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
peopleyears13703213023905new cases7

Practice exercise (made-up values, not clinical guidance). In a cohort, groups of people were followed for different lengths of time (see the table: number of people and years of follow-up for each). During follow-up, 7 new cases occurred. Find the incidence rate IR per 1000 person-years. Round to one decimal place.

370 × 3 years130 × 2 years90 × 5 years7 new cases
IR = per 1000 person-years
2.Evidence-based practice statisticsHard
CER14%EER?RRR 13%

Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 13% in relative terms (RRR = 13%). 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 = 13%CER = 14%
NNT =
3.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 13, 39, 35, 55, 47.

n = 5
s =
4.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 72.5.

μ = 60σ = 10x = 72.5
z =
5.Confidence intervalsMedium
n = 25, s = 11LUx̄ = 47.5E = ?99%

A sample of 25 items has mean 47.5 and standard deviation s = 11. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 25x̄ = 47.5s = 1199%t(24) = 2.797
E =
6.Confidence intervalsHard
σ = 37, n = ?x̄ − Ex̄ + Ex̄E = 199%

The population standard deviation is 37. 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 1?

σ = 37E = 199%
n =
7.Hypothesis tests & confidence intervalsMedium
−2.2012.201t = ?df = 11

12 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 1.99 and the standard deviation of the differences is 6.9. Find the paired test statistic t for H₀: μd = 0.

n = 12d̄ = 1.99sd = 6.9
t =
8.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ < 50 is done at α = 0.05 with a sample of 25; the standard deviation is known to be 5. In reality the mean is 48.9. Find the probability of a Type II error β in percent.

μ₀ = 50μ = 48.9σ = 5n = 25α = 0.05
β = %

3Solutions

1.Evidence-based practice statistics

Answer: IR = 3.8 per 1000 person-years

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.

Person-years

PY = 370·3 + 130·2 + 90·5 = 1820

Incidence rate

IR = 7/1820·1000 = 3.8 per 1000 person-years

Answer

IR = 3.8 per 1000 person-years
2.Evidence-based practice statistics

Answer: NNT = 55

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 = 0.14·0.13 = 0.0182

Number needed to treat

NNT = 1/0.0182 = 54.95 → 55

Answer

NNT = 55

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

3.Descriptive statistics

Answer: s = 15.85

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 = 189/5 = 37.8

Squared deviations

Σ(x − x̄)² = (−24.8)² + 1.2² + (−2.8)² + 17.2² + 9.2² = 1005

Standard deviation

s = √(1005/4) = 15.85

Answer

s = 15.85
4.Descriptive statistics

Answer: z = 1.25

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 = (72.5 − 60)/10 = 1.25

Answer

z = 1.25
5.Confidence intervals

Answer: E = 6.153

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.

Critical value

df = n − 1 = 24, t = 2.797

Margin of error

E = 2.797·11/√25 = 6.153

Answer

E = 6.153
6.Confidence intervals

Answer: n = 9085

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 = (2.576·37/1)² = 9084

Round up

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

n = 9085

Answer

n = 9085
7.Hypothesis tests & confidence intervals

Answer: t = 0.9991

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 of the mean difference

SE = 6.9/√12 = 1.992

Test statistic

t = 1.99/1.992 = 0.9991

Answer

t = 0.9991

With df = 11, the two-sided 5% critical value is 2.201; |t| is smaller, so H₀ is not rejected.

8.Hypothesis tests & confidence intervals

Answer: β = 70.71 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 5/√25 = 1
x̄c = 50 − 1.645·1 = 48.35

z under the true mean

z = (48.35 − 48.9)/1 = −0.545

Type II error

power = P(Z < −0.545) = 29.29%
β = 100% − 29.29% = 70.71%

Answer

β = 70.71%

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