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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, 820 patients received a new treatment and 290 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: 31/820control: 15/290
RR =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 72% and a specificity of 82%. It is used in a population where the prevalence is 20%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 72%Sp = 82%p = 20%
PPV = %
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 47, 45, 46, 21, 49.
n = 5
s =
4.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 40. Find the z-score z of a score of 550.
μ = 500σ = 40x = 550
z =
5.Confidence intervalsMedium
The blood pressure of 20 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −2.9 and standard deviation 2.4. Find the upper limit of the 99% confidence interval for the mean difference.
n = 20d̄ = −2.9sd = 2.499%t(19) = 2.861
upper limit = mmHg
6.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 5 plots, mean yield 82.5 kg, standard deviation 2 kg. Plot group 2: 7 plots, mean 80.5 kg, standard deviation 2 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.
The mean of a population is claimed to be 100; the population standard deviation is known to be 24. A random sample of 36 observations has a mean of 95.92. Test H₀: μ = 100 against H₁: μ < 100. Find the p-value p.
μ₀ = 100σ = 24n = 36x̄ = 95.92H₁: μ < μ₀
p =
8.Hypothesis tests & confidence intervalsHard
In a trial, 60 of 100 patients on treatment A recover, and 147 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 60n₁ = 100x₂ = 147n₂ = 300
z =
3Solutions
1.Evidence-based practice statistics
Answer: RR = 0.73
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 = 31/820 = 0.0378
CER = 15/290 = 0.05172
Relative risk
RR = 0.0378/0.05172 = 0.7309 ≈ 0.73
Answer
RR = 0.73
2.Evidence-based practice statistics
Answer: PPV = 50 %
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.
True and false positives (per person)
Se·p = 0.72·0.2 = 0.144
(1 − Sp)·(1 − p) = 0.18·0.8 = 0.144
Positive predictive value
PPV = 0.144/(0.144 + 0.144) = 0.5 = 50%
Answer
PPV = 50%
The lower the prevalence, the lower the PPV of the same test.
3.Descriptive statistics
Answer: s = 11.61
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 = (550 − 500)/40 = 1.25
Answer
z = 1.25
5.Confidence intervals
Answer: upper limit = −1.365 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 = 19, t = 2.861
Margin of error
E = 2.861·2.4/√20 = 1.535
Interval
−2.9 ± 1.535 → [−4.435, −1.365]
Answer
upper limit = −1.365 mmHg
6.Confidence intervals
Answer: upper limit = 4.609 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 = (4·2² + 6·2²)/10 = 4
sp = 2
Margin of error
E = 2.228·2·√(1/5 + 1/7) = 2.609
Interval
2 ± 2.609 → [−0.6092, 4.609]
Answer
upper limit = 4.609 kg
7.Hypothesis tests & confidence intervals
Answer: p = 0.1539
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 = 24/√36 = 4
z = (95.92 − 100)/4 = −1.02
Tail area
Φ(1.02) = 0.8461
1 − Φ(1.02) = 0.1539
p-value
p = 0.1539
Answer
p = 0.1539
p > 0.05, so H₀ is not rejected at the 5% level.
8.Hypothesis tests & confidence intervals
Answer: z = 1.906
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 = 60/100 = 0.6
p̂₂ = 147/300 = 0.49
p̂ = 207/400 = 0.5175
Standard error under H₀
SE = √(0.5175·0.4825·(1/100 + 1/300)) = 0.0577
Test statistic
z = (0.6 − 0.49)/0.0577 = 1.906
Answer
z = 1.906
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).