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
eventno eventtotaltreatment33207240abcontrol166554720cd

Practice exercise (made-up values, not clinical guidance). In a randomised trial, 240 patients received a new treatment and 720 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the relative risk RR of the event with treatment. Round to two decimal places.

treatment: 33/240control: 166/720
RR =
2.Evidence-based practice statisticsHard
p (%)NPV (%)20%NPV = ?100

Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 80% and a specificity of 94%. It is used in a population where the prevalence is 20%. Find the negative predictive value NPV as a percentage. Round to one decimal place.

Se = 80%Sp = 94%p = 20%
NPV = %
3.Descriptive statisticsMedium
10152025303540455055

Find the sample standard deviation s of the data: 10, 14, 48, 32, 45, 48, 55.

n = 7
s =
4.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 60.4.

μ = 70σ = 8x = 60.4
z =
5.Confidence intervalsMedium
n = 21, s = 20L = ?Ux̄ = 38E90%

A sample of 21 items has mean 38 and standard deviation s = 20. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 21x̄ = 38s = 2090%t(20) = 1.725
lower limit =
6.Confidence intervalsHard
σ = 29, n = ?x̄ − Ex̄ + Ex̄E = 590%

The population standard deviation is 29. 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 5?

σ = 29E = 590%
n =
7.Hypothesis tests & confidence intervalsMedium
−2.0452.045t = ?df = 29

Two teaching methods are compared. Group 1: 13 students, mean score 90.5, standard deviation 12. Group 2: 18 students, mean score 87.8, standard deviation 3.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 13x̄₁ = 90.5s₁ = 12n₂ = 18x̄₂ = 87.8s₂ = 3.5
t =
8.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 1.5α = 0.01βσ = 6, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 1.5 units with a power of 90%. The standard deviation is 6. Find the minimum sample size n.

α = 0.01power = 90%δ = 1.5σ = 6zα = 2.326zβ = 1.282
n =

3Solutions

1.Evidence-based practice statistics

Answer: RR = 0.6

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.

Event rates

EER = 33/240 = 0.1375
CER = 166/720 = 0.2306

Relative risk

RR = 0.1375/0.2306 = 0.5964 ≈ 0.6

Answer

RR = 0.6
2.Evidence-based practice statistics

Answer: NPV = 94.9 %

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.

True and false negatives (per person)

Sp·(1 − p) = 0.94·0.8 = 0.752
(1 − Se)·p = 0.2·0.2 = 0.04

Negative predictive value

NPV = 0.752/(0.752 + 0.04) = 0.9495 ≈ 94.9%

Answer

NPV = 94.9%
3.Descriptive statistics

Answer: s = 17.82

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 = 252/7 = 36

Squared deviations

Σ(x − x̄)² = (−26)² + (−22)² + 12² + (−4)² + 9² + 12² + 19² = 1906

Standard deviation

s = √(1906/6) = 17.82

Answer

s = 17.82
4.Descriptive statistics

Answer: z = −1.2

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 = (60.4 − 70)/8 = −1.2

Answer

z = −1.2
5.Confidence intervals

Answer: lower limit = 30.47

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 = 20, t = 1.725

Margin of error

E = 1.725·20/√21 = 7.529

Interval

38 ± 7.529 → [30.47, 45.53]

Answer

lower limit = 30.47
6.Confidence intervals

Answer: n = 92

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 = (1.645·29/5)² = 91.03

Round up

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

n = 92

Answer

n = 92
7.Hypothesis tests & confidence intervals

Answer: t = 0.9078

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Pooled variance

sp² = (12·12² + 17·3.5²)/29 = 66.77
sp = 8.171

Standard error

SE = 8.171·√(1/13 + 1/18) = 2.974

Test statistic

t = (90.5 − 87.8)/2.974 = 0.9078

Answer

t = 0.9078

With df = 29, the two-sided 5% critical value is 2.045; |t| is smaller, so H₀ is not rejected.

8.Hypothesis tests & confidence intervals

Answer: n = 209

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 = ((2.326 + 1.282)·6/1.5)² = 208.3

Round up

n = 209

Answer

n = 209

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