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
casescontrolsexposed38110abnot exposed122180cd

Practice exercise (made-up values, not clinical guidance). A case-control study compares 160 patients with a disease (cases) and 290 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 = 38b = 110c = 122d = 180
OR =
2.Evidence-based practice statisticsHard
p (%)NPV (%)15%NPV = ?100

Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 70% and a specificity of 83%. It is used in a population where the prevalence is 15%. Find the negative predictive value NPV as a percentage. Round to one decimal place.

Se = 70%Sp = 83%p = 15%
NPV = %
3.Descriptive statisticsMedium
253035404550

Find the sample standard deviation s of the data: 23, 32, 52, 40, 39.

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

Exam scores are normally distributed with mean 500 and standard deviation 75. Find the z-score z of a score of 410.

μ = 500σ = 75x = 410
z =
5.Confidence intervalsMedium
n = 16, s = 8LUx̄ = 28.5E = ?90%

A sample of 16 items has mean 28.5 and standard deviation s = 8. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 16x̄ = 28.5s = 890%t(15) = 1.753
E =
6.Confidence intervalsHard
μ₁ − μ₂, t(19) = 1.7290L = ?Ux̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 5 plots, mean yield 60.5 kg, standard deviation 4.5 kg. Plot group 2: 16 plots, mean 52 kg, standard deviation 2.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 5x̄₁ = 60.5s₁ = 4.5n₂ = 16x̄₂ = 52s₂ = 2.590%t(19) = 1.729
lower limit = kg
7.Hypothesis tests & confidence intervalsMedium
−2.3062.306t = ?df = 8

9 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 1.69 and the standard deviation of the differences is 4.6. Find the paired test statistic t for H₀: μd = 0.

n = 9d̄ = 1.69sd = 4.6
t =
8.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2.5α = 0.01βσ = 21, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 2.5 units with a power of 80%. The standard deviation is 21. Find the minimum sample size n.

α = 0.01power = 80%δ = 2.5σ = 21zα = 2.326zβ = 0.842
n =

3Solutions

1.Evidence-based practice statistics

Answer: OR = 0.51

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: 38/122 = 0.3115
controls: 110/180 = 0.6111

Odds ratio

OR = (38·180)/(110·122) = 0.5097 ≈ 0.51

Answer

OR = 0.51

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

2.Evidence-based practice statistics

Answer: NPV = 94 %

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) = 0.83·0.85 = 0.7055
(1 − Se)·p = 0.3·0.15 = 0.045

Negative predictive value

NPV = 0.7055/(0.7055 + 0.045) = 0.94 ≈ 94%

Answer

NPV = 94%
3.Descriptive statistics

Answer: s = 10.71

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 = 186/5 = 37.2

Squared deviations

Σ(x − x̄)² = (−14.2)² + (−5.2)² + 14.8² + 2.8² + 1.8² = 458.8

Standard deviation

s = √(458.8/4) = 10.71

Answer

s = 10.71
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 = (410 − 500)/75 = −1.2

Answer

z = −1.2
5.Confidence intervals

Answer: E = 3.506

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 = 15, t = 1.753

Margin of error

E = 1.753·8/√16 = 3.506

Answer

E = 3.506
6.Confidence intervals

Answer: lower limit = 5.813 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 = (4·4.5² + 15·2.5²)/19 = 9.197
sp = 3.033

Margin of error

E = 1.729·3.033·√(1/5 + 1/16) = 2.687

Interval

8.5 ± 2.687 → [5.813, 11.19]

Answer

lower limit = 5.813 kg
7.Hypothesis tests & confidence intervals

Answer: t = 1.102

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 = 4.6/√9 = 1.533

Test statistic

t = 1.69/1.533 = 1.102

Answer

t = 1.102

With df = 8, the two-sided 5% critical value is 2.306; |t| is smaller, so H₀ is not rejected.

8.Hypothesis tests & confidence intervals

Answer: n = 709

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 + 0.842)·21/2.5)² = 708.2

Round up

n = 709

Answer

n = 709

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