In the print window, choose “Save as PDF” to keep a copy on your device.
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
Draw the 2 × 2 table and label a, b, c and d before you calculate.
Read sensitivity and specificity down the columns, predictive values along the rows.
For another prevalence, imagine 10 000 people and fill the table with Se, Sp and the prevalence.
For a likelihood ratio, convert the probability to odds, multiply, and convert back.
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
Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
Find the critical value for the confidence level.
Margin of error: E = critical value × standard deviation/√n.
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
State H₀ and H₁ and decide whether the test is one- or two-sided.
Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
Compare with the critical value, or find the p-value.
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
Practice exercise (made-up values, not clinical guidance). A case-control study compares 320 patients with a disease (cases) and 130 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 = 122b = 22c = 198d = 108
OR =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 84% and a specificity of 93%. It is used in a population where the prevalence is 15%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 84%Sp = 93%p = 15%
PPV = %
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 23, 30, 23, 16, 52, 16.
n = 6
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is below 46?
μ = 70σ = 20x = 46
P =
5.Confidence intervalsMedium
The blood pressure of 6 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.2 and standard deviation 3.4. Find the lower limit of the 95% confidence interval for the mean difference.
n = 6d̄ = 6.2sd = 3.495%t(5) = 2.571
lower limit = mmHg
6.Confidence intervalsHard
The population standard deviation is 31. 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 = 31E = 1.599%
n =
7.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 6 students, mean score 42, standard deviation 9. Group 2: 8 students, mean score 45.7, standard deviation 3. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 6x̄₁ = 42s₁ = 9n₂ = 8x̄₂ = 45.7s₂ = 3
t =
8.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 80 against H₁: μ > 80 is done at α = 0.01 with a sample of 100; the standard deviation is known to be 10. In reality the mean is 81.7. Find the probability of a Type II error β in percent.
μ₀ = 80μ = 81.7σ = 10n = 100α = 0.01
β = %
3Solutions
1.Evidence-based practice statistics
Answer: OR = 3.02
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: 122/198 = 0.6162
controls: 22/108 = 0.2037
Odds ratio
OR = (122·108)/(22·198) = 3.025 ≈ 3.02
Answer
OR = 3.02
OR above 1: exposure is more common among the cases.
2.Evidence-based practice statistics
Answer: PPV = 67.9 %
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 = 0.84·0.15 = 0.126
(1 − Sp)·(1 − p) = 0.07·0.85 = 0.0595
Positive predictive value
PPV = 0.126/(0.126 + 0.0595) = 0.6792 ≈ 67.9%
Answer
PPV = 67.9%
The lower the prevalence, the lower the PPV of the same test.
3.Descriptive statistics
Answer: s = 13.47
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.
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 = 0.1151
Answer
P = 0.1151
5.Confidence intervals
Answer: lower limit = 2.631 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 = 5, t = 2.571
Margin of error
E = 2.571·3.4/√6 = 3.569
Interval
6.2 ± 3.569 → [2.631, 9.769]
Answer
lower limit = 2.631 mmHg
6.Confidence intervals
Answer: n = 2835
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 = (2.576·31/1.5)² = 2834
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 2835
Answer
n = 2835
7.Hypothesis tests & confidence intervals
Answer: t = −1.097
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.
Pooled variance
sp² = (5·9² + 7·3²)/12 = 39
sp = 6.245
Standard error
SE = 6.245·√(1/6 + 1/8) = 3.373
Test statistic
t = (42 − 45.7)/3.373 = −1.097
Answer
t = −1.097
With df = 12, the two-sided 5% critical value is 2.179; |t| is smaller, so H₀ is not rejected.
8.Hypothesis tests & confidence intervals
Answer: β = 73.43 %
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).