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
casescontrolsexposed560abnot exposed115540cd

Practice exercise (made-up values, not clinical guidance). A case-control study compares 120 patients with a disease (cases) and 600 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 = 5b = 60c = 115d = 540
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 79% and specificity 92%. The prevalence of the disease is 24%. How many positive results n₊ are expected in total?

N = 10 000p = 24%Se = 79%Sp = 92%
n₊ =
3.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 46, 43, 38, 21, 56, 50.

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

Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 50.

μ = 70σ = 20x = 50
z =
5.Confidence intervalsMedium
n = 8, s = 20L = ?Ux̄ = 36.5E95%

A sample of 8 items has mean 36.5 and standard deviation s = 20. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 8x̄ = 36.5s = 2095%t(7) = 2.365
lower limit =
6.Confidence intervalsHard
μ₁ − μ₂, t(20) = 2.0860LUx̄₁ − x̄₂E = ?95%

Two fertilisers are tested. Plot group 1: 14 plots, mean yield 36.5 kg, standard deviation 5 kg. Plot group 2: 8 plots, mean 42.5 kg, standard deviation 2.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 14x̄₁ = 36.5s₁ = 5n₂ = 8x̄₂ = 42.5s₂ = 2.595%t(20) = 2.086
E = kg
7.Hypothesis tests & confidence intervalsMedium
−2.1312.131t = ?df = 15

16 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 3.98 and the standard deviation of the differences is 5.9. Find the paired test statistic t for H₀: μd = 0.

n = 16d̄ = 3.98sd = 5.9
t =
8.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.05βσ = 30, power 80%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 3 units with a power of 80%. The standard deviation is 30. Find the minimum sample size n.

α = 0.05power = 80%δ = 3σ = 30zα = 1.645zβ = 0.842
n =

3Solutions

1.Evidence-based practice statistics

Answer: OR = 0.39

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: 5/115 = 0.04348
controls: 60/540 = 0.1111

Odds ratio

OR = (5·540)/(60·115) = 0.3913 ≈ 0.39

Answer

OR = 0.39

OR below 1: exposure is less common among the cases, which suggests a protective effect.

2.Evidence-based practice statistics

Answer: n₊ = 2504

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

10 000·0.24 = 2400
10 000 − 2400 = 7600

All positive results

TP = 2400·0.79 = 1896
FP = 7600·0.08 = 608
n₊ = 1896 + 608 = 2504

Answer

n₊ = 2504

Of the 2504 positives only 1896 have the disease (PPV 75.7%).

3.Descriptive statistics

Answer: s = 12.11

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 = 254/6 = 42.33

Squared deviations

Σ(x − x̄)² = 3.667² + 0.6667² + (−4.333)² + (−21.33)² + 13.67² + 7.667² = 733.3

Standard deviation

s = √(733.3/5) = 12.11

Answer

s = 12.11
4.Descriptive statistics

Answer: z = −1

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 = (50 − 70)/20 = −1

Answer

z = −1
5.Confidence intervals

Answer: lower limit = 19.78

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 = 7, t = 2.365

Margin of error

E = 2.365·20/√8 = 16.72

Interval

36.5 ± 16.72 → [19.78, 53.22]

Answer

lower limit = 19.78
6.Confidence intervals

Answer: E = 3.97 kg

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.

Pooled standard deviation

sp² = (13·5² + 7·2.5²)/20 = 18.44
sp = 4.294

Margin of error

E = 2.086·4.294·√(1/14 + 1/8) = 3.97

Answer

E = 3.97 kg
7.Hypothesis tests & confidence intervals

Answer: t = 2.698

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 = 5.9/√16 = 1.475

Test statistic

t = 3.98/1.475 = 2.698

Answer

t = 2.698

With df = 15, the two-sided 5% critical value is 2.131; |t| is larger, so H₀ is rejected.

8.Hypothesis tests & confidence intervals

Answer: n = 619

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 + 0.842)·30/3)² = 618.5

Round up

n = 619

Answer

n = 619

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