Biostatistical Methods MCQ Quiz in தமிழ் - Objective Question with Answer for Biostatistical Methods - இலவச PDF ஐப் பதிவிறக்கவும்

The correct answer is: (A) - (II), (B) - (III), (C) - (IV), (D) - (I)

Explanation:

• (A) Median (II):
  The median is the value of the middle item in a series when it is arranged in ascending or descending order of magnitude.
  Example: In the series 3, 5, 8, the median is 5.

• (B) Mean (III):
  The mean is the arithmetic average of a set of numbers, calculated by dividing the total value of all items in the series by the number of items.
  Formula: Mean = Sum of all valuesNumber of items \frac{\text{Sum of all values}}{\text{Number of items}}" id="MathJax-Element-1-Frame" role="presentation" style="position: relative;" tabindex="0">Sum of all valuesNumber of items$\frac{\text{Sum of all values}}{\text{Number of items}}$ $\frac{\text{Sum of all values}}{\text{Number of items}}$

• (C) Mode (IV):
  The mode is the most frequently occurring value in a dataset.
  Example: In the series 2, 2, 3, 5, 2, the mode is 2.

• (D) Harmonic mean (I):
  The harmonic mean is the reciprocal of the average of the reciprocals of the values in a series.
  Formula: Harmonic mean = n∑1xi \frac{n}{\sum \frac{1}{x_i}}" id="MathJax-Element-2-Frame" role="presentation" style="position: relative;" tabindex="0">n∑1xi$\frac{n}{\sum \frac{1}{x_i}}$ $\frac{n}{\sum \frac{1}{x_i}}$ , where n n" id="MathJax-Element-3-Frame" role="presentation" style="position: relative;" tabindex="0">n$n$ $n$ is the total number of items.

The correct answer is Histogram

Explanation:

• A histogram is a type of graph that provides a visual representation of the distribution of numerical data. It is particularly useful for summarizing the frequency of data points within specified ranges or intervals, known as bins.
• It allows one to quickly understand the underlying distribution of the data, including its shape, central tendency, and variability.
• Histograms are widely used in various fields such as statistics, data analysis, and quality control to identify patterns, detect anomalies, and make informed decisions.

Other Options:

• Stem and leaf plot: This is a method of displaying quantitative data in a graphical format, similar to a histogram, to show the frequency distribution. 
• Scatter diagram: Also known as a scatter plot, this type of graph is used to display the relationship between two continuous variables. Each point on the scatter plot represents an individual data point, making it useful for identifying correlations.
• Time-series plot: This graph shows data points at successive time intervals. It is used to visualize trends, cycles, and patterns over time

The correct answer is 27

Concept:

• The median is the middle value in a list of numbers sorted in ascending or descending order.
• If the list has an odd number of observations, the median is the middle number.
• If the list has an even number of observations, the median is the average of the two middle numbers.

Explanation:

• By arranging the given data in ascending order: 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52.
• Since there are 11 observations (an odd number), the median is the 6th number in this sorted list.
• The 6th number in the sorted list is 27.

The correct answer is 52.5, 2.5

Concept:

• The mean (average) and standard deviation are two fundamental measures in statistics used to describe the central tendency and dispersion of a set of data.
• The mean weight is calculated by summing all the individual weights and dividing by the number of fishes.
• The standard deviation measures the amount of variation or dispersion of a set of values.
• If a constant value is added to or subtracted from each data point, the mean will change by that constant, but the standard deviation will remain unchanged.

Explanation:

The mean that was initially calculated is 50 grams. Since the scale underreported each fish by 2.5 grams, the correct mean is

• \(\text{Adjusted Mean} = \text{Original Mean} + 2.5\text{ gm} \)
• \( \text{Adjusted Mean} = 50\text{ gm} + 2.5\text{ gm} = 52.5\text{ gm}\)

The standard deviation is a measure of the dispersion of the data and does not change with a consistent shift in all values it changes only with changes in the data's spread or distribution.

Since each fish's weight was uniformly adjusted, the standard deviation remains the same:

• \(\text{Adjusted Standard Deviation} = \text{Original Standard Deviation} \)
• Adjusted Standard Deviation = 2.5 gm

The correct answer is Mann-Whitney U test

Concept:

• Statistical tests are used to make inferences about populations based on sample data.
• They can be broadly categorized into parametric and non-parametric tests.
• Parametric tests assume that the data follows a certain distribution, usually a normal distribution.
• Non-parametric tests do not assume a specific distribution and are often used when data does not meet the assumptions required for parametric tests.

Explanation:

• ANOVA (Analysis of Variance): This is a parametric test used to compare means across multiple groups. It assumes normality and equal variances among groups.
• Paired t-test: This is a parametric test used to compare means from two related groups. Assumptions include normality of the differences between paired observations.
• Mann-Whitney U test: This is a non-parametric test used to compare differences between two independent groups. It does not assume a normal distribution and is used when data is ordinal or not normally distributed.
• Linear Regression: This is a parametric method used to model the relationship between a dependent variable and one or more independent variables. It assumes normality of residuals, linearity, and homoscedasticity.

Additional Information:

• Non-parametric Tests: Other examples of non-parametric tests include the Wilcoxon signed-rank test, Kruskal-Wallis test, and the Spearman rank correlation. These tests are more flexible as they do not require the data to fit a normal distribution.
• Use Cases: Non-parametric tests are particularly useful for small sample sizes, ordinal data, or when the data violates the assumptions necessary for parametric tests.

The correct answer is Probability of the t-test being effective in detecting significant differences in the mean annual temperatures of the two time periods

Concept:

• The p-value is a fundamental concept in statistical hypothesis testing. It helps to determine the significance of the results obtained from a test.
• In hypothesis testing, a null hypothesis (H₀) represents a default position that there is no effect or no difference. The alternative hypothesis (H_A) represents a position that there is an effect or a difference.
• The p-value measures the probability of obtaining test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is correct.
• A low p-value (typically less than 0.05) indicates that the observed data is unlikely under the null hypothesis, leading researchers to reject the null hypothesis in favor of the alternative hypothesis.

Explanation:

​1) Ratio of the mean temperatures of the two time periods tested:

• This is not correct. The p-value does not represent a ratio of the mean values.
• The p-value directly relates to the probability associated with the observed data under the null hypothesis.

2) Probability of the error of rejecting a true null hypothesis:

• This describes a Type I error (false positive).
• The common interpretation is that the p-value corresponds to the probability of observing the data, or something more extreme, assuming the null hypothesis is true.
• By convention, it helps inn deciding whether to reject the null hypothesis.

3) Probability of the error of accepting a false null hypothesis:

• This describes a Type II error (false negative) and is not directly represented by the p-value.
• Instead, it relates to the power of the test (1 - Type II error rate).

4) Probability of the t-test being effective in detecting significant differences in the mean annual temperatures of the two time periods:

• This relates to the power of the test, the likelihood that the test will detect a difference when there is one.
• While the p-value itself does not directly represent the power, achieving a p-value less than the chosen alpha level (e.g., 0.05) indicates that the test has detected a statistically significant difference.

The correct answer is 15 male 25 female

Explanation:

The null hypothesis (H0) states that there is no significant difference in the offspring sex ratio between the control group and the diet-restricted group.

Control Group Data:
Males: 22

• Females: 18
• Total offspring in the control group: 40
• Expected proportion of males in the offspring \(\frac{22}{40} = 0.55\) 
• Expected proportion of females in the offspring \(\frac{18}{40} = 0.45\)

Expected Values for Diet-Restricted Group:
Given that the diet-restricted females laid a total of 40 eggs:

• Expected number of males = 0.55 x 40 = 22 
• Expected number of females = 0.45 x 40 = 18 

The chi-square test statistic is calculated as \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \)

where O_i is the observed frequency and E_i is the expected frequency.

By calculating the chi-square value for each option to find the correct answer i.e. chi-square statistic greater than 3.84 (from the chi-square distribution table for (\(\alpha\)= 0.05) and df = 1).

Option 1: 18 males, 22 females

• \( \chi^2 = \frac{(18 - 22)^2}{22} + \frac{(22 - 18)^2}{18} \)
• \(\chi^2 = \frac{(-4)^2}{22} + \frac{4^2}{18} \)
• \(\chi^2 = \frac{16}{22} + \frac{16}{18} \)
• \(\chi^2 \approx 0.727 + 0.889 \)
• \(\chi^2 \approx 1.616 \)

Option 2: 20 males, 20 females

• \( \chi^2 = \frac{(20 - 22)^2}{22} + \frac{(20 - 18)^2}{18}\) 
• \( \chi^2 = \frac{(-2)^2}{22} + \frac{2^2}{18} \)
• \( \chi^2 = \frac{4}{22} + \frac{4}{18} \)
• \(\chi^2 \approx 0.182 + 0.222\)
• \(\chi^2 \approx 0.404\)

Option 3: 15 males, 25 females

•  \(\chi^2 = \frac{(15 - 22)^2}{22} + \frac{(25 - 18)^2}{18} \)
• \( \chi^2 = \frac{(-7)^2}{22} + \frac{7^2}{18} \)
• \( \chi^2 = \frac{49}{22} + \frac{49}{18}\)
• \(\chi^2 \approx 2.227 + 2.722\)
• \( \chi^2 \approx 4.949 \)

Option 4: 10 males, 30 females

• \( \chi^2 = \frac{(10 - 22)^2}{22} + \frac{(30 - 18)^2}{18}\)
• \( \chi^2 = \frac{(-12)^2}{22} + \frac{12^2}{18}\)
• \(\chi^2 = \frac{144}{22} + \frac{144}{18} \)
• \( \chi^2 \approx 6.545 + 8\)
• \( \chi^2 \approx 14.545 \)

Option 3 (15 males, 25 females) yields a chi-square value of approximately 4.949, which is greater than 3.84.
Option 4 (10 males, 30 females) yields a chi-square value of approximately 14.545, which is much greater than 3.84.
Therefore, the minimum deviation to conclude a significantly female-biased sex ratio is 15 males, 25 females

The correct answer is A non-parametric comparison of the two groups

Explanation:

The provided frequency distributions indicate that the data might not follow a normal distribution, as the distributions are skewed rather than bell-shaped. This observation justifies the use of a non-parametric test.

• Non-parametric tests, such as the Mann-Whitney U test or Kruskal-Wallis test, do not assume normality in the data and are suitable for comparing medians between two groups when the data distribution is non-normal.
• The height distributions differ between low and high rainfall areas, and a non-parametric test can help assess whether these differences are statistically significant.

Other Options:

• t-test for comparison of means (Option 1): A t-test assumes the data is normally distributed. Since the provided histograms indicate skewed data, this assumption is incorrect.
• Correlation analysis of rainfall and mean tree heights (Option 3): Correlation measures the strength and direction of the relationship between two variables. It is not suitable for directly testing group differences (low vs. high rainfall) in tree heights.
• Regression of tree heights on rainfall (Option 4): Regression analysis examines the relationship between a dependent variable (tree height) and an independent variable (rainfall). While regression can assess trends, it is not specifically designed to compare the distributions of two groups.

The correct answer is Needleman-Wunsch algorithm

Explanation:

The Needleman-Wunsch algorithm is a foundational computational technique used for the global alignment of two sequences, such as protein or DNA sequences. The algorithm was first described by Saul B. Needleman and Christian D. Wunsch in 1970.

Global Alignment:-

• Global alignment aims to align sequences from start to finish, considering the entire length of each sequence. This is particularly useful when the sequences are of comparable length and are expected to have regions of similarity throughout their entire lengths.

Needleman-Wunsch Algorithm:-

• The Needleman-Wunsch algorithm is widely used in bioinformatics for aligning entire protein or DNA sequences, making it fundamental for genome comparisons, phylogenetic studies, and structural prediction.
• While the algorithm is computationally intensive (with a time complexity of (O(n m)), where n and m are the lengths of the sequences), it guarantees finding the optimal global alignment by considering all possible alignments.

Comparison with Smith-Waterman Algorithm:-

• Needleman-Wunsch: Global alignment, which means it aligns the entire sequences.
• Smith-Waterman: Local alignment, which focuses on finding local regions of similarity within larger sequences.

Other Methods Mentioned:-

• Chou Fasman method: This is used for predicting secondary structures of proteins, not for sequence alignment.
• Garnier Osguthorpe Robson (GOR) method: This is also used for predicting protein secondary structures, not for sequence alignment.

Conclusion:

In summary, the Needleman-Wunsch algorithm is the method used for global alignment of two protein sequences due to its ability to consider the entire length of the sequences to find the optimal alignment.

Explanation:

A 95% confidence interval (CI) for the estimated mean indicates the range within which we expect the true population mean to lie, based on our sample data, with a 95% level of confidence.

They are limits between which, in the long run, 95% of observations fall.

• This statement is incorrect. The confidence interval pertains to the population mean, not individual observations.

They are limits within which, the sample mean falls with probability 0.95.

• This statement is incorrect. The sample mean is a fixed value once calculated; the confidence interval refers to where we expect the population mean to be located, not where the sample mean lies.

They are a way of measuring the variability of a set of observations.

• This statement is incorrect. Variability is typically measured by statistics like range, variance, or standard deviation, not by confidence intervals.

Thus, the statement that correctly describes the confidence interval is that it measures the precision of the estimate of the mean.