08. Statistics Flashcards
What is the outcome result of a parametric test?
P-value
What type of test would be appropriate to compare two groups of patients for incidence of hypotension?
The incidence of hypotension, ie it did/did not occur, would be categorical (or qualitative) data thus requiring a non-parametric test,
eg Chi squared test, McNemars
Which is more powerful, parametric or non-parametric?
Parametric
They are more powerful, ie they are more likely to show a difference that really exists.
How are confidence intervals derived?
Are they proportional to sample size?
CI’s are derived from the Standard Error.
Standard Error is inversely proportional to the sample size.
- What do confidence intervals indicate?
- What is the meaning if confidence intervals overlap?
- How do rare complications affect the confidence interval?
Confidence Intervals indicate the range within which the true value will plausibly lie.
If the confidence intervals of 2 groups overlap, the true value may be identical (whereas the mean of each group may be very different).
For rare complications, the CI’s will again give a range in which the true incidence will plausibly lie.
Can the ASA status of a group of patients be best described by median and standard deviation?
ASA is ordinal data (a subtype of qualitative or categorical data).
Median and Standard Deviation are reserved for quantiative data that has a normal distribution.
Can ASA status be compared by chi-squared test?
Yes because it is non-parametric
Does SD increase as sample size increase?
No
Standard deviation is calculated as?
The square root of the variance
If p = 0.05 for a comparison of treatments: T/F
A. there is a 95% chance that there is no difference between the treatments.
False.
A p-value of 0.05 would be considered significant; thus there is evidence that the treatments are not equal.
If p = 0.05 for a comparison of treatments: T/F
The null hypothesis is incorrect.
False.
As the p value decreases the null hypothesis will be increasingly questioned, but at no point can we say it is definitely false.
If p = 0.05 for a comparison of treatments: T/F
The chance of a false negative is 5%
The chance of a false negative is 1 - Power.
If p = 0.05 for a comparison of treatments: T/F
We can conclude that one treatment is more effective
False.
Although this result is statistically significant, we have no information as to whether it is clinically significant.
If p=0.05 for a comparison of two treatments: T/F
The data we have observed would only occur 5% of the time or less if the treatments were equally effective
TRUE
How can you show the spread in non-normal data?
Using interquartile range.
Non-normal data can be divided into 4 Quartiles which can be used to show the range of spread (whereas SD is used for normal data).
The range of values that include the 2nd and 3rd quartiles represents the half of the data lying closest to the central value - this is the Interquartile Range.
In a study comparing two airway devices, the incidence of sore throat is 20% with the standard device and 15% with the new device.
How is relative risk reduction calculated?
Relative Risk Reduction =
Absolute Risk Reduction / Control Incidence.
In this case 5/20 = 25%.
In a study comparing two airway devices, the incidence of sore throat is 20% with the standard device and 15% with the new device.
What is the absolute risk reduction?
Absolute risk reduction is the Incidence in the treatment group minus the incidence in the control group.
(20 - 15) = a reduction by 5%.
What is the power of a study?
The Power of a study represents the chance of detecting a difference that does exist.
ie if the Power was 80% then you would have an 80% chance of detecting that difference (or a 20% chance of missing it and producing a false -ve.
T/F: The power of a study depends upon the statistical test used to compare the data?
True.
What type of data are t-tests used for?
Parametric and normally distributed data.
Comparing two sets of urine output volumes. T/F?
The standard deviations should be similar in order to do a statistical comparison.
False
Comparing two sets of urine output volumes. T/F?
Can be done using parametric and non-parametric tests
True.
Urine output is likely to be normally distributed, therefore a parametric or non-parametric test could be used.
The following measures of central tendency could correctly describe the matched type of data:
A. Variance and normally distributed data
B. Standard error of the mean and normally distributed data
C. Mode and positively skewed data
D. Median and negatively skewed data
E. Mode and categorical data
A. False. Variance is not a measure of central tendency.
B. False. SEM is not a measure of central tendency.
C. False. Median is used for non-normally distributed data.
D. True. Median is used for non-normally distributed data.
E. True. It tells you the group with the most pices of data in.
Regarding the correlation coefficient: T/F
A positive value implies that a rise in one variable causes a rise in the other.
False.
It implies there is an association between them, but not that one rise in one causes the other to rise
Regarding the correlation coefficient: T/F
Can be calculated for parametric and non-parametric data.
TRUE
The following are non-parametric tests
Select true or false for each of the following statements.
A. The Chi-squared test B. ANOVA C. The Mann-Whitney U test D. The Student's t-test E. The Wilcoxon signed rank sum test
Chi-Squared
Mann-Whitney U Test
Wilcoxon signed rank sum test
How is SEM calculated?
Standard error of the mean is the Standard Deviation divided by the square root of (n - 1)
Variance is calculated how?
Variance is the sum of the differences divided by the degrees of freedom.
The power of a study will be greater or lower if you are trying to detect a small difference?
Detecting a small difference is harder, so if looking at a given number of patients, the power will be lower.
The following are parametric tests
Select true or false for each of the following statements.
A. Spearman's rank correlation B. Fisher's exact test C. ANOVA D. Kruskal-Wallis E. Forest plots
ANOVA
