Problem 1: Health psychologists conducted a study to examine the relationship between the number of close friends an elderly person (65 years and above) has and their general self-reported health condition. General health was measured on a scale from O to 100 with higher numbers indicating better health. Use the data from this study to answer questions 1 through 4.
Participant Number of Friends Health Score
Participant | X=number of close friends | Y= Health Score | |||
A | 2 | 45 | |||
B | 1 | 30 | |||
C | 3 | 65 | |||
D | 0 | 30 | |||
E | 4 | 58 | |||
F | 2 | 50 | |||
G | 4 | 70 | |||
H | 1 | 24 | |||
I | 0 | 25 | |||
J | 3 | 67 | |||
Participant | X=number of close friends | X^2 | Y= Health Score | y^2 | xy | |
A | 2 | 4 | 45 | 2025 | 90 | |
B | 1 | 1 | 30 | 900 | 30 | |
C | 3 | 9 | 65 | 4225 | 195 | |
D | 0 | 0 | 30 | 900 | 0 | |
E | 4 | 16 | 58 | 3364 | 232 | |
F | 2 | 4 | 50 | 2500 | 100 | |
G | 4 | 16 | 70 | 4900 | 280 | |
H | 1 | 1 | 24 | 576 | 24 | |
I | 0 | 0 | 25 | 625 | 0 | |
J | 3 | 9 | 67 | 4489 | 201 | |
sum of variables | 20 | 60 | 464 | 24504 | 1152 |
What are the research and null hypotheses?
Hr: there are relationship between close friends and health score
Ho: there are no relationship between close friends and health score
Does the scatterplot indicate a linear relationship between the two variables? if so, state whether the relationship is positive or negative. Explain the reason /for your answer.
No. its doesn’t indicate linear relationship
If the scatterplot in problem 1 indicated a linear relationship between the two variables, compute the correlation coefficient. Use the computational formula.
N = 10
2240/ 29133.07399
0.076888556
r calculated is 0.08
Use Table R to find the critical value for the correlation coefficient at p=.05 for a two tailed test.
df= 8
Critical value of r from the r table = 0.576
Compare the r (correlation coefficient) to the critical value of r from the table and make a decision about the null hypothesis.
Circle one:
Reject null Fail to reject null
Significant? No
If yes, “reject Ho/yes significant” then compute the coefficient of determination. 0.00591185
Interpret the findings of this study. Be sure to compute the means and standard deviations of each variable and include them in the interpretation.
Since calculated r is less than critical value of r, then we reject the null hypothesis and say that there is no significant relationship between the close friends and health score.
The standard deviation is too away from the mean and the wide deviation indicates very weak association between the two variable,
Problem 2: Developmental psychologists are concerned with the weekly number of hours young children watch violent television and its relationship to aggressive play behavior. They conducted a study and collected the following data. Use this data to answer problems 6 through 11.
Child Initials | X
Hours watching TV |
Y Aggressive Behaviors | |||
MJ | 12 | 16 | |||
BL | 10 | 30 | |||
CJ | 3 | 5 | |||
SV | 5 | 7 | |||
EK | 18 | 42 | |||
FN | 7 | 10 | |||
GG | 22 | 51 |
What are the research and null hypotheses for this relationship?
Hr: to investigate if there is relationship between hours watching TV and aggressive behavior
Ho: to investigate if there is no relationship between hours watching TV and aggressive behavior
Compute the correlation coefficient. Use the computational formula.
N = 7
0.000781093
Use Table R to find the critical value for the correlation coefficient at p= .05 for a two tailed test.
df= 5
Critical value =0.754
Compare the r (correlation coefficient) to the critical value of r from the table and make a decision about the null hypothesis.
Circle one:
Reject null Fail to reject null
Significant? Yes No
If yes, “reject Ho” then compute the coefficient of determination.
6.1E-07
Interpret the findings of this study. Be sure to compute the means and standard deviations of each variable and include them in the interpretation.
Since the r calculated is less than critical value, we accept null hypothesis and conclude that there is no relationship between hours of watching and aggressive behavior. The standard deviation is wide deviated away from the mean of both variables which likewise indicate that we have very weak relationship between the two variables.
Problem 3- Educational researchers predict there will be a correlation between SAT scores and GPA’s. To test this prediction they gathered a sample of the following data and did a correlational analysis.
The following data were collected from 8 college students. Use this data to answer questions 10 through 15.
Student | x-SAT Score | Y= GPA | |||
1 | 540 | 3.7 | |||
2 | 430 | 2.6 | |||
3 | 320 | 1.9 | |||
4 | 690 | 3.8 | |||
5 | 550 | 2.8 | |||
6 | 530 | 2.5 | |||
7 | 510 | 2.0 | |||
8 | 370 | 1.5 | |||
Child Initials | X | x^2 | Y=GPA | y^2 | xy |
STAT | |||||
MJ | 540 | 291600 | 3.7 | 13.69 | 1998 |
BL | 430 | 184900 | 2.6 | 6.76 | 1118 |
CJ | 320 | 102400 | 1.9 | 3.61 | 608 |
SV | 690 | 476100 | 3.8 | 14.44 | 2622 |
EK | 550 | 302500 | 2.8 | 7.84 | 1540 |
FN | 530 | 280900 | 2.5 | 6.25 | 1325 |
GG | 510 | 260100 | 2 | 4 | 1020 |
370 | 136900 | 1.5 | 2.25 | 555 | |
sum | 3940 | 2035400 | 20.8 | 58.84 | 10786 |
mean | 492.5 | 2.6 | |||
standard deviation | 105.4243127 | 0.694585 |
What are the research and null hypotheses for this relationship?
Hr: there is relationship between Stat score and GPA
Ho: there is no relationship between stat score and GPA
Compute the correlation coefficient. Use the computational formula.
= 1.97579E-07
Use Table R to find the critical value for the correlation coefficient at p= .05 for a two tailed test.
df= 6
Critical value = 0.707
Compare the r (correlation coefficient) to the critical value of r from the table and make a decision about the null hypothesis.
Circle one:
Reject null Fail to reject null
Significant? Yes No
If yes, “reject Ho” then compute the coefficient of determination.
3.90E-14
Interpret the findings of this study. Be sure to compute the means and standard deviations of each variable and include them in the interpretation.
Since the r calculated is less than critical value, we accept null hypothesis and conclude that there is no relationship between hours of watching and aggressive behavior. The standard deviation is too away from the mean of both variables which also indicates very weak relationship between the two value.
Problem 4 A group of students conducted a study examining the number of absences in a ciass and finai class grades. When the study was done, they analyzed the data and reported that poor grades were caused because students missed too many days of school. Use this information to answer questions 16 and 17.
Describe the pattern of the relationship between number of absences and final class grades. In other words, was the relationship positive or negative? Explain.
The relationship was positive. Since there was significant effect from both the relationship. Absenteeism has significantly influenced the students’ performance. Having potential mediating to influence other factors being taken into consideration and where else this is true for all cases of absenteeism. The fact is that the more days a student misses’ school, the more student weakens on performance irrespective of whether a female or male, poverty status, ethnicity, race and disability. Hence missing schools have hugely affected the results or performance of the student.
Review the conclusion reached by the students. Identify the error and explain why this conclusion was incorrect.
Type I error. The students committed this error due to the fact they are not giving exact evidence about the relationship and significance of the relationship between the students and missing to attend to school. Like the students did not bother the situation where null hypothesis is null and must be accepted.