Amsra's Internal Education Exam - February 2016
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Which of the following variables is a continuous quantitative variable?
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Favorite fruit
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Gender
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Decade of birth
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Age at first birth
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Occupation
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AMSRA's Internal Education Exam - February 2016 assesses knowledge in statistics and epidemiology, focusing on quantitative variables, data analysis, study designs, and statistical measures. It is designed to enhance analytical skills relevant to scientific research.
- 2.
Refer to this histogram for sleep duration (hours per night).Which of the following statements is TRUE?Note that sleep duration was measured in hours and minutes per night.
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More than half of the students slept 7.0 hours per night
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More than half of the students slept between 6.5 and 7.5 hours per night
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At least one student slept 10 or more hours per night
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25 percent of students slept 6.0 hours or fewer per night
Correct Answer
A. More than half of the students slept between 6.5 and 7.5 hours per nightExplanation
Based on the given histogram, the bar representing the range of 6.5 to 7.5 hours per night is taller than the bar representing 7.0 hours per night. This indicates that more students fall within the range of 6.5 to 7.5 hours, making the statement "More than half of the students slept between 6.5 and 7.5 hours per night" true.Rate this question:
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- 3.
Calculate the standard deviation of the following data points (round to the nearest whole number):-8, -10, -12, -16, -18, -20, -21, -24, -26, -30
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94
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50
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14
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10
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7
Correct Answer
A. 7Explanation
The standard deviation is a measure of how spread out the data points are from the mean. To calculate it, we need to find the mean of the data points first. Adding up all the data points and dividing by the total number of data points, we get a mean of -18. Then, we subtract the mean from each data point and square the result. Next, we find the average of these squared differences. Taking the square root of this average gives us the standard deviation, which in this case is 7.Rate this question:
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- 4.
What is the primary benefit of randomization?
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Improves patient recruitment
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Reduces costs
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Minimizes confounding
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Prevents recall bias
Correct Answer
A. Minimizes confoundingExplanation
Randomization minimizes confounding in a study. Confounding occurs when an extraneous variable is associated with both the exposure and the outcome, leading to a false association. By randomly assigning participants to different groups, the distribution of confounding variables is expected to be similar between groups, reducing the potential for confounding. This allows researchers to confidently attribute any differences observed in the outcome to the exposure being studied, increasing the internal validity of the study.Rate this question:
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- 5.
Which of the following is a key benefit of case-control studies?
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They avoid confounding
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Risk factors are measured before the outcome has occurred
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They are efficient for studying rare diseases
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None of the above
Correct Answer
A. They are efficient for studying rare diseasesExplanation
Case-control studies are efficient for studying rare diseases because they allow researchers to identify individuals with the disease and compare them to a control group without the disease. This design is particularly useful when studying rare diseases because it is often difficult to find a sufficient number of cases for a traditional cohort study. By selecting cases and controls from a defined population, researchers can efficiently investigate the potential risk factors associated with the rare disease.Rate this question:
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- 6.
What measure of disease frequency can be calculated from case-control studies?
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Incidence rate
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Cumulative risk
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Prevalence
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You cannot calculate disease frequency from this study design
Correct Answer
A. You cannot calculate disease frequency from this study designExplanation
Case-control studies are observational studies that compare individuals with a specific disease (cases) to individuals without the disease (controls). These studies are primarily used to examine the association between exposures and diseases, rather than calculating disease frequency. Disease frequency measures such as incidence rate, cumulative risk, and prevalence are typically calculated using cohort or cross-sectional study designs, where individuals are followed over time or at a specific point in time, respectively. Therefore, it is not possible to calculate disease frequency directly from case-control studies.Rate this question:
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- 7.
In a study that enrolled 1000 women for 0.5 years, 10 women developed breast cancer. What was the incidence rate for breast cancer?
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2%
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1%
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2 events per 1000 women-years
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2 events per 100 women-years
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5 events per 1000 women-years
Correct Answer
A. 2 events per 100 women-yearsExplanation
The incidence rate for breast cancer is calculated by dividing the number of new cases (10 women) by the total person-time at risk (100 women-years). In this case, the incidence rate is 10/100 = 0.1 or 10%. To convert this to events per 100 women-years, we multiply by 100, resulting in 10 events per 100 women-years. Therefore, the answer "2 events per 100 women-years" is incorrect.Rate this question:
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- 8.
If 40% of a treated group has a positive response versus just 10% of the placebo group, what are the risk ratio and odds ratio (for treatment vs. placebo)?
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RR=4.0; OR=4.0
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RR=4.0; OR=6.0
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RR=0.5; OR=0.6
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RR=2.0; OR=4.1
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RR=1.0; OR=1.0
Correct Answer
A. RR=4.0; OR=6.0Explanation
The risk ratio (RR) is calculated by dividing the probability of a positive response in the treated group by the probability of a positive response in the placebo group. In this case, the probability of a positive response in the treated group is 40% (0.4) and the probability of a positive response in the placebo group is 10% (0.1). Therefore, RR = 0.4 / 0.1 = 4.0.
The odds ratio (OR) is calculated by dividing the odds of a positive response in the treated group by the odds of a positive response in the placebo group. The odds of a positive response in the treated group is 0.4 / (1 - 0.4) = 0.4 / 0.6 = 0.67 and the odds of a positive response in the placebo group is 0.1 / (1 - 0.1) = 0.1 / 0.9 = 0.11. Therefore, OR = 0.67 / 0.11 = 6.0.Rate this question:
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- 9.
In a cross-sectional study about sleep quality and duration and hypertension (Sleep.2009; 32:491-7.), researchers found that about 25% of normal sleepers had hypertension. They also found that insomniacs with short sleep duration (<5 hours per night) were significantly more likely to have hypertension than normal sleepers. The odds ratio for this association was 5.12. Which of the following statements is correct?
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Insomniacs with short sleep duration have a 5-fold higher risk of having hypertension than normal sleepers.
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People with hypertension have a 5-fold higher risk of having insomnia and short sleep duration.
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Insomniacs with short sleep duration have a 5-fold higher odds of having hypertension, which corresponds to a 2.5-fold higher risk.
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Insomniacs with short sleep duration have a 5-fold higher odds of having hypertension, which corresponds to a 4.1-fold higher risk.
Correct Answer
A. Insomniacs with short sleep duration have a 5-fold higher odds of having hypertension, which corresponds to a 2.5-fold higher risk.Explanation
The correct answer is "Insomniacs with short sleep duration have a 5-fold higher odds of having hypertension, which corresponds to a 2.5-fold higher risk." This answer accurately reflects the information provided in the study. The odds ratio of 5.12 indicates that insomniacs with short sleep duration have 5 times higher odds of having hypertension compared to normal sleepers. However, the corresponding risk is 2.5 times higher, as odds ratios overestimate the risk ratio in this case.Rate this question:
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- 10.
What is statistical power?
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The probability of getting a statistically significant result when the null hypothesis is true
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The probability of getting a statistically significant result when the null hypothesis is false
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The probability of getting a non-significant result when the null hypothesis is true
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The probability of getting a non-significant result when the null hypothesis is false
Correct Answer
A. The probability of getting a statistically significant result when the null hypothesis is falseExplanation
Statistical power refers to the probability of obtaining a statistically significant result when the null hypothesis is false. In other words, it measures the likelihood of correctly rejecting the null hypothesis when there is actually a true effect or relationship in the population. A high statistical power indicates a greater ability to detect true effects, while a low power increases the chances of missing important findings.Rate this question:
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- 11.
In a study that looked at 100,000 children, researchers found a highly significant inverse correlation (p<.0001) between blood sugar and IQ. What conclusions can we draw?
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Eating too much sugar lower IQ in children
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There is a strong association between sugar levels and IQ in children, but it might not be causal
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This study provides strong evidence of an important relationship between blood sugar and IQ in kids
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We should not draw any conclusions without knowing the size of the effect
Correct Answer
A. We should not draw any conclusions without knowing the size of the effect -
- 12.
Which of the following would increase the width of a confidence interval?
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Changing from a 99% to 95% confidence level
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Increasing the variability of the outcome
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Increasing the sample size
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Removing an outlier from the data
Correct Answer
A. Increasing the variability of the outcomeExplanation
Increasing the variability of the outcome would increase the width of a confidence interval. A confidence interval is a range of values within which the true population parameter is estimated to lie. The width of the confidence interval is affected by the level of confidence chosen, the variability of the outcome, the sample size, and the presence of outliers. Increasing the variability of the outcome means that the data points will be more spread out, resulting in a wider range of possible values for the population parameter, thus increasing the width of the confidence interval.Rate this question:
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- 13.
The most useful of the following ratios in term of helping in evidence-based diagnosis is:
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Specificity
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Sensitivity
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Likelihood
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Predictive value
Correct Answer
A. LikelihoodExplanation
The likelihood ratio is the most useful ratio in terms of helping in evidence-based diagnosis. It provides information about how much a particular diagnostic test result changes the odds of a disease being present. It takes into account both the sensitivity (the ability of the test to correctly identify those with the disease) and the specificity (the ability of the test to correctly identify those without the disease). Therefore, the likelihood ratio helps in determining the probability of a disease based on the test results, making it a valuable tool in evidence-based diagnosis.Rate this question:
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- 14.
A young patient came to you complaining of painful joints and you want to screen for lupus erythematosus, what is the feature of the ideal test that you should start with?
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Highly sensitive test
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Highly specific test
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Test of high positive predictive value
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Test of high negative predictive value
Correct Answer
A. Highly sensitive testExplanation
When screening for lupus erythematosus in a young patient with painful joints, the ideal test should be highly sensitive. A highly sensitive test will correctly identify a high proportion of patients who have the condition, minimizing the chances of false negatives. This is important in the initial screening stage to ensure that patients who may have lupus erythematosus are not missed and can receive further diagnostic evaluation and appropriate treatment if necessary.Rate this question:
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- 15.
A 40 years old male with moderately-controlled allergic asthma who takes inhaled corticosteroid (ICS), heard that other asthmatic patients tried a new drug called omalizumab and found it good and useful. The patient asked you if adding omalizumab to ICS would decrease his symptoms and the likelihood of exacerbation and keep him away from the hospital.What is the best PICO question for this case?
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P: patients with asthma. I: ICS + omalizumab. C: ICS. O: improvement in asthma control and reduction of exacerbation
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P: patients with allergic asthma who takes ICS. I: omalizumab. C: not applicable. O: improvement in asthma control and reduction of exacerbation
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P: patients with allergic asthma. I: ICS. C: omalizumab. O: improvement in asthma control and reduction of exacerbation
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P: patients with allergic asthma.I: ICS + omalizumab. C: ICS. O: improvement in asthma control and reduction of exacerbation
Correct Answer
A. P: patients with allergic asthma.I: ICS + omalizumab. C: ICS. O: improvement in asthma control and reduction of exacerbationExplanation
The best PICO question for this case is "In patients with allergic asthma, does the addition of omalizumab to ICS (inhaled corticosteroid) compared to ICS alone, result in improvement in asthma control and reduction of exacerbation?" This question specifically addresses the patient's concern about adding omalizumab to their current treatment and whether it would lead to improved asthma control and fewer exacerbations.Rate this question:
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- 16.
A study to test the hypothesis that volcanic dust causes silicosis is to be tested by comparing the population of an island very close to a volcanic eruption with the population of an island not so exposed. After 5 and 10 years the incidence of silicosis will be measured and compared in both populations. This study is
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A cross sectional study
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A case control study
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A prospective study
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A randomised controlled trial
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A sequential trial
Correct Answer
A. A prospective studyExplanation
A prospective study is the correct answer because it involves following a group of individuals over a period of time to observe the development of a disease or condition. In this case, the study plans to measure and compare the incidence of silicosis in two populations after 5 and 10 years. This approach allows for the evaluation of cause and effect relationships and can help determine if volcanic dust exposure is indeed a risk factor for silicosis.Rate this question:
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- 17.
In 1945, there were 1,000 women who worked in a factory painting radium dials on watches. The incidence of bone cancer in these women up to 1975 was compared to that of 1,000 women who worked as telephone operators in 1945. Twenty of the radium dial painters and four of the telephone operators developed bone cancer between 1945 and 1975. What study design is this?
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Cohort study
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Experimental study
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Clinical trial
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Cross-sectional study
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Case-control study
Correct Answer
A. Cohort studyExplanation
This study design is a cohort study. In a cohort study, a group of individuals who share a common characteristic or exposure (in this case, women who worked in a factory painting radium dials) is followed over time to compare their outcomes (in this case, the incidence of bone cancer) with a control group (in this case, telephone operators). The study compares the incidence of bone cancer in the two groups between 1945 and 1975 to determine if there is a relationship between the exposure (working as a radium dial painter) and the outcome (developing bone cancer).Rate this question:
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- 18.
Researchers set out to demonstrate that a new drug is more effective in lowering systolic blood pressure than beta-blockers. They select two groups from among a number of previously uncontrolled hypertensive patients to receive either the new drug or a beta-blocker using a computer program to make the allocation purely by chance. What is this method of assignment called?
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Random selection
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Randomization
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Blinding
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Cross-over
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Factorial
Correct Answer
A. RandomizationExplanation
The method of assignment described in the question is called randomization. Randomization involves using a computer program to allocate participants into different groups purely by chance. This helps to eliminate any bias or potential confounding factors that could influence the results. By randomly assigning participants to either receive the new drug or a beta-blocker, researchers can compare the effectiveness of the two treatments in lowering systolic blood pressure in a more unbiased manner.Rate this question:
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- 19.
A community assesses a random sample of its residents by telephone questionnaire. Obesity is strongly associated with diagnosed diabetes. This study design is best described as which one of the following:
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Case control
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Cohort
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Cross sectional
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Experimental
Correct Answer
A. Cross sectionalExplanation
This study design is best described as cross-sectional because it involves assessing a random sample of residents at a specific point in time through a telephone questionnaire. It aims to gather information about the prevalence of obesity and its association with diagnosed diabetes in the community. A cross-sectional study design allows for the collection of data from different individuals at a single time point, providing a snapshot of the relationship between variables without establishing causality or following individuals over time.Rate this question:
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- 20.
The initial studies establishing maternal diethylstilbesterol (DES) intake as a cause of vaginal adenocarcinoma in female offspring were case-control studies. This was probably largely because:
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A couple of decades ago cohort studies hadn't been invented
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A woman taking DES was always rare
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The disease outcome is rare
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The investigators had probably just happened to have a number of cases in their practices
Correct Answer
A. The disease outcome is rareExplanation
The correct answer is that the disease outcome is rare. This suggests that the reason why the initial studies establishing maternal DES intake as a cause of vaginal adenocarcinoma in female offspring were case-control studies is because the disease outcome is rare. Case-control studies are often used when studying rare diseases or outcomes because they allow researchers to efficiently compare cases (individuals with the disease) to controls (individuals without the disease) and identify potential risk factors. In this case, using a case-control design would have been more feasible and practical than conducting a cohort study.Rate this question:
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- 21.
In a case-control study of alcohol intake and bladder cancer, cases and matched controls are each interviewed by interviewers who are not blinded as to whether the subject is a case or a control. Many of the interviewers are in fact convinced that drinking alcohol is a cause of bladder cancer. Is this likely to represent a bias?
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No, because the interviewers can't affect whether the subjects are considered cases or controls; that's already decided
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Yes, but it's hard to predict the direction of the bias
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Yes, and would predispose to a rejection of the null hypothesis
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Yes, and would predispose to an acceptance of the null hypothesis
Correct Answer
A. Yes, and would predispose to a rejection of the null hypothesisExplanation
Yes, this is likely to represent a bias because the interviewers' beliefs about alcohol intake and bladder cancer could influence the way they ask questions or interpret the responses. This could lead to a higher likelihood of finding a significant association between alcohol intake and bladder cancer, predisposing to a rejection of the null hypothesis.Rate this question:
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- 22.
The purpose of random sampling is to ensure
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A sufficient sample size.
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A clearly defined target population
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That the sample is representative of the target population
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Representation of specific subgroups in the population
Correct Answer
A. That the sample is representative of the target populationExplanation
The purpose of random sampling is to ensure that the sample is representative of the target population. Random sampling involves selecting individuals or items from the population in a way that each member has an equal chance of being included in the sample. This helps to minimize bias and increase the generalizability of the findings from the sample to the larger population. By ensuring representativeness, random sampling allows researchers to make accurate inferences about the population based on the characteristics observed in the sample.Rate this question:
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- 23.
Which of the following is a logical reason for using non-random sampling?
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Students have already been assigned to classes
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Non-random sampling assures the researcher of selecting equivalent groups
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Representation is a concern
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Generalization is a concern
Correct Answer
A. Students have already been assigned to classesExplanation
Using non-random sampling is a logical reason when students have already been assigned to classes. In this situation, random sampling may not be feasible or practical, as the researcher cannot randomly assign students to different classes. Therefore, non-random sampling allows the researcher to study the specific groups that have already been formed, providing insights into the characteristics and behaviors of those particular classes.Rate this question:
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- 24.
Ms. Landry has begun to hear the same things from each additional participant she interviews. What would you advise her to do in terms of her sample size?
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Stop sampling
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Select a few more participants
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Select at least 10 more participants
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Continue to sample until her quota is reached
Correct Answer
A. Stop samplingExplanation
The correct answer is to stop sampling. Ms. Landry has already started hearing the same things from each additional participant she interviews, which indicates that she has reached saturation point in her data collection. Continuing to sample more participants would not provide any new or additional information, as the same patterns and themes are being repeated. Therefore, it is advisable for Ms. Landry to stop sampling and proceed with the data analysis phase.Rate this question:
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- 25.
Which of the following characteristics clearly differentiates probability and purposive sampling?
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Sample sizes are typically the same
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Probability sampling starts with a defined population and selects a sample from it, while non-probability sampling starts with a sample and defines the population relative to the characteristic of that sample
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Probability sampling makes it difficult to generalize from the sample to populations, while non-probability sampling makes it easy to do so.
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More than one technique can be used to select a sample with either approach
Correct Answer
A. Probability sampling starts with a defined population and selects a sample from it, while non-probability sampling starts with a sample and defines the population relative to the characteristic of that sampleExplanation
The correct answer explains the key difference between probability and non-probability sampling. Probability sampling involves starting with a defined population and selecting a sample from it, while non-probability sampling begins with a sample and defines the population based on the characteristics of that sample. This means that probability sampling is more representative of the entire population, while non-probability sampling may introduce bias by only including certain characteristics in the population.Rate this question:
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Current Version
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Mar 15, 2023Quiz Edited by
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Feb 18, 2016Quiz Created by
Arc16amsra
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