Understand By The Term Reliability

1. Explain in a short paragraph what you understand by the term “reliability”. (2)

a. Reliability means that a rater or instrument is consistent and reliable in their measurements from trial to trial. When a measurement is reliable it can be reproduced over and over again and still consistently be free from error. However, it is impossible for a measurement to be completely free of error. Therefore, it is important that the measurement error is calculated. This is the difference between the observed and true values. A reliable measure will have a small measurement error.

2. Explain why the intraclass correlation coefficient is preferred to Pearson’s correlation coefficient as a measure of reliability? (2)

a. The Intraclass Correlation (ICC)

is used when quantitative measurements are organized into groups. The ICC describes how strongly values in the same group resemble each other. In ICC the data are organized using a group mean and standard deviation. In the Pearson correlation, each variable is organized by its own mean and standard deviation. This grouped organization for the ICC is the most appropriate because all measurements are of the same quantity. For example, in a paired data set where each "pair" represents a single measurement made for each of the two groups (weighing each person in a group) rather than two different measurements taken for a single group (measuring height and weight for each person), the ICC is a more appropriate measure of association than Pearson's correlation.
3. You wish to determine whether there is a difference in the “Timed up and go (mean)” defined as the LABEL on your SPSS spreadsheet, which is named the “Pre_TUG” VARIABLE between fallers versus non-fallers. (5)
a) Which test is the appropriate one to conduct? Why?
a. Independent T Test
b. The Independent T test is used when two independent groups are compared. The two groups in this question are fallers and non-fallers. Each of these groups is composed of independent sets of subjects.
b) What do the results from the Levene’s test tell you?
a. Levene’s is not sig (p=.063) so equal variance is assumed
b. Sig (2tailed)=.527 which is not sig and is in equal variance row
c. Conclusion: w/ equal variances assumed, there is no difference between fallers and non fallers for timed up and go mean as p=.527 which is greater than 0.05
c) What is the 95th % Confidence Interval?
a. (-1.34 to 2.619)
b. We are 95% confident the difference between the means is between -1.3 and 2.619
4. You wish to determine if there is a difference in trial 1 versus trial 2 for TREs. In the data set, they are written like this; Pre_TRE1 and Pre_TRE2.(5)
a) What is the Mean for Pre_TRE1?
a. Mean= 3.53
b) What is the mean difference between Pre_TRE1 and Pre_TRE2?
a. Mean difference= .06154

c) Which test is the appropriate one to conduct?
a. Paired T test
b. The Paired T test is used when there is a definite relationship b/w each pair of data points or the measurements are taken from the same subject.
d) Perform the test. What is your conclusion?
a. There is not a difference between the Pre_TRE1 and Pre_TRE2 because p=.807 which is not sig
5. Explain, using an example, what you understand by the Known Groups Method of construct validation? (2)
a.

Known groups validity is when the validity of a measurement is determined by the degree to which the instrument can detect different scores between the groups. This test is used when the groups are already known to vary on the variables being tested and determines how sensitive the instruments detection of difference is.

Example: Known groups validity may be used to validate an instrument that is created to measure the amount of body image anxiety that exists between groups. It would be expected that a group of people diagnosed with anorexia would score higher on body image anxiety than a group of healthy individuals. Therefore, if the measurements gathered by the instrument followed this pattern we would conclude that the measure was valid. A finding of no difference between the group diagnosed with anorexia and the healthy group would lead to the conclusion that the measurement was not valid because the instrument was not able to detect this difference.

6. In order to compute the 95% confidence interval for Functional reach test (mean) or [Pre_FRT], you need to know 2 pieces of information about the Pre_FRT score in this sample. (3)

a. What are these 2 pieces of information?

i. Sample mean and SEM(standard error of the

mean)

b. Compute the 95% confidence interval (lower and upper boundaries) for Pre_FRT in this sample. Write out in words what this means.
i. Confidence Interval: ±0.88 (9.81 to 11.57)
ii. We are 95% confident that the population mean falls within this range of values
7. To test whether a variable is normally distributed we use the _ 1 sample K-S test (3)
a) Conduct this test to evaluate whether Pre_ABC6 conforms to the normal distribution.
b) What do you conclude?
a. Pre_ABC6 is normally distributed because after running the 1 sample K-S test Asymp. Sig (2 tailed) p=.614 which is greater than .05. Therefore you can use parametric statistics
8. What is a Z score?
a. A z score shows how many standard deviations away from the mean a score is.
b. z = (X - μ) / σ
c. Where z: z-score, X: value of the element, μ: population mean, and σ: standard deviation.
9. When is it most useful to use z scores in a data set? Use an example to explain your answer. (3)
a. It is most useful to use a z score when the data is normally distributed. The z score can then be used to help throw out outliers in a data set. If a data point is 2-3 standard deviations away from the mean then that data point can be considered an outlier.
b. Example: If I was looking at a normal distribution of NPTE scores I could tell how a score related to the sample mean by its z score. If the Z score was 2-3 I would know that this score may be an outlier.

10. In a short paragraph, referring to the normal curve, explain why the Z score for the 95% confidence interval is 1.96. (2)
a. 1.96 is the approximate value of the 97.5 percentile point of the normal distribution. This means that 95% of the area under a normal curve lies within roughly 1.96 standard deviations of the mean. This also means that in order to score in above the 95th percentile your score would have to be over 1.96 standard deviations above the mean.
11. If we wish to compute the 95% confidence interval for a variable, will the confidence interval be wider or narrower than for the 90% interval? (2)
a. 95% will be wider than a 90% interval
12. Using an example, explain what is meant by the term “internal validity” of a study? (2)
a. Internal Validity refers to how well a study is conducted. How well a study is conducted depends on how well confounding variables are controlled. If there are too many confounding variables then you cannot say that there was only one possible cause for the outcome and therefore there would be low internal validity. However, if the study did a good job of controlling confounding variables by following strict standards then the study would be considered to have high internal validity. Some examples of confounding variables are a lack of structure and obedience to the protocol, the instruments not being used in the same manner consistently, selection bias of the participants and many more.
i. An example of a study that does not have good internal validity would be if the participants were selected based on characteristics that were desirable to the investigators. This would suggest selection bias and would not allow the study to have good internal validity.
ii. Another example of a study that does not have good internal validity would be if every time the person was tested a different protocol were used. This would introduce many new confounding variables and you would not be able to decipher which one cause the outcome of the study.

13. The test that one should use to compare Timed up and go (mean) or Pre_TUG score among 3 independent groups of physical activity levels is known as a __ANOVA________. (3)
a) Perform this test to determine if TUG score differs among the 3 activity level groups. What is the p value?
a. P value= .155
b) What can you conclude?
a. There is not a significant trend toward a difference b/w the groups because p=.155 which is greater than .05