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
casesperson-yearsexposed16700not exposed483200

Practice exercise (made-up values, not clinical guidance). In a cohort study, exposed people contributed 700 person-years with 16 new cases, and unexposed people 3200 person-years with 48 new cases. Find the incidence rate ratio IRR (exposed versus unexposed). Round to two decimal places.

exposed: 16/700 PYnot exposed: 48/3200 PY
IRR =
2.Evidence-based practice statisticsHard
CER37%EER?RRR 38%

Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 38% in relative terms (RRR = 38%). In a patient group the risk without treatment (CER) is 37% over five years. Find the number needed to treat NNT for five years to prevent one event. Round up to a whole number of patients.

RRR = 38%CER = 37%
NNT =
3.Descriptive statisticsMedium
152025303540

Find the sample standard deviation s of the data: 26, 36, 28, 40, 15.

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

A quantity is normally distributed with mean 170 and standard deviation 20. What proportion P of values is below 195?

μ = 170σ = 20x = 195
P =
5.Confidence intervalsMedium
n = 25, s = 2.5LU = ?x̄ = 99.5E99%

A sample of 25 items has mean 99.5 and standard deviation s = 2.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 25x̄ = 99.5s = 2.599%t(24) = 2.797
upper limit =
6.Confidence intervalsHard
σ = 34, n = ?x̄ − Ex̄ + Ex̄E = 599%

The population standard deviation is 34. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 5?

σ = 34E = 599%
n =
7.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 121.2 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 121.2s = 5n = 25
t =
8.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ < 50 is done at α = 0.05 with a sample of 100; the standard deviation is known to be 10. In reality the mean is 46.5. Find the probability of a Type II error β in percent.

μ₀ = 50μ = 46.5σ = 10n = 100α = 0.05
β = %

3Solutions

1.Evidence-based practice statistics

Answer: IRR = 1.52

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.

Incidence rates

exposed: 16/700 = 22.86 per 1000 person-years
not exposed: 48/3200 = 15 per 1000 person-years

Rate ratio

IRR = 22.86/15 = 1.524 ≈ 1.52

Answer

IRR = 1.52
2.Evidence-based practice statistics

Answer: NNT = 8

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.

Absolute risk reduction

ARR = 0.37·0.38 = 0.1406

Number needed to treat

NNT = 1/0.1406 = 7.112 → 8

Answer

NNT = 8

The same relative effect gives a smaller NNT when the baseline risk is higher.

3.Descriptive statistics

Answer: s = 9.695

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 = 145/5 = 29

Squared deviations

Σ(x − x̄)² = (−3)² + 7² + (−1)² + 11² + (−14)² = 376

Standard deviation

s = √(376/4) = 9.695

Answer

s = 9.695
4.Descriptive statistics

Answer: P = 0.8944

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 0.8944

Answer

P = 0.8944
5.Confidence intervals

Answer: upper limit = 100.9

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 = 24, t = 2.797

Margin of error

E = 2.797·2.5/√25 = 1.399

Interval

99.5 ± 1.399 → [98.1, 100.9]

Answer

upper limit = 100.9
6.Confidence intervals

Answer: n = 307

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 = (2.576·34/5)² = 306.8

Round up

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

n = 307

Answer

n = 307
7.Hypothesis tests & confidence intervals

Answer: t = 1.2

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 = 5/√25 = 5/5 = 1

Test statistic

t = (121.2 − 120)/1 = 1.2

Answer

t = 1.2
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: β = 3.18 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 10/√100 = 1
x̄c = 50 − 1.645·1 = 48.35

z under the true mean

z = (48.35 − 46.5)/1 = 1.855

Type II error

power = P(Z < 1.855) = 96.82%
β = 100% − 96.82% = 3.18%

Answer

β = 3.18%

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