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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, 970 patients received a new treatment and 880 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: 290/970control: 202/880
NNH =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 25% for a disease. A test with sensitivity 98% and specificity 81% comes back negative. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 25%Se = 98%Sp = 81%negative
p′ = %
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 49, 17, 36, 29, 44.
n = 5
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is below 155.6?
μ = 170σ = 12x = 155.6
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 0.8 and standard deviation 1.1. Find the upper limit of the 90% confidence interval for the mean difference.
n = 6d̄ = 0.8sd = 1.190%t(5) = 2.015
upper limit = mmHg
6.Confidence intervalsHard
The diameters of 10 machined shafts have a sample standard deviation of 7.9 µm. Assuming the diameters are normally distributed, find the upper 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 = 7.9 µmχ² = 2.7, 19.023
σ = µm
7.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 120; the population standard deviation is known to be 7.5. A random sample of 25 observations has a mean of 122.94. Test H₀: μ = 120 against H₁: μ ≠ 120. Find the p-value p.
μ₀ = 120σ = 7.5n = 25x̄ = 122.94H₁: μ ≠ μ₀
p =
8.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 1 units with a power of 80%. The standard deviation is 8. Find the minimum sample size n.
α = 0.05power = 80%δ = 1σ = 8zα = 1.645zβ = 0.842
n =
3Solutions
1.Evidence-based practice statistics
Answer: NNH = 15
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 = 290/970 = 0.299
CER = 202/880 = 0.2295
Number needed to harm
ARI = 0.299 − 0.2295 = 0.06942
NNH = 1/0.06942 = 14.4 → 15
Answer
NNH = 15
About one in every 15 treated patients has the side effect because of the treatment.
2.Evidence-based practice statistics
Answer: p′ = 0.8 %
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 − 0.98)/0.81 = 0.02469
Pre-test odds
0.25/(1 − 0.25) = 0.3333
Post-test odds
0.3333·0.02469 = 0.00823
Post-test probability
p′ = 0.00823/(1 + 0.00823) = 0.008163 ≈ 0.8%
Answer
p′ = 0.8%
3.Descriptive statistics
Answer: s = 12.63
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 = 175/5 = 35
Squared deviations
Σ(x − x̄)² = 14² + (−18)² + 1² + (−6)² + 9² = 638
Standard deviation
s = √(638/4) = 12.63
Answer
s = 12.63
4.Descriptive statistics
Answer: P = 0.1151
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: upper limit = 1.705 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 = 2.015
Margin of error
E = 2.015·1.1/√6 = 0.9049
Interval
0.8 ± 0.9049 → [−0.1049, 1.705]
Answer
upper limit = 1.705 mmHg
6.Confidence intervals
Answer: σ = 14.42 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 = 9·7.9² = 561.7
Limits for σ²
561.7/19.023 = 29.53
561.7/2.7 = 208
Limit for σ
σ = √208 = 14.42 µm
Answer
σ = 14.42 µm
95% interval for σ: [5.434, 14.42] µm.
7.Hypothesis tests & confidence intervals
Answer: p = 0.05
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.
Test statistic
SE = σ/√n = 7.5/√25 = 1.5
z = (122.94 − 120)/1.5 = 1.96
Tail area
Φ(1.96) = 0.9750
1 − Φ(1.96) = 0.0250
p-value
p = 2·0.0250 = 0.05
Answer
p = 0.05
p ≤ 0.05, so H₀ is rejected at the 5% level.
8.Hypothesis tests & confidence intervals
Answer: n = 396
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.