Exercise 27
5. In the legend beneath Figure 2[->0], the authors give an equation indicating that systolic blood pressure is SBP = 43.2 + 0.17x. If the value of x is postnatal age of 30 hours, what is the value for Ŷ or SBP for neonates ≤1,000 grams? Show your calculations. Y = a + bx43.2 + 0.17(30) = 48.3

The SBP for neonates ≤1,000 grams at 30 hours is 48.3.
6. In the legend beneath Figure 2[->1], the authors give an equation indicating that systolic blood pressure is SBP = 50.3 + 0.12x. If the value of x is postnatal age of 30 hours, what is the value for Ŷ or SBP for neonates 1,001–1,500 grams? Show your calculations. Y = a + bx 50.3 + 0.12(30) = 53.9

The SBP for neonates between 1,001-1,500 grams at 30 hours is 53.9. 7. Compare the SBP readings you found in Questions 5 and 6. Explain the difference in these two readings. The y-value or systolic blood pressure increases with gestational weight. The neonates at a gestational weight of ≤1,000 grams will have a lower systolic blood pressure than those neonates between 1,001-1,500 grams. 8. In the legend beneath Figure 2[->2], the authors give an equation indicating that diastolic blood pressure is DBP = 25.8 + 0.13x. If the value of x is postnatal age of 30 hours, what is the value for Ŷ for neonates ≤ 1,000 grams? Show your calculations. Y = a + bx 25.8 + 0.13(30) = 29.7

The DBP for neonates ≤1,000 grams at 30 hours is 53.9.
9. In the legend beneath Figure 3[->3], the authors give an equation indicating that diastolic blood pressure is DBP = 30.4 + 0.11x. If the value of x is postnatal age of 30 hours, what is the value for Ŷ for neonates 1,001–1,500 grams? Show your calculations. Y = a + bx 30.4 + 0.11(30) = 33.7

The DBP for neonates between 1,001-1,500 grams at 30 hours is 53.9. 10. In the legend beneath Figure 3[->4], the authors give an equation indicating that diastolic blood pressure is DBP = 30.4 + 0.11x. How different is the DBP when the value of x is postnatal age of...

...PEARSON’S PRODUCT-MOMENT CORRELATION COEFFICIENT
ANSWERS TO EXERCISE 23
Question 1
The r value for the relationship between Hamstring strength index 60o and the Shuttle run test is -0.149. This r value shows a weak correlation between the two variables, as it is less than the 0.3 threshold for significance. Therefore, the r value is not significant.
Question 2
Between r=1.00 and r=-1.00, there is no difference in terms of strength. Both values are on the extreme ends of the spectrum and signify the maximum significance within the r value scale. A value of 1.00, whether negative or positive, shows that the two variables have a perfect linear relationship, and as such, the independent variable can be used to accurately predict the value of the dependent variable. The only difference is that the negative value signifies that a rise in one variable causes the corresponding variable to drop while the positive value signifies that the rise in one variable causes the corresponding variable to increase in value as well. But strength wise, they are similar.
Question 3
The relationship between the hamstring strength index 60o and shuttle run test index is a negative one. This is signified by the negative nature of the r value (-0.498). A negative relationship occurs when a rise in one variable causes the corresponding variable to decrease in value.
Question 4
This research study had the primary objective of measuring the relationship between muscle strength and...

...of 1000 flights and proportions of three routes in the sample. He divides them into different sub-groups such as satisfaction, refreshments and departure time and then selects proportionally to highlight specific subgroup within the population. The reasons why Mr Kwok used this sampling method are that the cost per observation in the survey may be reduced and it also enables to increase the accuracy at a given cost.
TABLE 1: Data Summaries of Three Routes
Route 1
Route 2
Route 3
Normal(88.532,5.07943)
Normal(97.1033,5.04488)
Normal(107.15,5.15367)
Summary Statistics
Mean
88.532
Std Dev
5.0794269
Std Err Mean
0.2271589
Upper 95% Mean
88.978306
Lower 95% Mean
88.085694
N
500
Sum
44266
Summary Statistics
Mean
97.103333
Std Dev
5.0448811
Std Err Mean
0.2912663
Upper 95% Mean
97.676525
Lower 95% Mean
96.530142
N
300
Sum
29131
Summary Statistics
Mean
107.15
Std Dev
5.1536687
Std Err Mean
0.3644194
Upper 95% Mean
107.86862
Lower 95% Mean
106.43138
N
200
Sum
21430
From the table above, the total number of passengers for route 1 is 44,266, route 2 is 29,131 and route 3 is 21,430 and the total numbers of passengers for 3 routes are 94,827.
Although route 1 has the highest number of passengers and flights but it has the lowest means of passengers among the 3 routes. From...

...alpha level of .05 is a good compromise between the likelihoods of making Type I and Type II errors.
An experiment where you may want a lower alpha level (e.g., 0.01) would be for example a drug study for coagulation times. You would want to be certain the drug is effective, therefore a lower alpha level would be prudent. Within this same drug study, you would accept a higher alpha level when looking for drug side-effects. (University of Texas-Houston Health Science Center , 2013)
References
Price, I. (2000). What Alpha Level? In I. Price, Inferential Statistics (p. Chapter 5). New England: University of New England.
University of Texas-Houston Health Science Center . (2013). Hypothesis Testing . Retrieved March 21, 2013, from Biostatistics for the Clinician : http://www.uth.tmc.edu/uth_orgs/educ_dev/oser/L2_2.HTM
4DQ1
How would you explain the analysis of variance, assuming that your audience has not had a statistics class before?
When examining the differences between two or more groups, you can use the analysis of variance which is known as ANOVA. This is a statistical technique that is used to compare the means or averages of more than two groups. There are three uses of ANOVA which are the one-way, the two-way and N-way Multivariate ANOVA. (Solutions, 2013) The determining factor when to use one of the “ways” is dependent upon how many “treatments” are used in the study. We use the term treatment because ANOVA originated...

...EXERCISE 36
6. Can ANOVA be used to test proposed relationships or predicted correlations between variables in a single group? Provide rationale for your answer.
ANOVA cannot be used to test proposed relationships or predicted correlations between variables in a single group because it is designed to test for correlations and interactions amongst groups, i.e. in the test group of patients with OA you are testing the correlations between those who do not use GI and PMR and those that do. Although all participants suffered from OA they were separated into two groups for comparison.
7. If a study had a result of F(2, 147) = 4.56, p=0.003, how many groups were in the study, and what was the sample size?
Number of groups = K
Group degrees of freedom= K-1+=2 = K=2+1=3
K=3 which means the number of groups = 3
Sample size = N
Error degrees of freedom = N-K = 147=N-3 = N=147+3 = 150
N=150 which means the sample size is 150
8. The researchers state that the sample of their study was 28 women with a diagnosis of OA, and that 18 were randomly assigned to the intervention group and 10 were randomly assigned to the control group. Discuss the studies strengths and/or weaknesses in this statement.
One of the biggest weaknesses in this study is the number of participants. There are only 18 women participating. With a larger group of participants you can obtain more credible and concrete results. One of the strengths is that there is in fact a control group, so...

...Exercise 36 Answers
1. Since the F value is significant, based on the p-value of 0.005 which is less than 0.05 which is sufficient to reject the null hypothesis. This suggests that there is a difference in the control and treatment groups.
2. Since the p- value is less than 0.05 and therefor the null hypothesis can be rejected. This presents that the mean, difficulty and mobility scores, must be different
3. The result was statistically significant with a probability score of p < 0.001.
4. Yes, because 0.001 < 0.01 and would still be significant.
5. The 0.04 > 0.01 would indicate that there is no statistical significance and except the null and conclude that there is no difference between the groups.
6. NOVA cannot be used to test proposed relationships or predicted correlations between variables in a single group. This is because ANOVA is tests relationships within various groups and among the groups.
7. The study had 149 subjects and 2 groups
8. The strength of the study where that they include a control group to test the dependent variable to examine the differences over time. The weakness of the study comes from the low number of subjects in the study. More subjects would have made the study more creditable.
9. The study results indicated a significant improvement in the pain scores of women with OA who received the treatment of guided imagery (F(1, 26) =4.406, p = 0.046). Thus, the null hypothesis was rejected. But in my opinion...

...1. Were the groups in this study independent or dependent? Provide a rationale for your answer.
Independently due to the data being collected independently
2. t = −3.15 describes the difference between women and men for what variable in this study? Is this value significant? Provide a rationale for your answer.
t = -3.15 describes mental health in this study. This is significant because mental health plays a big role in the outcome of MI patients.
3. Is t = −1.99 significant? Provide a rationale for your answer. Discuss the meaning of this result in this study.
t = -1.99 describes health functioning. This is very significant. It is used to describe decline from illness or disease. It could solely explain change in quality of life post MI.
4. Examine the t ratios in Table VI. Which t ratio indicates the largest difference between the males and females post MI in this study? Is this t ratio significant? Provide a rationale for your answer.
Mental health revealed the largest difference between the males and females post MI in this study. This is significant because it truly reveals that men and women cope differently. I also believe this value would be inaccurate due to men being less likely to admit that they have mental health issues.
5. Consider t = −2.50 and t = −2.54. Which t ratio has the smaller p value? Provide a rationale for your answer. What does this result mean?
-2.54 has the smaller p value at 0.007. It shows the health outcome between...

...Oct. 10, 2012
1. Multiple and True/False Questions (10 points) Please circle the right answer for the questions below. Each question is assigned 2.5 points.
1. The sample mean of population 1 is smaller than that of population 2. If we are interested in testing whether the mean of population 1 is significantly smaller than the mean of population 2, the a. null hypothesis should state µ1 − µ2 < 0 b. null hypothesis should state µ1 − µ2 ≤ 0 c. alternative hypothesis should state µ1 − µ2 < 0 d. alternative hypothesis should state µ1 − µ2 > 0 ANSWER: c
2.
A Type I error is committed when a. a true alternative hypothesis is not accepted b. a true null hypothesis is rejected c. the critical value is greater than the value of the test statistic d. sample data contradict the null hypothesis ANSWER: b In determining an interval estimate of a population mean when σ is unknown, we use a t distribution with a. n − 1 degrees of freedom
3.
b. c. d. ANSWER: 4.
n degrees of freedom
n − 1 degrees of freedom n degrees of freedom c
The purpose of statistical inference is to provide information about the a. sample based upon information contained in the population b. population based upon information contained in the sample c. population based upon information contained in the population d. mean of the sample based upon the mean of the population ANSWER: b
FMBA SQA Final Exam
Prof. Kihoon Kim Oct. 10, 2012
2. (10 points) A researcher is...

...Trajico, Maria Liticia D.
BSEd III-A2
REFLECTION
The first thing that puffs in my mind when I heard the word STATISTIC is that it was a very hard subject because it is another branch of mathematics that will make my head or brain bleed of thinking of how I will handle it. I have learned that statistic is a branch of mathematics concerned with the study of information that is expressed in numbers, for example information about the number of times something happens. As I examined on what the statement says, the phrase “number of times something happens” really caught my attention because my subconscious says “here we go again the non-stop solving, analyzing of problems” and I was right. This course of basic statistic has provided me with the analytical skills to crunch numerical data and to make inference from it. At first I thought that I will be alright all along with this subject but it seems that just some part of it maybe it is because I don’t pay much of my attention to it but I have learned many things. I have learned my lesson.
During our every session in this subject before having our midterm examination I really had hard and bad times in coping up with this subject. When we have our very first quiz I thought that I would fail it but it did not happen but after that, my next quizzes I have taken I failed. I was always feeling down when in every quiz I failed because even though I don’t like this...