## Answers to Practice Questions - Simple Biostatistics (Third Edition)

**ANSWERS TO PRACTICE QUESTIONS AT ADVANCE LEVEL FOR USMLE**

Page 187 SIMPLE BIOSTATISTICS (Third Edition)

BY A. INDRAYAN AND L.S. SATYANARAYANA,

ACADEMA PUBLISHERS, DELHI

Ans1. (e) In an RCT, allocation of subjects to control and test groups is done in a random manner which helps in equal distribution of unaccounted factors or confounding factors where as in choices a, b and d randomization and matching were advised together which is not possible. Option is c is also inappropriate because matching cannot be done in RCT.

Ans2. (d) Side effects and blinding (options a and e) considerations are not related to external validity. A statistically significant result may not be clinically significant. Thus whatever P-value is obtained from an investigation, it may not have any clinical relevance. Generalisation does not depend on whether the result is significant or not.

Ans3. (b) As compared to 10 years ago the life span for the patients suffering from disease A could be longer today due to advances in treatment strategies. This reduces death and increases the cases with disease--thus also the prevalence of disease. Options a, d and e are wrong because increased prevalence cannot be explained by odds ratio, or relative risk or attributable risk. If option c was correct, prevalence would have been low today.

Ans4. (b) Relative risk (RR) is a measure of strength of association. This is basically ratio of incidence among exposed to unexposed. Even if the ratio is reverse the strength of association does not change. RR 20.0 inversely is 1/20 which is 0.05 and this has same strength of association. Interchanging rows and columns does not affect th strength of association so long as properly interpreted--thus option a is incorrect. Option d is incorrect becauseboth odds ratio and relative risk have same kind of interpretation. A causal relationship has to satisfy several other conditions apart from significant association.

Ans5. (a) Because in thes data we have qualitative characteristics, viz. disease and gender, the test to compare various grades of disease between males and females is chi-square. These data are represented in the form of contingency table whereas all other options deal with quantitative data tests that require a measure such as mean.

Ans6. (e) For a skewed distribution to the right, mean lies to the left of median.Option a is incorrect as there is no such a relationship between mode and median. Skewed distribution cannot be corrected by Student’s t test, and the correlation coefficient can never be more than 1 in absolute value. Thus options c and d are incorrect. The option e is partially correct as chi-square is a nonparametric test.

Ans7. (c) S.E. of mean is obtained by dividing S.D. by the square root of sample size. Obvious S.E. will be smaller. Options a and b are not always correct for the measures to fall at the mid-point. Option d is incorrect as the S.E. does depends on the number of observations or sample size. Option e is not valid for mean, it is true for median.

Ans8. (a) P=0.06 is interpreted as not significant as the general threshhold value is 0.05. Options b is incorrect as the sample size is reasonably large. Option c is incorrect as P–value is not the same as how many more patients respond. Option d is incorrect as unequal numbers do not invalidate any trial. Before-after measurements in this case are helpful for studying the change.

Ans9. (b) Median divides the number of observations into two equal halves. Mean may be more than mode or less than mode depending on the skewness. The S.D. is square root of variance and not the the converse so the option c is incorrect. Options d and e arecorrect sometimes but not always.

Ans10. (d) There is no direct relationship as the deciles divide the distribution into 10 parts and tertiles into 3 parts. Option b is incorrect because calculation of positive predictivity requires both sensitivity as well as disease prevalence. In a Gaussian distribution, the probability below Mean – 2S.D. is nearly 2.5% and not 5%. Regression is not a method measure for strength of relationship but it is only for the nature of relationship. Thus option e is incorrect. Option d is correct since test of hypothesis can be done by seeing if the null value is in the CI or not.

Ans11. (b) Blood groups have no fixed graded order so these are not ordinal data. Option c is incorrect as ANOVA is not applicable for data on proportions. Both the options d and e are incorrect as is no transformation of data is involved. S.E. of mean is incorrect in this case as there are no quantities.

Ans12. (d) The options a, b and c are incorrect, respectively, as the correlation is not used for relationship between population and sample means, chi-square test is used for proportions and not for means and S.D. has the same units as of the values themselves. Mean, median and mode in normal distribution are equal. Thus option e is also incorrect.

Ans13. (a) Options b is incorrect as multiplicationve rule is for mutually exclusive and not for inclusive events. The probability of disease given complaints and of complaints given disease are different. Blinding does not help in generalization and bias is not a random error as it is a systematic error. Option a is correct as reference range for some medical parameters (e.g., blood pressure) is based on health consequence and not ±2SD limits.

Ans14. (b) Option a is incorrect because P-value is the actual value of probability of Type I error and level of significance is the predetermined threshold value such as 0.05. These are not same. Cluster sampling generally requires larger sample as compared to SRS. Standard error is not the square root of SD. For large samples parametric methods are always preferred so the options c, d and e are incorrect.

Ans15. (e) Options a, b, c and d are correct statements because the distributions generally are symmetric or skewed, 30th percentile is not 30 percent score, randomization is a preferred method in clinical trials, and some positive value of P is always obtained. P is the probability of error and this can never be zero in actual-life setup. Option e is a wrong statement because normal range is calculated for healthy people and not for the general population.