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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
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). In a randomised trial, 940 patients received a new treatment and 260 received the control treatment. The table shows how many in each group had a side effect. Find the number needed to harm NNH: how many patients must be treated for one extra patient to have the side effect? Round up to a whole number of patients.
treatment: 211/940control: 39/260
NNH =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 78% and specificity 88%. The prevalence of the disease is 24%. How many positive results n₊ are expected in total?
N = 10 000p = 24%Se = 78%Sp = 88%
n₊ =
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 54, 19, 40, 58, 37, 24, 19.
n = 7
s =
4.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 480.
μ = 500σ = 25x = 480
z =
5.Confidence intervalsMedium
The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 3.4 and standard deviation 1.8. Find the margin of error E of the 90% confidence interval for the mean difference.
n = 10d̄ = 3.4sd = 1.890%t(9) = 1.833
E = mmHg
6.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 6 plots, mean yield 83 kg, standard deviation 9.5 kg. Plot group 2: 6 plots, mean 88 kg, standard deviation 6 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.
Two teaching methods are compared. Group 1: 13 students, mean score 101, standard deviation 3. Group 2: 18 students, mean score 107.8, standard deviation 8. 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 = 13x̄₁ = 101s₁ = 3n₂ = 18x̄₂ = 107.8s₂ = 8
t =
8.Hypothesis tests & confidence intervalsHard
A sample of 36 measurements has mean 53. The population standard deviation is 20. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 53σ = 20n = 3690%
L =
3Solutions
1.Evidence-based practice statistics
Answer: NNH = 14
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 = 211/940 = 0.2245
CER = 39/260 = 0.15
Number needed to harm
ARI = 0.2245 − 0.15 = 0.07447
NNH = 1/0.07447 = 13.43 → 14
Answer
NNH = 14
About one in every 14 treated patients has the side effect because of the treatment.
2.Evidence-based practice statistics
Answer: n₊ = 2784
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.
With and without the disease
10 000·0.24 = 2400
10 000 − 2400 = 7600
All positive results
TP = 2400·0.78 = 1872
FP = 7600·0.12 = 912
n₊ = 1872 + 912 = 2784
Answer
n₊ = 2784
Of the 2784 positives only 1872 have the disease (PPV 67.2%).
3.Descriptive statistics
Answer: s = 16.06
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 = (480 − 500)/25 = −0.8
Answer
z = −0.8
5.Confidence intervals
Answer: E = 1.043 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 = 9, t = 1.833
Margin of error
E = 1.833·1.8/√10 = 1.043
Answer
E = 1.043 mmHg
6.Confidence intervals
Answer: E = 8.312 kg
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.
Pooled standard deviation
sp² = (5·9.5² + 5·6²)/10 = 63.13
sp = 7.945
Margin of error
E = 1.812·7.945·√(1/6 + 1/6) = 8.312
Answer
E = 8.312 kg
7.Hypothesis tests & confidence intervals
Answer: t = −2.909
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 = (12·3² + 17·8²)/29 = 41.24
sp = 6.422
Standard error
SE = 6.422·√(1/13 + 1/18) = 2.337
Test statistic
t = (101 − 107.8)/2.337 = −2.909
Answer
t = −2.909
With df = 29, the two-sided 5% critical value is 2.045; |t| is larger, so H₀ is rejected.
8.Hypothesis tests & confidence intervals
Answer: L = 47.52
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).