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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). A case-control study compares 190 patients with a disease (cases) and 380 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 = 23b = 99c = 167d = 281
OR =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 40% for a disease. A test with sensitivity 86% and specificity 64% comes back negative. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 40%Se = 86%Sp = 64%negative
p′ = %
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 22, 59, 32, 43, 12, 53.
n = 6
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is above 56?
μ = 60σ = 5x = 56
P =
5.Confidence intervalsMedium
A sample of 13 items has mean 87.5 and standard deviation s = 14.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 13x̄ = 87.5s = 14.599%t(12) = 3.055
upper limit =
6.Confidence intervalsHard
The diameters of 15 machined shafts have a sample standard deviation of 5.5 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.
n = 15s = 5.5 µmχ² = 5.629, 26.119
σ = µm
7.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 100 parts has mean length 140 mm, and the line's known standard deviation is 7 mm. Line 2: a sample of 40 parts has mean 138 mm, known standard deviation 3 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A one-sided z test of H₀: μ = 80 against H₁: μ < 80 is done at α = 0.10 with a sample of 36; the standard deviation is known to be 9. In reality the mean is 77.45. Find the power 1 − β of the test in percent.
μ₀ = 80μ = 77.45σ = 9n = 36α = 0.10
1 − β = %
3Solutions
1.Evidence-based practice statistics
Answer: OR = 0.39
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: 23/167 = 0.1377
controls: 99/281 = 0.3523
Odds ratio
OR = (23·281)/(99·167) = 0.3909 ≈ 0.39
Answer
OR = 0.39
OR below 1: exposure is less common among the cases, which suggests a protective effect.
2.Evidence-based practice statistics
Answer: p′ = 12.7 %
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 − 0.86)/0.64 = 0.2188
Pre-test odds
0.4/(1 − 0.4) = 0.6667
Post-test odds
0.6667·0.2188 = 0.1458
Post-test probability
p′ = 0.1458/(1 + 0.1458) = 0.1273 ≈ 12.7%
Answer
p′ = 12.7%
3.Descriptive statistics
Answer: s = 18.17
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.
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 = −0.8
Standard normal table
Φ(−0.8) = 0.2119
P = 1 − 0.2119 = 0.7881
Answer
P = 0.7881
5.Confidence intervals
Answer: upper limit = 99.79
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 = 12, t = 3.055
Margin of error
E = 3.055·14.5/√13 = 12.29
Interval
87.5 ± 12.29 → [75.21, 99.79]
Answer
upper limit = 99.79
6.Confidence intervals
Answer: σ = 4.027 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 = 14·5.5² = 423.5
Limits for σ²
423.5/26.119 = 16.21
423.5/5.629 = 75.24
Limit for σ
σ = √16.21 = 4.027 µm
Answer
σ = 4.027 µm
95% interval for σ: [4.027, 8.674] µm.
7.Hypothesis tests & confidence intervals
Answer: z = 2.365
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.
Standard error of the difference
SE = √(7²/100 + 3²/40) = √0.715 = 0.8456
Test statistic
z = (140 − 138)/0.8456 = 2.365
Answer
z = 2.365
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
8.Hypothesis tests & confidence intervals
Answer: 1 − β = 66.2 %
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).