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-yearsexposed3536200not exposed784100

Practice exercise (made-up values, not clinical guidance). In a cohort study, exposed people contributed 6200 person-years with 353 new cases, and unexposed people 4100 person-years with 78 new cases. Find the incidence rate ratio IRR (exposed versus unexposed). Round to two decimal places.

exposed: 353/6200 PYnot exposed: 78/4100 PY
IRR =
2.Evidence-based practice statisticsHard
CER29%EER?RRR 29%

Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 29% in relative terms (RRR = 29%). In a patient group the risk without treatment (CER) is 29% 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 = 29%CER = 29%
NNT =
3.Descriptive statisticsMedium
1015202530354045

Find the sample standard deviation s of the data: 10, 46, 26, 45, 33, 11.

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

A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is above 185?

μ = 170σ = 15x = 185
P =
5.Confidence intervalsMedium
n = 11, s = 15.5LU = ?x̄ = 85.5E99%

A sample of 11 items has mean 85.5 and standard deviation s = 15.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 11x̄ = 85.5s = 15.599%t(10) = 3.169
upper limit =
6.Confidence intervalsHard
n = 11Lσ = ?s = 0.895%

The diameters of 11 machined shafts have a sample standard deviation of 0.8 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.

n = 11s = 0.8 µmχ² = 3.247, 20.483
σ = µm
7.Hypothesis tests & confidence intervalsMedium
−0.82p = ?

The mean of a population is claimed to be 500; the population standard deviation is known to be 8. A random sample of 64 observations has a mean of 499.18. Test H₀: μ = 500 against H₁: μ < 500. Find the p-value p.

μ₀ = 500σ = 8n = 64x̄ = 499.18H₁: μ < μ₀
p =
8.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 0.590%
n =

3Solutions

1.Evidence-based practice statistics

Answer: IRR = 2.99

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.

Incidence rates

exposed: 353/6200 = 56.94 per 1000 person-years
not exposed: 78/4100 = 19.02 per 1000 person-years

Rate ratio

IRR = 56.94/19.02 = 2.993 ≈ 2.99

Answer

IRR = 2.99
2.Evidence-based practice statistics

Answer: NNT = 12

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 = 0.29·0.29 = 0.0841

Number needed to treat

NNT = 1/0.0841 = 11.89 → 12

Answer

NNT = 12

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

3.Descriptive statistics

Answer: s = 15.83

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 = 171/6 = 28.5

Squared deviations

Σ(x − x̄)² = (−18.5)² + 17.5² + (−2.5)² + 16.5² + 4.5² + (−17.5)² = 1254

Standard deviation

s = √(1254/5) = 15.83

Answer

s = 15.83
4.Descriptive statistics

Answer: P = 0.1587

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

Standard normal table

Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
5.Confidence intervals

Answer: upper limit = 100.3

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 = 10, t = 3.169

Margin of error

E = 3.169·15.5/√11 = 14.81

Interval

85.5 ± 14.81 → [70.69, 100.3]

Answer

upper limit = 100.3
6.Confidence intervals

Answer: σ = 1.404 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 10·0.8² = 6.4

Limits for σ²

6.4/20.483 = 0.3125
6.4/3.247 = 1.971

Limit for σ

σ = √1.971 = 1.404 µm

Answer

σ = 1.404 µm

95% interval for σ: [0.559, 1.404] µm.

7.Hypothesis tests & confidence intervals

Answer: p = 0.2061

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.

Test statistic

SE = σ/√n = 8/√64 = 1
z = (499.18 − 500)/1 = −0.82

Tail area

Φ(0.82) = 0.7939
1 − Φ(0.82) = 0.2061

p-value

p = 0.2061

Answer

p = 0.2061

p > 0.05, so H₀ is not rejected at the 5% level.

8.Hypothesis tests & confidence intervals

Answer: n = 1559

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).

Solve for n

n = (1.645·12/0.5)² = 1559

Round up

n = 1559

Answer

n = 1559

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