Evidence-based practice statistics worksheet with answers

Free printable Evidence-based practice statistics worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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Name: ____________________   Date: ____________

1.Easy
01 yearN = 28 000cases: 548new cases: 159CI = ?

Practice exercise (made-up values, not clinical guidance). A town has 28 000 inhabitants. At the start of a year, 548 of them have a chronic disease. During the year, 159 more people are newly diagnosed (assume nobody moves away or dies). Find the cumulative incidence CI over the year per 1000 people at risk. Round to one decimal place.

N = 28 000existing cases = 548new cases = 159
CI = per 1000
2.Easy
01 yearN = 50 000cases: 928new cases: 840CI = ?

Practice exercise (made-up values, not clinical guidance). A town has 50 000 inhabitants. At the start of a year, 928 of them have a chronic disease. During the year, 840 more people are newly diagnosed (assume nobody moves away or dies). Find the cumulative incidence CI over the year per 1000 people at risk. Round to one decimal place.

N = 50 000existing cases = 928new cases = 840
CI = per 1000
3.Medium
casescontrolsexposed3644abnot exposed15476cd

Practice exercise (made-up values, not clinical guidance). A case-control study compares 190 patients with a disease (cases) and 120 people without it (controls). The table shows how many in each group were exposed to a risk factor. Find the odds ratio OR for the exposure. Round to two decimal places.

a = 36b = 44c = 154d = 76
OR =
4.Medium
eventno eventtotaltreatment15205220abcontrol69421490cd

Practice exercise (made-up values, not clinical guidance). In a randomised trial, 220 patients received a new treatment and 490 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the number needed to treat NNT to prevent one event. Round up to a whole number of patients.

treatment: 15/220control: 69/490
NNT =
5.Hard
testdisease presentdisease absenttotalpositivenegative?total10 000

Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 89% and specificity 98%. The prevalence of the disease is 24%. How many people with the disease are missed by the test (false negatives, FN)?

N = 10 000p = 24%Se = 89%Sp = 98%
FN =
6.Hard
p (%)PPV (%)0.5%PPV = ?100

Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 92% and a specificity of 98%. It is used in a population where the prevalence is 0.5%. Find the positive predictive value PPV as a percentage. Round to one decimal place.

Se = 92%Sp = 98%p = 0.5%
PPV = %

Answer key

  1. 1. CI = 5.8 per 1000
  2. 2. CI = 17.1 per 1000
  3. 3. OR = 0.4
  4. 4. NNT = 14
  5. 5. FN = 264
  6. 6. PPV = 18.8 %

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