means" from the drop down menu. Tukey's HSD lowers the comparison-wise error rate by making p-values larger, rather than by making the significance level smaller. d. Run this function in JMP, and identify which pairwise comparisons are significantly different, using the 0.05 significance level. (you have three attempts) μM_S - μM_M μM_S - μS_S μM_S - μS_M μM_S - μM_M(new) μM_M - μS_S μM_M - μS_M μM_M - μM_M(new) μS_S - μS_M μS_S - μM_M(new) μS_M - μM_M(new) 4. Now you'll interpret the results in light of the study's hypotheses. a. The researchers predicted that morphine tolerant rats receiving saline when expecting morphine would be extra sensitive to pain. The scientific hypothesis was that the expectation of morphine would cause rats to compensate by increasing their pain sensitivity. What statistical result would constitute evidence for this hypothesis? (you have three attempts) Observing a significant difference in the means of M_M(new) and S_S Observing a significant difference in the means of M_S and S_M Observing a significant difference in the means of M_S and S_S Observing a non-significant difference in the means of M_M and S_S Observing a significant difference in the means of M_M(new) and S_M Observing a non-significant difference in the means of M_M and M_S Observing a non-significant difference in the means of M_M(new) and S_M Observing a non-significant difference in the means of M_M(new) and S_S b. The researchers also predicted that placing morphine tolerant rats in a new environment before administering more morphine would "undo" the effect of morphine tolerance. What two statistical results would be consistent with this hypothesis? (There are two correct answers; select them both) (you have five attempts) Observing a significant difference in the means of S_S and M_M(new) Observing a non-significant difference in the means of S_S and M_M(new) Observing a significant difference in the means of M_M and M_M(new) Observing a significant difference in the means of S_M and M_M(new) Observing a non-significant difference in the means of M_M and M_M(new) Observing a non-significant difference in the means of S_M and M_M(new) c. Which of the following are appropriate interpretations for the comparison of the S_S and M_M groups (There are four correct answers; select them all)? (you have five attempts) The observed data are not inconsistent with the hypothesis that population mean tolerances for the M_M and S_S treatments are equal We FTR H0: μM_M = μS_S The difference in means between morphine tolerant rats who received more morphine and rats who never received any morphine is no bigger than what we'd expect to see by chance alone, if their population means were equal. We have significant evidence that μM_M and μS_S are not equal. We have significant evidence that μM_M and μS_S are equal. Morphine tolerant rats who received more morphine did not have a significantly different average pain tolerance than rats who never received any morphine. The difference in means between morphine tolerant rats who received more morphine and rats who never received any morphine is smaller than what we'd expect to see by chance alone, if their population means were not equal. The observed data are inconsistent with the hypothesis that population mean tolerances for the M_M and S_S treatments are not equal.

Morphine data set:

latency,Treatment

3,M_S

5,M_S

1,M_S

8,M_S

1,M_S

1,M_S

4,M_S

9,M_S

2,M_M

12,M_M

13,M_M

6,M_M

10,M_M

7,M_M

11,M_M

19,M_M

14,S_S

6,S_S

12,S_S

4,S_S

19,S_S

3,S_S

9,S_S

21,S_S

29,S_M

20,S_M

36,S_M

21,S_M

25,S_M

18,S_M

26,S_M

17,S_M

24,M_M(new)

26,M_M(new)

40,M_M(new)

32,M_M(new)

20,M_M(new)

33,M_M(new)

27,M_M(new)

30,M_M(new)

Answer 1

One-way ANOVA: Latency versus C2

Source DF SS MS F P
C2 4 3232.7 808.2 26.20 0.000
Error 35 1079.6 30.8
Total 39 4312.2

1. a. SSG=3232.7 b. SSE=1079.6 c. MSG=808.2 d. MSE=30.8 e. F=26.20

2. Since p-value<0.05, We reject H0 and conclue that at least one group mean differs from the rest.

3.

Grouping Information Using Fisher Method

C2 N Mean Grouping
M_M(new) 8 27.938 A
S_M 8 24.087 A
S_S 8 10.588 B
M_M 8 9.825 B C
M_S 8 4.575 C

Means that do not share a letter are significantly different.

Fisher 95% Individual Confidence Intervals
All Pairwise Comparisons among Levels of C2

Simultaneous confidence level = 72.67%

C2 = M_M subtracted from:

C2 Lower Center Upper ---------+---------+---------+---------+
M_M(new) 12.475 18.113 23.750 (---*---)
M_S -10.887 -5.250 0.387 (---*--)
S_M 8.625 14.262 19.900 (---*--)
S_S -4.875 0.763 6.400 (---*--)
---------+---------+---------+---------+
-15 0 15 30

C2 = M_M(new) subtracted from:

C2 Lower Center Upper ---------+---------+---------+---------+
M_S -29.000 -23.363 -17.725 (--*---)
S_M -9.487 -3.850 1.787 (--*---)
S_S -22.987 -17.350 -11.713 (--*---)
---------+---------+---------+---------+
-15 0 15 30

C2 = M_S subtracted from:

C2 Lower Center Upper ---------+---------+---------+---------+
S_M 13.875 19.512 25.150 (---*---)
S_S 0.375 6.013 11.650 (---*---)
---------+---------+---------+---------+
-15 0 15 30

C2 = S_M subtracted from:

C2 Lower Center Upper ---------+---------+---------+---------+
S_S -19.137 -13.500 -7.863 (---*---)
---------+---------+---------+---------+
-15 0 15 30

c.

Level N Mean
M_M 8 9.825  
M_M(new) 8 27.938  
M_S 8 4.575  
S_M 8 24.087  
S_S 8 10.588  

d.

Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of C2

Individual confidence level = 99.32%

C2 = M_M subtracted from:

C2 Lower Center Upper +---------+---------+---------+---------
M_M(new) 10.121 18.113 26.104 (----*----)
M_S -13.242 -5.250 2.742 (----*----)
S_M 6.271 14.262 22.254 (----*----)
S_S -7.229 0.763 8.754 (----*----)
+---------+---------+---------+---------
-32 -16 0 16

C2 = M_M(new) subtracted from:

C2 Lower Center Upper +---------+---------+---------+---------
M_S -31.354 -23.363 -15.371 (----*----)
S_M -11.842 -3.850 4.142 (----*----)
S_S -25.342 -17.350 -9.358 (----*----)
+---------+---------+---------+---------
-32 -16 0 16

C2 = M_S subtracted from:

C2 Lower Center Upper +---------+---------+---------+---------
S_M 11.521 19.512 27.504 (----*----)
S_S -1.979 6.013 14.004 (----*----)
+---------+---------+---------+---------
-32 -16 0 16

C2 = S_M subtracted from:

C2 Lower Center Upper +---------+---------+---------+---------
S_S -21.492 -13.500 -5.508 (----*----)
+---------+---------+---------+---------
-32 -16 0 16

***Please give thumbs up to my answer***