Evidence-Based Practice: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Evidence-based practice statistics

Sensitivity and specificity
Se = a/(a + c), Sp = d/(b + d)
Predictive values
PPV = a/(a + b), NPV = d/(c + d)
Accuracy
(a + d)/n
Likelihood ratios
LR+ = Se/(1 − Sp), LR− = (1 − Se)/Sp
PPV at prevalence p
Se·p / (Se·p + (1 − Sp)(1 − p))
Odds
odds = p/(1 − p); p = odds/(1 + odds); post-test odds = pre-test odds × LR
Risk measures
RR = EER/CER, ARR = CER − EER, RRR = ARR/CER
Number needed to treat
NNT = 1/ARR, rounded up; NNH = 1/(EER − CER)
Odds ratio
OR = ad/(bc)
Incidence rate
new cases / person-time

Method

  1. Draw the 2 × 2 table and label a, b, c and d before you calculate.
  2. Read sensitivity and specificity down the columns, predictive values along the rows.
  3. For another prevalence, imagine 10 000 people and fill the table with Se, Sp and the prevalence.
  4. For a likelihood ratio, convert the probability to odds, multiply, and convert back.
  5. For a trial, find EER and CER first; then RR = EER/CER, ARR = CER − EER and NNT = 1/ARR rounded up.

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Confidence intervals

σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up

Method

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. Interval: x̄ ± E.

Hypothesis tests & confidence intervals

z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n

Method

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.

2Practice problems

1.Evidence-based practice statisticsMedium
casescontrolsexposed110139abnot exposed90281cd

Practice exercise (made-up values, not clinical guidance). A case-control study compares 200 patients with a disease (cases) and 420 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 = 110b = 139c = 90d = 281
OR =
2.Evidence-based practice statisticsHard
CER14%EER?RRR 21%

Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 21% in relative terms (RRR = 21%). In a patient group the risk without treatment (CER) is 14% over five years. Find the number needed to treat NNT for five years to prevent one event. Round up to a whole number of patients.

RRR = 21%CER = 14%
NNT =
3.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 28, 32, 55, 42, 44, 55.

n = 6
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 100 and standard deviation 5. What proportion P of values is below 107.5?

μ = 100σ = 5x = 107.5
P =
5.Confidence intervalsMedium
x̄E₁ = 999%, n = 100x̄E₂ = ?90%, n = 900

A 99% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 9. What is the margin of error E₂ of a 90% interval based on 900 observations?

99%n₁ = 100E₁ = 990%n₂ = 900
E₂ =
6.Confidence intervalsHard
σ = 38, n = ?x̄ − Ex̄ + Ex̄E = 399%

The population standard deviation is 38. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 3?

σ = 38E = 399%
n =
7.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 98 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 98s = 20n = 81
t =
8.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 64 measurements has mean 50.4. The population standard deviation is 5. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 50.4σ = 5n = 6495%
L =

3Solutions

1.Evidence-based practice statistics

Answer: OR = 2.47

Likelihood ratios and risk measures: LR+ = sensitivity/(1 − specificity), LR− = (1 − sensitivity)/specificity. In a trial, EER and CER are the event rates with and without treatment: RR = EER/CER, ARR = CER − EER, RRR = ARR/CER and NNT = 1/ARR, rounded up. In a case-control study only the odds ratio OR = ad/(bc) can be used. An incidence rate = new cases/person-time.

Odds of exposure

cases: 110/90 = 1.222
controls: 139/281 = 0.4947

Odds ratio

OR = (110·281)/(139·90) = 2.471 ≈ 2.47

Answer

OR = 2.47

OR above 1: exposure is more common among the cases.

2.Evidence-based practice statistics

Answer: NNT = 35

Predictive values depend on prevalence: PPV = Se·p/(Se·p + (1 − Sp)(1 − p)) and NPV = Sp(1 − p)/(Sp(1 − p) + (1 − Se)p). With likelihood ratios: pre-test odds = p/(1 − p), post-test odds = pre-test odds × LR, probability = odds/(1 + odds). For a population of N, true positives = N·p·Se and false positives = N·(1 − p)·(1 − Sp). NNT = 1/(CER × RRR), rounded up.

Absolute risk reduction

ARR = 0.14·0.21 = 0.0294

Number needed to treat

NNT = 1/0.0294 = 34.01 → 35

Answer

NNT = 35

The same relative effect gives a smaller NNT when the baseline risk is higher.

3.Descriptive statistics

Answer: s = 11.27

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Mean

x̄ = 256/6 = 42.67

Squared deviations

Σ(x − x̄)² = (−14.67)² + (−10.67)² + 12.33² + (−0.6667)² + 1.333² + 12.33² = 635.3

Standard deviation

s = √(635.3/5) = 11.27

Answer

s = 11.27
4.Descriptive statistics

Answer: P = 0.9332

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Standardise

z = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 0.9332

Answer

P = 0.9332
5.Confidence intervals

Answer: E₂ = 1.916

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Ratio of margins

E₂/E₁ = (1.645/2.576)·√(100/900) = 0.2129

New margin

E₂ = 9·0.2129 = 1.916

Answer

E₂ = 1.916
6.Confidence intervals

Answer: n = 1065

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Solve for n

n = (2.576·38/3)² = 1065

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 1065

Answer

n = 1065
7.Hypothesis tests & confidence intervals

Answer: t = −0.9

Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).

Standard error

SE = 20/√81 = 20/9 = 2.222

Test statistic

t = (98 − 100)/2.222 = −0.9

Answer

t = −0.9
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

8.Hypothesis tests & confidence intervals

Answer: L = 49.17

Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).

Margin of error

E = 1.96·5/√64 = 1.225

Interval limit

L = 50.4 − 1.225 = 49.17

Answer

L = 49.17

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