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
eventno eventtotaltreatment103847950abcontrol257693950cd

Practice exercise (made-up values, not clinical guidance). In a randomised trial, 950 patients received a new treatment and 950 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the relative risk reduction RRR as a percentage. Round to one decimal place.

treatment: 103/950control: 257/950
RRR = %
2.Evidence-based practice statisticsHard
pre-test50%post-test?p′LR−

Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 50% for a disease. A test with sensitivity 82% and specificity 92% comes back negative. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.

pre-test 50%Se = 82%Sp = 92%negative
p′ = %
3.Descriptive statisticsMedium
26283032343638404244

Find the sample standard deviation s of the data: 31, 29, 42, 30, 45, 25.

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

A quantity is normally distributed with mean 50 and standard deviation 20. What proportion P of values is below 80?

μ = 50σ = 20x = 80
P =
5.Confidence intervalsMedium
n = 12, sd​ = 3.60LU = ?d̄ = 1E90%

The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 1 and standard deviation 3.6. Find the upper limit of the 90% confidence interval for the mean difference.

n = 12d̄ = 1sd = 3.690%t(11) = 1.796
upper limit = mmHg
6.Confidence intervalsHard
σ = 16, n = ?x̄ − Ex̄ + Ex̄E = 2.599%

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 2.5?

σ = 16E = 2.599%
n =
7.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 199.1 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 199.1s = 5n = 25
t =
8.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ > 100 is done at α = 0.01 with a sample of 36; the standard deviation is known to be 12. In reality the mean is 108. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 108σ = 12n = 36α = 0.01
1 − β = %

3Solutions

1.Evidence-based practice statistics

Answer: RRR = 59.9 %

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.

Event rates

EER = 103/950 = 0.1084
CER = 257/950 = 0.2705

Relative risk reduction

ARR = 0.2705 − 0.1084 = 0.1621
RRR = 0.1621/0.2705 = 0.5992 ≈ 59.9%

Answer

RRR = 59.9%
2.Evidence-based practice statistics

Answer: p′ = 16.4 %

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.

Likelihood ratio

LR− = (1 − 0.82)/0.92 = 0.1957

Pre-test odds

0.5/(1 − 0.5) = 1

Post-test odds

1·0.1957 = 0.1957

Post-test probability

p′ = 0.1957/(1 + 0.1957) = 0.1636 ≈ 16.4%

Answer

p′ = 16.4%
3.Descriptive statistics

Answer: s = 7.941

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 = 202/6 = 33.67

Squared deviations

Σ(x − x̄)² = (−2.667)² + (−4.667)² + 8.333² + (−3.667)² + 11.33² + (−8.667)² = 315.3

Standard deviation

s = √(315.3/5) = 7.941

Answer

s = 7.941
4.Descriptive statistics

Answer: P = 0.9332

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 0.9332

Answer

P = 0.9332
5.Confidence intervals

Answer: upper limit = 2.866 mmHg

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.

Critical value

df = 11, t = 1.796

Margin of error

E = 1.796·3.6/√12 = 1.866

Interval

1 ± 1.866 → [−0.8665, 2.866]

Answer

upper limit = 2.866 mmHg
6.Confidence intervals

Answer: n = 272

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 = (2.576·16/2.5)² = 271.8

Round up

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

n = 272

Answer

n = 272
7.Hypothesis tests & confidence intervals

Answer: t = −0.9

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 = (199.1 − 200)/1 = −0.9

Answer

t = −0.9
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: 1 − β = 95.29 %

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 = 12/√36 = 2
x̄c = 100 + 2.326·2 = 104.7

z under the true mean

z = (104.7 − 108)/2 = −1.674

Power

power = P(Z > −1.674) = 95.29%
1 − β = 95.29%

Answer

1 − β = 95.29%

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