An investigator wants to study the association between maternal iron supplements intake (Present / Absent) and birth weight (in gms) of new born babies. She collects this data from 1000 pregnant women and their newborns. How will you help the investigator in testing his hypothesis?

a. Chi Square Test
b. Independent t Test
c. Mc Nemar Test
d. ANOVA










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Ans: b

Step 1: Identify type of data to be compared. Here, we will compare baby birth weight(in gms) which is an Interval/ Ratio (continous) type of data

Step 2: Identify the number of groups across which data is to be compared. Here we want to compare baby birth weight across 2 groups namely Iron Supplements Yes and No


Step 3: Identify if the group is unpaired or paired.

Then using the following table we will see that the correct answer here will be Independent t test.

Note that if data to be compared was Low birth weight and Normal Birth Weight (and not Birth weight in grams) then it will be a Nominal Type of data and then correct answer will be Chi Square test.

A screening test is used in the same way in two similar populations but the proportions of false positive results of population A is lower than that in population B. What is the likely explanation ?

a) specificity of the test is lowr in population A
b) the prevalence of the disease is lower in population A
c) the prevalence of the disease is higher in population A
d) the specificity of the test is higher in population A

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Ans: c

Sensitivity and Specificity are intrinsic properties of a test and do not change with change in prevalence. Similarly “FALSE POSITIVE RATE = 1 – Specificity” also does not change with prevalence.

However, when the prevalence is low the NUMBER OF FALSE POSITIVES will increase.  Thus the answer here is c – prevalence of disease is higher in population A

Q: Residents of 3 villages with 3 different types of water supply were asked to participate in a study to identify cholera carriers. Because several cholera deaths had occurred in the recent past, virtually everyone present at the time submitted to examination. The proportion of residents in each village who were carriers was computed and compared. This study is a:

a. Cross-Sectional Study                b. Case-control stuy    

c. Con-current Cohort Study          d. Non-concurrent cohort study

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Ans: d

If all people in a village are asked for disease present / absent only, it is cross-sectional study.

If diseased people in one village are compared with neighbouring village, it has to be an analytic design – Cohort or Case Control.

The Cause and effect diagram here will be :

Residence in a particular village ---> Cholera carrier state

If those with carrier state are compared with those without carrier state for village of residence, then it will be case-control study.

If those residing in one village are compared with another village for number of people with carrier state, then it will be cohort study.

Now cohort study can be of 2 types:

1. Concurrent (Prospective): Study starts when cause variable has occurred but effect variable is yet to occur

2. Non-concurrent (retrospective): Study starts after both cause and effect have occurred.

Thus, I will mark option d – Non concurrent Cohort study as the best answer.

Q: In a group of 100 people, Avg GFR = 85ml/min with a SD of 2.5. Range for 90% confidence interval is:

a. 80-90

b. 81-89

c. 75-95

d. 70-100

  

 

 

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Ans: b.

 

 

By the property of Normal Distribution Curve:

Mean ± 1 SD covers 68% of the observations

Mean ± 2 SD covers 95% of the observations

Mean ± 3 SD covers 99% of the observations

Also,

Mean ± 1.65 SD covers 90% of the observations

 

Method 1: (Direct Method)

90% of observations will lie witin thge range of Mean ± 1.65 SD = 85 ± (1.65) (2.5) = 85 ± 4.125 = 80.875 to 89.125

Rounding this will give the answer as 81 to 89 (option b)

 

Method 2: (Indirect Method)

68% of people will lie within Mean +/- 1SD = 85 +/- 2.5 = 82.5 to 87.5

95% people will lie within Mean +/- 2SD = 85 +/- 5 = 80 to 90

 

Now 90% of people will lie between the ranges for 68% and 95%.

There is only one option given which has lower limits between 80 and 82.5… and upper limit between 87.5 and 90… So Answer is b

 

Q: A psychiatrist devised a short screening test for depression. An independent blind comparison was made with a gold standard for diagnosis of depression among 200 psychiatric outpatients. Among the 50 outpatients found to be depressed according to the gold standard, 35 patients were positive for the test. Among 150 patients found not to be depressed according to the gold standard, 30 patients were found to be positive for the test.

a. The sensitivity was 80%
b. The specificity was 80%
c. The positive predictive value was 80%
d. The negative predictive value was 80%

 

 

 

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Ans: b.

 

Step 1: Make a 2x2 table in this format:

	Gold Std +	Gold Std -	Total
New Test +
New Test  -
Total

 

Step 2: Start filling this table based on data given:

	Gold Std +	Gold Std -	Total
New Test +	35	30
New Test  -
Total	50	150	200

 

Step 3: Complete remaining table :

	Gold Std +	Gold Std -	Total
New Test +	35	30	65
New Test  -	15	120	135
Total	50	150	200

Step 4: Use definitions to calculate indices:

• 
  

  Sensitivity = True Positive / Total Diseased           =  35/50 =  0.7 (or 70%)

• 
  

  Specificity = True Negative / Total Not diseased    = 120/150 = 0.8  (or 80%)

• 
  

  PPV = True Positive / All Positives                        = 35/65 = 0.54 (or 54%)

• 
  

  NPV = True Negative / All Negatives                     = 120 / 135 = 0.89 (or 89%)

 

Q: A case control study is done to see association of smoking and oral cancer. 30 patients of oral cancer smoked while 20 did not smoke. Amongst the controls, 20 smoked and 30 were non-smokers. Odds ratio of getting oral cancer in smokers to that of non smokers is.

a) 0.44     

b) 1           

c) 1.5        

d) 2.25

 

 

 

 

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Ans: d) 2.25

 

Step 1: Make a 2x2 table with the given data. It will look as follows:

	Oral CA	Control	Total
Smoker	30	20
Non-Smoker	20	30
Total

 

Step 2: Use the formula for Odds Ratio:

Odds Ratio = ad/ bc  =  30x30 / 20x20        = 900 / 400        = 9/4     = 2.25

In a Normal Distribution Curve, the area to the left of 1SD is

a. 0.16
b. 0.34
c. 0.45
d. 0.84

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Ans: d. 0.84

• Total area = 1;
• Area to left of median = 0.5;
• Area between median and +1SD = 0.34;
• So total area to left of +1SD = 0.5 + 0.34 = 0.84