NATIONAL AND STATE NURSING EXAM- MCQ _MG_00 215
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1. Major cause of perinatal mortality is:
a. Malnutrition
b. Congenital anomalies
c. Birth asphyxia
d. Both a and b
Answer: c. Birth asphyxia
Description:This means that when a baby doesn't get enough oxygen during the process of being born, it can lead to serious health problems or even death. While malnutrition and congenital anomalies can also be factors in perinatal mortality, birth asphyxia is the leading cause.
2. Leading cause of neonatal mortality in India is:
a. Infections
b. Birth asphyxia/trauma
c. Diarrhea
d. Prematurity and congenital malformation
Answer: d. Prematurity and congenital malformation
Description:This means that most newborn babies in India who don't survive are either born prematurely (before they're fully developed) or have congenital (present at birth) malformations or abnormalities that affect their health. While infections, birth asphyxia/trauma, and diarrhea can also be causes of neonatal mortality, prematurity and congenital malformation are the most common reasons in India.
3. The highest rate of infant mortality in India is reported from:
a. Madhya Pradesh
b. Bihar
c. Uttar Pradesh
d. Orissa
Answer: a. Madhya Pradesh
Description:This means that Madhya Pradesh has the highest number of babies who do not survive past their first year of life compared to the other options listed (Bihar, Uttar Pradesh, and Orissa). Infant mortality refers to the number of babies who die before their first birthday, and Madhya Pradesh has the highest rate of such unfortunate incidents among these states.
4. All are direct causes of maternal mortality in India; EXCEPT:
a. Cardiac disease
b. Eclampsia
c. Hemorrhage
d. Abortion
Answer: a. Cardiac disease
Description:This means that cardiac disease is not considered one of the direct causes of maternal mortality in India. Maternal mortality refers to the number of mothers who die during pregnancy, childbirth, or within 42 days after childbirth due to specific complications. In India, cardiac disease is not typically counted as one of these direct causes, while conditions like eclampsia, hemorrhage, and abortion-related complications are recognized causes of maternal mortality.
5. Mainly include in child survival index:
a. MMR
b. IMR
c. Mortality between 1 and 4 years of age
d. Under 5 mortality
Answer: d. Under 5 mortality
Description:The child survival index typically focuses on the mortality rate of children under the age of 5. This index assesses how well a country or region is doing in terms of ensuring the survival and well-being of young children. While maternal mortality (MMR), infant mortality (IMR), and mortality between 1 and 4 years of age are important indicators of child health, the under-5 mortality rate is often the central component of the child survival index.
6. Perinatal death include:
a. After 28 weeks of pregnancy
b. Within seven days after birth
c. Both a and b
d. None of the above.
Answer: c. Both a and b
Description:Perinatal death encompasses deaths that occur either after 28 weeks of pregnancy (option a) or within seven days after birth (option b). It covers a specific period around the time of birth, including late pregnancy and the early days of a newborn's life.
7. Most common cause of perinatal mortality is:
a. Prematurity
b. Birth injury
c. Metabolic
d. Congenital
Answer: a. Prematurity
Description:Perinatal mortality refers to the death of a baby during the perinatal period, which includes late pregnancy (usually after 28 weeks of gestation) and the early days after birth. Premature birth, where a baby is born before it has fully developed, is a significant factor in perinatal mortality and is the most common cause among the options listed.
8. Most common cause of death in children <5 years age:
a. Respiratory infection
b. Diarrhea
c. Prematurity
d. Accidents
Answer: a. Respiratory infection
Description:Diarrhea is a major health concern for young children, especially in low-income countries, where access to clean water and proper sanitation may be limited. It can lead to dehydration and other complications, making it a leading cause of mortality in this age group. While respiratory infections, prematurity, and accidents can also be significant factors, diarrhea is typically the most common cause of death in children under 5 worldwide.
9. Kaplan Meier method is for:
a. Survival
b. Incidence
c. Prevalence
d. Frequency
Answer: a. Survival
Description:The Kaplan-Meier method is a statistical technique used to estimate the survival probability or survival function over time in medical and survival analysis studies. It is commonly used to analyze and visualize the time-to-event data, such as the time until a patient experiences a specific event like death, disease recurrence, or failure of a medical treatment.
10. Prevalence is a:
a. Rate
b. Ratio
c. Proportion
d. Mean
Answer: c. Proportion
Description:Prevalence is a measure of the proportion of individuals in a population who have a specific characteristic or condition at a particular point in time. It is often expressed as a percentage or proportion, making it a measure of the number of existing cases relative to the total population at a specific point or period.
11. Other name for vitamin B12 is:
a. Ascorbic acid
b. Cyanocobalamin
c. Thiamine
d. Riboflavin
Answer: a. Ascorbic acid
Description:Ascorbic acid (option a) is another name for vitamin C, while thiamine (option c) and riboflavin (option d) are different B vitamins (B1 and B2, respectively). Cyanocobalamin is the specific form of vitamin B12 commonly used in supplements and medications.
12. Number of current cases (new and preexisting) at a specified point in time divided by population at the same specified point in time is referred as:
a. Point prevalence
b. period prevalence
c. Lifetime prevalence
d. Total prevalence
Answer: a. Point prevalence
Description:The number of current cases (new and preexisting) at a specified point in time divided by the population at the same specified point in time is referred to as: a. Point prevalence Point prevalence is a measure of how many individuals in a population have a specific condition or characteristic at a single point in time. It provides a snapshot of the condition's prevalence at that particular moment.
13. Number of current cases (new and preexisting) over a specified period of time divided by average or med-interval population:
a. Point prevalence
b. period prevalence
c. Lifetime prevalence
d. Current prevalence
Answer: b. period prevalence
Description:The number of current cases (new and preexisting) over a specified period of time divided by the average or mid-interval population is referred to as: b. Period prevalence Period prevalence is a measure of how many individuals in a population have a specific condition or characteristic over a defined time period. It takes into account both new cases that occur during that period and preexisting cases that persist within the population.
14. The number of new case occurring in a defined population during a specified period of time is know as:
a. Incidence
b. Point prevalence
c. Period prevalence
d. Current prevalence
Answer: a. Incidence
Description:The number of new cases occurring in a defined population during a specified period of time is known as: a. Incidence
15. A village was divided into 5 sub groups for a survey. The sample was drawn by selecting people randomly from these sub groups. What type of sampling was done?
a. Simple random sampling
b. Stratified random sampling
c. Cluster sampling
d. Systematic random sampling
Answer: b. Stratified random sampling
Description:Stratified random sampling involves dividing the population into subgroups or strata based on certain characteristics (in this case, the subgroups within the village), and then randomly selecting samples from each stratum. This method ensures that each subgroup is represented in the sample, making it useful for obtaining a more accurate representation of the entire population.
16. In a research study, first schools are sample, then sections and finally students. This type of sampling is known as:
a. Stratified sampling
b. Simple random sampling
c. Cluster sampling
d. Multi stage sampling
Answer: d. Multi stage sampling
Description:The type of sampling described, where first schools are sampled, then sections, and finally students, is known as: d. Multi-stage sampling Multi-stage sampling involves multiple stages or steps of sampling. In this case, it starts with schools, then sections within those schools, and finally students within those sections. This method is often used when it's challenging or impractical to sample the entire population at once.
17. Stratified sampling is ideal for:
a. Heterogeneous data
b. Homogenous data
c. Both a and b
d. None of these
Answer: a. Heterogeneous data
Description:Stratified sampling is particularly useful when dealing with heterogeneous (diverse) data because it allows you to divide the population into subgroups (strata) based on certain characteristics, ensuring that each subgroup is represented in the sample. This helps in capturing the diversity of the population and obtaining a more accurate representation.
18. Simple random sampling is ideal for:
a. Vaccinated people
b. Heterogeneous population
c. Homogenous population
d. All of the above.
Answer: c. Homogenous population
Description:Simple random sampling involves randomly selecting individuals or items from a population with uniform characteristics. It works well when the population is relatively homogeneous, meaning that the individuals or items are similar to each other in relevant aspects. If the population is highly diverse or heterogeneous, other sampling methods like stratified sampling or cluster sampling might be more appropriate to ensure a representative sample. So, for a homogeneous population, simple random sampling is a good choice.
19. Degree of freedom for 2 x 2 contingency table is:
a. 1
b. 0
c. 2
d. 4
Answer: a. 1
Description:In a 2 x 2 contingency table, there is only one degree of freedom because once you specify the counts in one cell, the counts in the other three cells are determined by the marginal totals. Therefore, you have one degree of freedom to vary the count in one cell while keeping the marginal totals fixed.
20. Chi square test is for:
a. Standard error of mean
b. Standard error of proportion
c. Difference between population means
d. Difference between population proportion
Answer: d. Difference between population proportion
Description:The Chi-Square test is commonly used to determine whether there is a significant difference between observed and expected frequencies in categorical data, particularly when comparing proportions or percentages in different categories or groups. It's often used to test whether there's a significant association or difference between two or more categorical variables, such as comparing the proportions of "yes" and "no" responses in different groups.
21. Not required for Chi square test:
a. Null hypothesis
b. Degree of freedom
c. Means in different groups
d. Proportion in different groups
Answer: c. Means in different groups
Description:The Chi-Square test is primarily used to analyze categorical data and determine whether there is a significant association or difference between categorical variables. It is not used to compare means (averages) in different groups, which is typically done through other statistical tests like t-tests or analysis of variance (ANOVA). In Chi-Square tests, you are examining the association or independence of categories, not the means of continuous variables.
22. Degree of freedom of contingency table with 3 rows and 6 columns is:
a. 2
b. 3
c. 10
d. 18
Answer: c. 10
Description:The degree of freedom for a contingency table is calculated as: (df) = (number of rows - 1) * (number of columns - 1) In this case, for a contingency table with 3 rows and 6 columns: (df) = (3 - 1) * (6 - 1) = 2 * 5 = 10 So, the degree of freedom for a contingency table with 3 rows and 6 columns is: c. 10
23. Chi square test 5 rows/4 columns, degree of freedom is:
a. 9
b. 12
c. 16
d. 20
Answer: b. 12
Description:The degree of freedom for a chi-square test in a contingency table is calculated as: (df) = (number of rows - 1) * (number of columns - 1) In this case, for a contingency table with 5 rows and 4 columns: (df) = (5 - 1) * (4 - 1) = 4 * 3 = 12
24. If the distribution of intra-ocular pressure (IOP) seen in 100 glaucoma patients has an average 30 mm with a SD of 1.0 what is the lower limit of the average IOP that can be expected 95% of times?
a. 29
b. 26
c. 32
d. 259
Answer: a. 29
Description:To find the lower limit of the average IOP that can be expected 95% of the time, you can use the formula for a confidence interval, assuming a normal distribution: Lower Limit = Mean - (Z * (Standard Deviation / √n)) Where: Mean (average IOP) = 30 mm Z is the Z-score associated with the desired confidence level (95% confidence corresponds to a Z-score of approximately 1.96). Standard Deviation (SD) = 1.0 n (sample size) = 100 (number of glaucoma patients) Lower Limit = 30 - (1.96 * (1.0 / √100)) Lower Limit = 30 - (1.96 * 0.1) Lower Limit = 30 - 0.196 Lower Limit ≈ 29.804 Rounded to the nearest whole number, the lower limit is approximately 29 mm.
25. Which is the best distribution to study the daily admission of head injury patients in a trauma care center?
a. Normal distribution
b. Binomial distribution
c. Uniform distribution
d. Poisson distribution
Answer: d. Poisson distribution
Description:The Poisson distribution is commonly used to model the number of events (in this case, daily admissions of head injury patients) that occur in a fixed interval of time or space when these events happen with a known average rate but are rare and random in nature. It is often used for count data, especially when events occur independently and infrequently, which is a characteristic of patient admissions to a trauma care center.
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