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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 cohort study, exposed people contributed 5100 person-years with 260 new cases, and unexposed people 900 person-years with 15 new cases. Find the incidence rate ratio IRR (exposed versus unexposed). Round to two decimal places.
exposed: 260/5100 PYnot exposed: 15/900 PY
IRR =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 96% and specificity 96%. The prevalence of the disease is 3%. How many false-positive results FP are expected?
N = 10 000p = 3%Se = 96%Sp = 96%
FP =
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 39, 54, 40, 59, 11.
n = 5
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is above 167.5?
μ = 170σ = 5x = 167.5
P =
5.Confidence intervalsMedium
A sample of 25 items has mean 98 and standard deviation s = 2. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 25x̄ = 98s = 290%t(24) = 1.711
lower limit =
6.Confidence intervalsHard
The diameters of 10 machined shafts have a sample standard deviation of 7.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.
n = 10s = 7.8 µmχ² = 2.7, 19.023
σ = µm
7.Hypothesis tests & confidence intervalsMedium
9 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 9.28 and the standard deviation of the differences is 8.7. Find the paired test statistic t for H₀: μd = 0.
n = 9d̄ = 9.28sd = 8.7
t =
8.Hypothesis tests & confidence intervalsHard
In a trial, 38 of 150 patients on treatment A recover, and 39 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 38n₁ = 150x₂ = 39n₂ = 300
z =
3Solutions
1.Evidence-based practice statistics
Answer: IRR = 3.06
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: 260/5100 = 50.98 per 1000 person-years
not exposed: 15/900 = 16.67 per 1000 person-years
Rate ratio
IRR = 50.98/16.67 = 3.059 ≈ 3.06
Answer
IRR = 3.06
2.Evidence-based practice statistics
Answer: FP = 388
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
10 000·0.03 = 300
10 000 − 300 = 9700
False positives
FP = 9700·0.04 = 388
Answer
FP = 388
Of the 676 positives only 288 have the disease (PPV 42.6%).
3.Descriptive statistics
Answer: s = 18.69
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915
Answer
P = 0.6915
5.Confidence intervals
Answer: lower limit = 97.32
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 = 24, t = 1.711
Margin of error
E = 1.711·2/√25 = 0.6844
Interval
98 ± 0.6844 → [97.32, 98.68]
Answer
lower limit = 97.32
6.Confidence intervals
Answer: σ = 5.365 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 = 9·7.8² = 547.6
Limits for σ²
547.6/19.023 = 28.78
547.6/2.7 = 202.8
Limit for σ
σ = √28.78 = 5.365 µm
Answer
σ = 5.365 µm
95% interval for σ: [5.365, 14.24] µm.
7.Hypothesis tests & confidence intervals
Answer: t = 3.2
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 of the mean difference
SE = 8.7/√9 = 2.9
Test statistic
t = 9.28/2.9 = 3.2
Answer
t = 3.2
With df = 8, the two-sided 5% critical value is 2.306; |t| is larger, so H₀ is rejected.
8.Hypothesis tests & confidence intervals
Answer: z = 3.275
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.
Sample proportions
p̂₁ = 38/150 = 0.2533
p̂₂ = 39/300 = 0.13
p̂ = 77/450 = 0.1711
Standard error under H₀
SE = √(0.1711·0.8289·(1/150 + 1/300)) = 0.03766
Test statistic
z = (0.2533 − 0.13)/0.03766 = 3.275
Answer
z = 3.275
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).