Transcribed Questions
Section A: Essay Questions
A. In managing 250 hypertensive patients, a new drug was compared with a conventional drug. Using the information provided in the table below, determine if both treatments are similar or not.
New Drug: Controlled (60), Not Controlled (65)
Conventional Drug: Controlled (55), Not Controlled (70)
Tabulated Chi-Square value at 5% level of significance is 3.84.
B. Define Standard Error of Mean.
The BMI distribution of women attending ANC care in a health facility in Uyo is provided below:
15-18.4 (underweight): 12
18.5-24.9 (normal): 25
25-29.9 (overweight): 23
30-50 (obese): 20
I) Determine the cumulative frequency, relative frequency, and the relative cumulative frequency of the data.
II) Calculate the median and the inter-quartile range.
III) From the table, what percentage of the women have normal BMI and what percentage are above 24.9 kg/m^2?
A) Define standard error of mean.
B) State the conditions for using the following test statistics:
I) Paired t-test.
II) Pearson correlation test.
Section B: Multiple Choice Questions (MCQs)
Concerning measurement scale, which of these is false?
a) Sex is a nominal scale
b) Age is an interval scale
c) Weight is a ratio
d) Severity of pain [ordinal]
Interquartile range:
a) Its measure of location
b) It describes 50% of the values that are located at the extremes
c) Its usually values from Q1 to Q3
d) It is best used for parametric data
The characteristics of the mode include:
a) It usually exists in all set of values
b) It can be subjected to further statistical manipulation
c) It is not affected by outlier
d) The values have to be in array for you to determine the mode
Concerning data presentation:
a) Tables can only be used for continuous data
b) Frequency polygon can be used for continuous data
c) Bar chart is good for any type of data
d) Pie chart is appropriate for discrete data
Answers to All Questions
Section A: Essay
1. A. Chi-Square Test for Independence
To determine if the treatments are similar, we perform a Chi-Square test of independence.
Step 1: State the Hypotheses
Null Hypothesis (H0): There is no significant difference between the new drug and the conventional drug (treatments are similar).
Alternative Hypothesis (H1): There is a significant difference between the new drug and the conventional drug.
Step 2: Table of Observed Frequencies (O)
New Drug: 60 (Controlled), 65 (Not Controlled) | Total = 125
Conventional Drug: 55 (Controlled), 70 (Not Controlled) | Total = 125
Column Totals: 115 (Controlled), 135 (Not Controlled) | Grand Total (N) = 250
Step 3: Calculate Expected Frequencies (E)
Formula: E = (Row Total * Column Total) / Grand Total
E(New, Controlled) = (125 * 115) / 250 = 57.5
E(New, Not Controlled) = (125 * 135) / 250 = 67.5
E(Conv, Controlled) = (125 * 115) / 250 = 57.5
E(Conv, Not Controlled) = (125 * 135) / 250 = 67.5
Step 4: Calculate Chi-Square (X^2)
Formula: X^2 = Sum of ((O - E)^2 / E)
(60 - 57.5)^2 / 57.5 = 6.25 / 57.5 = 0.1087
(65 - 67.5)^2 / 67.5 = 6.25 / 67.5 = 0.0926
(55 - 57.5)^2 / 57.5 = 6.25 / 57.5 = 0.1087
(70 - 67.5)^2 / 67.5 = 6.25 / 67.5 = 0.0926
Total X^2 = 0.1087 + 0.0926 + 0.1087 + 0.0926 = 0.4026
Step 5: Conclusion
The calculated X^2 (0.4026) is much less than the tabulated X^2 (3.84). Therefore, we fail to reject the null hypothesis. We conclude that both treatments are similar in their effectiveness.
1. B. Definition of Standard Error of Mean (SEM)
The Standard Error of the Mean (SEM) is a statistical measure that quantifies the dispersion of sample means around the true population mean. It indicates how much the mean of a sample is likely to vary from the actual population mean. It is calculated by dividing the standard deviation by the square root of the sample size (n).
2. BMI Distribution Analysis
I) Frequency Table Construction
| BMI Range | Frequency (f) | Cumulative Frequency (CF) | Relative Frequency (RF) | Relative CF |
| 15 - 18.4 | 12 | 12 | 0.15 (15%) | 0.15 |
| 18.5 - 24.9 | 25 | 37 | 0.3125 (31.25%) | 0.4625 |
| 25 - 29.9 | 23 | 60 | 0.2875 (28.75%) | 0.75 |
| 30 - 50 | 20 | 80 | 0.25 (25%) | 1.00 |
| Total | 80 | 1.00 |
II) Median and Inter-quartile Range (IQR)
Median: The median position is (N/2) = 80/2 = 40th value. Looking at the CF, the 40th value falls in the 25 - 29.9 range.
IQR: This is the difference between the third quartile (Q3) and the first quartile (Q1).
Q1 position: 80/4 = 20th value (Falls in 18.5 - 24.9 range).
Q3 position: 3 * (80/4) = 60th value (Falls in 25 - 29.9 range).
III) Percentages
Percentage with Normal BMI: (25 / 80) * 100 = 31.25%
Percentage above 24.9 kg/m^2: These are the "overweight" and "obese" categories. (23 + 20) = 43. Percentage = (43 / 80) * 100 = 53.75%
3. A. Standard Error of Mean (Re-stated)
Standard Error of Mean represents the standard deviation of the sampling distribution of the mean. It serves as a measure of the precision of the sample mean as an estimate of the population mean.
3. B. Conditions for Test Statistics
I) Paired t-test:
The dependent variable must be continuous (interval or ratio scale).
The observations are paired (e.g., measurements on the same subject before and after an intervention).
The differences between the pairs should be approximately normally distributed.
The samples are obtained through random sampling.
II) Pearson Correlation Test:
Both variables must be continuous (interval or ratio scale).
There must be a linear relationship between the two variables.
There should be no significant outliers.
The variables should follow a bivariate normal distribution.
Homoscedasticity (the variance of residuals is constant across the range of variables).
Section B: Multiple Choice Questions
1. Concerning measurement scale, which of these is false?
Answer: b) Age is an interval scale.
Explanation: Age is typically a ratio scale because it has a true zero point (birth).
2. Interquartile range:
Answer: c) Its usually values from Q1 to Q3.
Explanation: The IQR represents the range of the middle 50% of the data, calculated as Q3 minus Q1.
3. The characteristics of the mode include:
Answer: c) It is not affected by outlier.
Explanation: The mode is simply the most frequent value; unlike the mean, it is resistant to extreme values.
4. Concerning data presentation:
Answer: b) Frequency polygon can be used for continuous data.
Explanation: Frequency polygons are ideal for representing the shape of a continuous distribution.