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
casescontrolsexposed227167abnot exposed93273cd

Practice exercise (made-up values, not clinical guidance). A case-control study compares 320 patients with a disease (cases) and 440 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 = 227b = 167c = 93d = 273
OR =
2.Evidence-based practice statisticsHard
testdisease presentdisease absenttotalpositive?negativetotal10 000

Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 80% and specificity 76%. The prevalence of the disease is 15%. How many positive results n₊ are expected in total?

N = 10 000p = 15%Se = 80%Sp = 76%
n₊ =
3.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 11, 20, 38, 24, 57, 13.

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

A quantity is normally distributed with mean 60 and standard deviation 20. What proportion P of values is above 68?

μ = 60σ = 20x = 68
P =
5.Confidence intervalsMedium
n = 13, sd​ = 2.30LU = ?d̄ = 5.9E99%

The blood pressure of 13 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 5.9 and standard deviation 2.3. Find the upper limit of the 99% confidence interval for the mean difference.

n = 13d̄ = 5.9sd = 2.399%t(12) = 3.055
upper limit = mmHg
6.Confidence intervalsHard
σ = 17, n = ?x̄ − Ex̄ + Ex̄E = 2.590%

The population standard deviation is 17. 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 2.5?

σ = 17E = 2.590%
n =
7.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 117.8 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 117.8s = 12n = 25
t =
8.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.05βσ = 16, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 95%. The standard deviation is 16. Find the minimum sample size n.

α = 0.05power = 95%δ = 4σ = 16zα = 1.645zβ = 1.645
n =

3Solutions

1.Evidence-based practice statistics

Answer: OR = 3.99

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: 227/93 = 2.441
controls: 167/273 = 0.6117

Odds ratio

OR = (227·273)/(167·93) = 3.99

Answer

OR = 3.99

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

2.Evidence-based practice statistics

Answer: n₊ = 3240

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.

With and without the disease

10 000·0.15 = 1500
10 000 − 1500 = 8500

All positive results

TP = 1500·0.8 = 1200
FP = 8500·0.24 = 2040
n₊ = 1200 + 2040 = 3240

Answer

n₊ = 3240

Of the 3240 positives only 1200 have the disease (PPV 37%).

3.Descriptive statistics

Answer: s = 17.5

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.

Mean

x̄ = 163/6 = 27.17

Squared deviations

Σ(x − x̄)² = (−16.17)² + (−7.167)² + 10.83² + (−3.167)² + 29.83² + (−14.17)² = 1531

Standard deviation

s = √(1531/5) = 17.5

Answer

s = 17.5
4.Descriptive statistics

Answer: P = 0.3446

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446

Answer

P = 0.3446
5.Confidence intervals

Answer: upper limit = 7.849 mmHg

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Critical value

df = 12, t = 3.055

Margin of error

E = 3.055·2.3/√13 = 1.949

Interval

5.9 ± 1.949 → [3.951, 7.849]

Answer

upper limit = 7.849 mmHg
6.Confidence intervals

Answer: n = 126

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 = (1.645·17/2.5)² = 125.1

Round up

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

n = 126

Answer

n = 126
7.Hypothesis tests & confidence intervals

Answer: t = −0.9167

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 = 12/√25 = 12/5 = 2.4

Test statistic

t = (117.8 − 120)/2.4 = −0.9167

Answer

t = −0.9167
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

8.Hypothesis tests & confidence intervals

Answer: n = 174

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)·16/4)² = 173.2

Round up

n = 174

Answer

n = 174

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