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
casescontrolsexposed301187abnot exposed99153cd

Practice exercise (made-up values, not clinical guidance). A case-control study compares 400 patients with a disease (cases) and 340 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 = 301b = 187c = 99d = 153
OR =
2.Evidence-based practice statisticsHard
control29%treatment?

Practice exercise (made-up values, not clinical guidance). Without a drug, 29% of patients have a certain side effect during a year. The drug raises this risk with a relative risk of RR = 1.66. Find the number needed to harm NNH: how many patients must take the drug for a year for one extra patient to have the side effect? Round up to a whole number of patients.

control 29%RR = 1.66
NNH =
3.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 58, 50, 21, 30, 49.

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

Exam scores are normally distributed with mean 50 and standard deviation 15. Find the z-score z of a score of 27.5.

μ = 50σ = 15x = 27.5
z =
5.Confidence intervalsMedium
x̄E₁ = 8.695%, n = 36x̄E₂ = ?95%, n = 72

A 95% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 8.6. What is the margin of error E₂ if the same study uses 72 observations instead?

95%n₁ = 36E₁ = 8.6n₂ = 72
E₂ =
6.Confidence intervalsHard
σ = 30, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

The population standard deviation is 30. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 0.5?

σ = 30E = 0.590%
n =
7.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 77.1 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 77.1s = 8n = 64
t =
8.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2α = 0.05βσ = 15, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 2 units with a power of 95%. The standard deviation is 15. Find the minimum sample size n.

α = 0.05power = 95%δ = 2σ = 15zα = 1.645zβ = 1.645
n =

3Solutions

1.Evidence-based practice statistics

Answer: OR = 2.49

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: 301/99 = 3.04
controls: 187/153 = 1.222

Odds ratio

OR = (301·153)/(187·99) = 2.488 ≈ 2.49

Answer

OR = 2.49

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

2.Evidence-based practice statistics

Answer: NNH = 6

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.

Risk with the drug

0.29·1.66 = 0.4814

Absolute risk increase

ARI = 0.4814 − 0.29 = 0.1914

Number needed to harm

NNH = 1/0.1914 = 5.225 → 6

Answer

NNH = 6
3.Descriptive statistics

Answer: s = 15.44

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 = 208/5 = 41.6

Squared deviations

Σ(x − x̄)² = 16.4² + 8.4² + (−20.6)² + (−11.6)² + 7.4² = 953.2

Standard deviation

s = √(953.2/4) = 15.44

Answer

s = 15.44
4.Descriptive statistics

Answer: z = −1.5

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 = (27.5 − 50)/15 = −1.5

Answer

z = −1.5
5.Confidence intervals

Answer: E₂ = 6.081

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 = (1.96/1.96)·√(36/72) = 0.7071

New margin

E₂ = 8.6·0.7071 = 6.081

Answer

E₂ = 6.081
6.Confidence intervals

Answer: n = 9742

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 = (1.645·30/0.5)² = 9742

Round up

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

n = 9742

Answer

n = 9742
7.Hypothesis tests & confidence intervals

Answer: t = 2.1

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 = 8/√64 = 8/8 = 1

Test statistic

t = (77.1 − 75)/1 = 2.1

Answer

t = 2.1
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

8.Hypothesis tests & confidence intervals

Answer: n = 609

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 1.645)·15/2)² = 608.9

Round up

n = 609

Answer

n = 609

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