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
casescontrolsexposed40242abnot exposed20198cd

Practice exercise (made-up values, not clinical guidance). A case-control study compares 60 patients with a disease (cases) and 440 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 = 40b = 242c = 20d = 198
OR =
2.Evidence-based practice statisticsHard
p (%)PPV (%)30%PPV = ?100

Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 77% and a specificity of 84%. It is used in a population where the prevalence is 30%. Find the positive predictive value PPV as a percentage. Round to one decimal place.

Se = 77%Sp = 84%p = 30%
PPV = %
3.Descriptive statisticsMedium
15202530354045

Find the sample standard deviation s of the data: 19, 12, 49, 20, 17, 20.

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

A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is above 42?

μ = 60σ = 15x = 42
P =
5.Confidence intervalsMedium
n = 9, s = 2L = ?Ux̄ = 54.5E90%

A sample of 9 items has mean 54.5 and standard deviation s = 2. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 9x̄ = 54.5s = 290%t(8) = 1.86
lower limit =
6.Confidence intervalsHard
σ = 16, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

The population standard deviation is 16. 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 1.5?

σ = 16E = 1.599%
n =
7.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 121 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 121s = 4n = 49
t =
8.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 25 measurements has mean 49.8. The population standard deviation is 4. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 49.8σ = 4n = 2595%
L =

3Solutions

1.Evidence-based practice statistics

Answer: OR = 1.64

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: 40/20 = 2
controls: 242/198 = 1.222

Odds ratio

OR = (40·198)/(242·20) = 1.636 ≈ 1.64

Answer

OR = 1.64

OR above 1: exposure is more common among the cases.

2.Evidence-based practice statistics

Answer: PPV = 67.3 %

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 positives (per person)

Se·p = 0.77·0.3 = 0.231
(1 − Sp)·(1 − p) = 0.16·0.7 = 0.112

Positive predictive value

PPV = 0.231/(0.231 + 0.112) = 0.6735 ≈ 67.3%

Answer

PPV = 67.3%

The lower the prevalence, the lower the PPV of the same test.

3.Descriptive statistics

Answer: s = 13.17

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 = 137/6 = 22.83

Squared deviations

Σ(x − x̄)² = (−3.833)² + (−10.83)² + 26.17² + (−2.833)² + (−5.833)² + (−2.833)² = 866.8

Standard deviation

s = √(866.8/5) = 13.17

Answer

s = 13.17
4.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
5.Confidence intervals

Answer: lower limit = 53.26

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 = 8, t = 1.86

Margin of error

E = 1.86·2/√9 = 1.24

Interval

54.5 ± 1.24 → [53.26, 55.74]

Answer

lower limit = 53.26
6.Confidence intervals

Answer: n = 756

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 = (2.576·16/1.5)² = 755

Round up

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

n = 756

Answer

n = 756
7.Hypothesis tests & confidence intervals

Answer: t = 1.75

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 = 4/√49 = 4/7 = 0.5714

Test statistic

t = (121 − 120)/0.5714 = 1.75

Answer

t = 1.75
t0.025; df = 48 = 2.011

|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).

8.Hypothesis tests & confidence intervals

Answer: L = 48.23

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).

Margin of error

E = 1.96·4/√25 = 1.568

Interval limit

L = 49.8 − 1.568 = 48.23

Answer

L = 48.23

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