This document contains Michael Parent's answers to 4 questions on a final exam for an ELMP course. The questions assess understanding of key statistical concepts like correlation coefficients, hypotheses testing, and interpreting statistical significance. Michael provides well-explained responses to each question, demonstrating knowledge of how to state hypotheses, perform calculations, make decisions, and draw conclusions based on statistical analyses.

1. Michael Parent
ELMP 6005
Final Exam
September 30, 2007
Red=Suggested improvements from QE Review session with Dr. Walsh
QUESTION 1
Explain what a correlation coefficient is
a) Strength of Relationship:
There is a strong relationship between the two variables. Why? Use the Rule of
Thumb chart. What does the Pearson r mean?
b) Direction of the Relationship:
There is a negative relationship between the two variables. An increase in one
variable causes a decrease in another variable. What variables? Use the
language.
c) Interpretation of the level of significance:
The reported level of significance is .002. This means that if I were to conduct a
similar analysis, 2 out of 1000 times the direction of the line would flip-flop.
Therefore, this correlation can be confidently applied. Anything greater than .05,
then it cannot be confidently applied.
d) Interpretation of r2:
R2=.737881, which translates as 74% of the variation in averages can be
explained by variations in hours of television viewed nightly.
What can you say about the policy and practice after this analysis?
QUESTION 2
1. Research Problem:
The purpose of this study is to determine if new Math GEPA preparation activities are
making a difference in an elementary school. Is there a difference, on average, between
the mean scores of Group 1 and the mean score of Group 2?
2. State the Null Hypothesis:
Ho: μ1 - μ2 = 0 (No Difference between two samples means) Do not try to use
language if you can’t make perfect sense of it. Use only the equation!
3. State the Alternate Hypothesis:
H1 : μ1 - μ2 ≠ 0 (there is a difference between the two sample means)
4. Decision Rule:
I’ll reject Ho at .05 level of significance (95% level of confidence) if t is equal or more
positive than the critical value of t (.05) or if t is equal or more negative than 1.97,given
df = 200+. I will also reject Ho if the calculated significance of a 2-tailed test is equal or
less than .05. Include the type of t-test being used. (single, paired, etc.) The p value

2. rule is NECESSARY, not the t value.
5. Calculations:
Degrees of freedom df are calculated as the sum of samples minus 2 (df = n1 + n2 – 2).
According to the Table, at .05 level of significance and 200+ degrees of freedom, the
critical value of t, or tcrit = ±1.97.
The t value calculated from the data is 2.51. The p value is .025. This value doesn’t fall
between my 95% probability or .05 significance level. If it doesn’t, it falls in the critical
area.
6. Decision:
There is enough evidence for suspecting the null hypothesis. I reject the Null
Hypothesis Ho at .05 level of significance because the calculated t value of 2.251 is
more negative or more positive than the t critical value of 1.97. The calculated t value of
2.251 falls in the critical or rejection area. The probability of the t value 2.251 beyond
tcrit = 1.97 equals .05. It happens about 5 times in 100; this is a rare outcome signifying
that something special is probably happening in the underlying populations.
Also, I reject Ho at .05 level of significance because the 2-tailed level of significance
of .025 is less than the .05 level of significance, it’s below the threshold .05 that I require
as stated in the decision rule.
7. Interpretative Statement:
I found strong evidence to suggest that there is significant difference between the mean
scores of students in Group 1 and students in Group 2.
8. PPR:
Based on the data given and the evidence of statistical significance between the mean
scores of the two paired groups, it is apparent that the GEPA preparation activities are
working well. We might consider this pilot project a success and expand it to the control
group. I recommend that we continue to monitor the mean scores of students and
possible expand the preparation program to the lower grades.
QUESTION 3
1. Research Problem:
The purpose of this study is to determine how the students’ mean scores on the
NDGKS compare with the mean scores of all students in the state who took the
NDGKS.
2. State the Null Hypothesis:
Ho: μ = 280
3. State the Alternate Hypothesis:

3. H1 : μ ≠ 280
4. Decision Rule:
To be 95% confident that the change in the one sample mean score didn’t happen by
random chance, I’ll reject Ho at .05 level of significance (95% level of confidence) if t
equals or is greater than the critical value of t 1.98, given df = n – 1: 109 – 1 = 100. I
will also reject Ho if the calculated significance of a 2-tailed test is equal or less than .
05. Include the type of t-test being used. (single, paired, etc.) The p value rule is
NECESSARY, not the t value.
5. Calculations:
The SPSS has provided the same t value = -15.827, and a two-tailed level of
significance (p) of .000. According to the Table at .05 level of significance and 100
degrees of freedom, the critical value of t is 1.98 .
This value doesn’t fall between my 95% probability. It falls in the critical area.
6. Decision:
Based on the data given, I reject the Null Hypothesis Ho at .05 level of significance
because the calculated t value is greater than the t critical value. I also reject Ho at .05
level of significance because the 2-tailed level of significance is less than the .05 level of
significance that it’s required as stated in the decision rule.
7. Interpretative Statement:
There is strong evidence to reject the null hypothesis. There is a significant difference
between the students’ mean score at this high school with the mean score statewide.
The 2-tailed level of significance suggests that a sample mean score of 280 is unlikely
to occur. The 2-tailed level of significance of .000 is below the threshold .05.
The difference between the hypothesized mean of 280 and the one-sample mean of
235.40 is significant; it didn’t happen by random chance, something happened to the
underlying population.
8. PPR:
The data provided is evidence to suggest that the students at this high school are
performing significantly poorer that other students in North Dakota. The principal might
consider implementing an improvement plan to improve students’ mean scores. I
suggest that the school continue to monitor students’ mean scores during any reform
measure.
QUESTION 4

4. 1. Research Problem:
Is there a difference between the doctor’s SCORES? Is there a significant difference
between those who took the pil and those who did not? The purpose of this study is to
determine if there is a difference between doctors’ abilities to focus during surgery after
having ingested the “Alert” pill.
2. State the Null Hypothesis:
Ho: μD = 0 (no difference between each pair of mean scores in two repeated samples)
The null hypothesis states that the doctors did not show any statistical difference in
ability to focus after having ingested the “Alert” pill. Do not try to use language if you
can’t make perfect sense of it. Use only the equation!
3. State the Alternate Hypothesis:
H1: μD ≠ 0
The alternate hypothesis states that the doctors did show a statistical difference in
ability to focus after having ingested the “Alert” pill. Do not try to use language if you
can’t make perfect sense of it. Use only the equation!
4. Decision Rule:
The reported mean is 5.02. The calculated t value provided in the SPSS output is 1.577
I’ll reject Ho at .05 level of significance (95% level of confidence) if t equals or is greater
than 2.01, the critical value of t, given df = n – 1 =50 – 1 = 49.
I will also reject Ho if the calculated significance of a 2-tailed test is equal or less than .
05. Include the type of t-test being used. (single, paired, etc.) The p value rule is
NECESSARY, not the t value.
5. Calculations:
The SPSS has provided me with the t value of 1.577 and the two-tailed level of
significance (p) of .121. This t value 1.577 does fall between 95% probability. It falls
within the range of acceptance.
6. Decision:
I fail to reject the Null Hypothesis Ho at .05 level of significance because the calculated t
value of 1.577 is not greater than the t critical value of 2.01. The calculated t value falls
in the normal area.
Also, I fail to reject Ho at .05 level of significance because the 2-tailed level of
significance of .121 is greater than the .05 level of significance that I require as stated in
the decision rule, and commonly accepted as the threshold of significance for social
science research.
7. Interpretative Statement:
I found evidence that there is no difference between the paired means of physician’s
who took the “Alert” pill pre-test and the pill’s post-test.

5. 8. PPR:
The company should consider further testing to determine if the “Alert” pill can have a
significant impact on surgeons. The company might reconsider its claim that their pill
improves focus. Because this is a medical test, the company should consider
establishing a level of significance at the .01 level on future studies.