PLEASE ANSWER ALL OF THE QUESTION IN THE PICTURE! FOR QUESTION 18: Determine the missing values (a) - (d). FOR THE TABLE IN THE 1ST PICTURE. THE SECOND PICTURE IS THE GRAPH WHERE YOU CAN FIND ALL OF THE INFORMATION. AND PLEASE ANSWER THE FOLLOWING QUESTIONS. THANK YOU. Answer the following questions assuming we randomly select one child that participated in the study. Show your work and it is ok to leave the answer as an unreduced fraction.  19. Find the probability that the individual had a successful outcome.   20. Find the probability the individual was in the medication treatment group and raised their math scores.    21. Find the probability the individual raised their math score or was in the combined treatment group.    Show your work for problem 22 - 24 , and write your answer rounded to the nearest percent (Note: the calculations for the proportion is not any different than if you were asked to find the probability). 22. What is the probability that a child raised their math score given they received the medication management treatment?  23. What is the probability that a child raised their math score given they received the behavioral treatment?  24. What is the probability that a child raised their math score given they received the combined treatment?  25. Write a few sentences summarizing what the proportions (in questions 22-24) indicate about which of the treatments in this study are most likely to result in success (a student raising their math score)?  26. Who could find this information useful in making decisions? Explain.

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PLEASE ANSWER ALL OF THE QUESTION IN THE PICTURE!

FOR QUESTION 18: Determine the missing values (a) - (d). FOR THE TABLE IN THE 1ST PICTURE.

THE SECOND PICTURE IS THE GRAPH WHERE YOU CAN FIND ALL OF THE INFORMATION.

AND PLEASE ANSWER THE FOLLOWING QUESTIONS. THANK YOU.


Answer the following questions assuming we randomly select one child that participated in the study. Show your work and it is ok to leave the answer as an unreduced fraction. 

19. Find the probability that the individual had a successful outcome.

 

20. Find the probability the individual was in the medication treatment group and raised their math scores. 

 

21. Find the probability the individual raised their math score or was in the combined treatment group. 

 

Show your work for problem 22 - 24 , and write your answer rounded to the nearest percent (Note: the calculations for the proportion is not any different than if you were asked to find the probability).

22. What is the probability that a child raised their math score given they received the medication management treatment? 

23. What is the probability that a child raised their math score given they received the behavioral treatment? 

24. What is the probability that a child raised their math score given they received the combined treatment? 

25. Write a few sentences summarizing what the proportions (in questions 22-24) indicate about which of the treatments in this study are most likely to result in success (a student raising their math score)? 

26. Who could find this information useful in making decisions? Explain.

S6 Calculating Probabilities
18. Below is a two-way table comparing the treatment they were on and whether they increased their math score or not.
Medication Management
Behavioral Treatment
Combine Treatment
Total
Success Failure Total
93
(a)
102
(c)
43
(b)
42
(d)
136
145
144
425
Directions
To analyze this data, you will create a contingency table (2-way table) by following the steps below. Once the contingency table is designed you will answer a
few questions, and then form an informal conclusion about which drug worked best (
Transcribed Image Text:S6 Calculating Probabilities 18. Below is a two-way table comparing the treatment they were on and whether they increased their math score or not. Medication Management Behavioral Treatment Combine Treatment Total Success Failure Total 93 (a) 102 (c) 43 (b) 42 (d) 136 145 144 425 Directions To analyze this data, you will create a contingency table (2-way table) by following the steps below. Once the contingency table is designed you will answer a few questions, and then form an informal conclusion about which drug worked best (
The Chi-Squared Test
Enter Data:
Individual Observations
Name of 1st Variable:
TREATMENT
326 M
327 M
328 M
329 M
330 M
331 M
Submit
Enter observations for each variable or copy and paste from
a spreadsheet: ?
Options:
TREATMENT
Stacked Barchart
Test of Independence/Homogeneity
Show Counts on Barchart
Show Conditional Distribution
Show Expected Cell Counts
Show Residuals
Name of 2nd Variable:
Show Standardized Residuals
OUTCOME
Download Chi-Square Distribution Plot
F
F
OUTCOME
Goodness of Fit
Observed Contingency Table:
F S Total
53
92
145
42
102
144
43
93
136
138
287
425
B
с
M
Overall
Conditional Distribution (in %):
FS Total
100.0
100.0
100.0
100.0
B
с
M
Overall
63.4
36.6
29.2 70.8
31.6 68.4
32.5 67.5
Pearson's Chi-Squared Test:
Null Hypothesis
Independence (or homogeneous distributions)
0.6061
0.3939
Percent (%)
60
40
20
0
H
B
OUTCOME:
Alternative Hypothesis
☐F ☐S
Side-By-Side Barchart
Association (or non-homogeneous distributions)
TREATMENT
Test Statistic X² df
Chi-Squared Distribution with df = 2
Ho: Independence, X² = 1.86, df = 2, P-value = 0.3939
1.86
M
P-value
0.3939
Transcribed Image Text:The Chi-Squared Test Enter Data: Individual Observations Name of 1st Variable: TREATMENT 326 M 327 M 328 M 329 M 330 M 331 M Submit Enter observations for each variable or copy and paste from a spreadsheet: ? Options: TREATMENT Stacked Barchart Test of Independence/Homogeneity Show Counts on Barchart Show Conditional Distribution Show Expected Cell Counts Show Residuals Name of 2nd Variable: Show Standardized Residuals OUTCOME Download Chi-Square Distribution Plot F F OUTCOME Goodness of Fit Observed Contingency Table: F S Total 53 92 145 42 102 144 43 93 136 138 287 425 B с M Overall Conditional Distribution (in %): FS Total 100.0 100.0 100.0 100.0 B с M Overall 63.4 36.6 29.2 70.8 31.6 68.4 32.5 67.5 Pearson's Chi-Squared Test: Null Hypothesis Independence (or homogeneous distributions) 0.6061 0.3939 Percent (%) 60 40 20 0 H B OUTCOME: Alternative Hypothesis ☐F ☐S Side-By-Side Barchart Association (or non-homogeneous distributions) TREATMENT Test Statistic X² df Chi-Squared Distribution with df = 2 Ho: Independence, X² = 1.86, df = 2, P-value = 0.3939 1.86 M P-value 0.3939
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