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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, 900 patients received a new treatment and 450 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the relative risk RR of the event with treatment. Round to two decimal places.
treatment: 74/900control: 50/450
RR =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 46% in relative terms (RRR = 46%). In a patient group the risk without treatment (CER) is 25% over five years. Find the number needed to treat NNT for five years to prevent one event. Round up to a whole number of patients.
RRR = 46%CER = 25%
NNT =
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 51, 11, 57, 18, 17, 31, 38.
n = 7
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is below 150?
μ = 170σ = 10x = 150
P =
5.Confidence intervalsMedium
The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 8.3 and standard deviation 2.3. Find the upper limit of the 90% confidence interval for the mean difference.
n = 12d̄ = 8.3sd = 2.390%t(11) = 1.796
upper limit = mmHg
6.Confidence intervalsHard
The diameters of 21 machined shafts have a sample standard deviation of 6.4 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.
n = 21s = 6.4 µmχ² = 9.591, 34.17
σ = µm
7.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 100; the population standard deviation is known to be 6. A random sample of 144 observations has a mean of 101.07. Test H₀: μ = 100 against H₁: μ > 100. Find the p-value p.
μ₀ = 100σ = 6n = 144x̄ = 101.07H₁: μ > μ₀
p =
8.Hypothesis tests & confidence intervalsHard
In a trial, 16 of 80 patients on treatment A recover, and 36 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 16n₁ = 80x₂ = 36n₂ = 150
z =
3Solutions
1.Evidence-based practice statistics
Answer: RR = 0.74
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 = 74/900 = 0.08222
CER = 50/450 = 0.1111
Relative risk
RR = 0.08222/0.1111 = 0.74
Answer
RR = 0.74
2.Evidence-based practice statistics
Answer: NNT = 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.
Absolute risk reduction
ARR = 0.25·0.46 = 0.115
Number needed to treat
NNT = 1/0.115 = 8.696 → 9
Answer
NNT = 9
The same relative effect gives a smaller NNT when the baseline risk is higher.
3.Descriptive statistics
Answer: s = 17.72
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 0.02275
Answer
P = 0.02275
5.Confidence intervals
Answer: upper limit = 9.492 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 = 11, t = 1.796
Margin of error
E = 1.796·2.3/√12 = 1.192
Interval
8.3 ± 1.192 → [7.108, 9.492]
Answer
upper limit = 9.492 mmHg
6.Confidence intervals
Answer: σ = 4.896 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 = 20·6.4² = 819.2
Limits for σ²
819.2/34.17 = 23.97
819.2/9.591 = 85.41
Limit for σ
σ = √23.97 = 4.896 µm
Answer
σ = 4.896 µm
95% interval for σ: [4.896, 9.242] µm.
7.Hypothesis tests & confidence intervals
Answer: p = 0.01618
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 = 6/√144 = 0.5
z = (101.07 − 100)/0.5 = 2.14
Tail area
Φ(2.14) = 0.9838
1 − Φ(2.14) = 0.0162
p-value
p = 0.01618
Answer
p = 0.01618
p ≤ 0.05, so H₀ is rejected at the 5% level.
8.Hypothesis tests & confidence intervals
Answer: z = −0.6907
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 = 16/80 = 0.2
p̂₂ = 36/150 = 0.24
p̂ = 52/230 = 0.2261
Standard error under H₀
SE = √(0.2261·0.7739·(1/80 + 1/150)) = 0.05791
Test statistic
z = (0.2 − 0.24)/0.05791 = −0.6907
Answer
z = −0.6907
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).