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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, 930 patients received a new treatment and 270 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: 171/930control: 62/270
RR =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 96% and a specificity of 80%. It is used in a population where the prevalence is 0.5%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 96%Sp = 80%p = 0.5%
PPV = %
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 11, 10, 33, 23, 11.
n = 5
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is above 122.5?
μ = 100σ = 15x = 122.5
P =
5.Confidence intervalsMedium
The blood pressure of 9 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 9.6 and standard deviation 5.5. Find the lower limit of the 95% confidence interval for the mean difference.
n = 9d̄ = 9.6sd = 5.595%t(8) = 2.306
lower limit = mmHg
6.Confidence intervalsHard
The diameters of 30 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 = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.
n = 30s = 5.5 µmχ² = 16.047, 45.722
σ = µm
7.Hypothesis tests & confidence intervalsMedium
10 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 1.62 and the standard deviation of the differences is 1.6. Find the paired test statistic t for H₀: μd = 0.
n = 10d̄ = 1.62sd = 1.6
t =
8.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 5 units with a power of 80%. The standard deviation is 21. Find the minimum sample size n.
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 = 171/930 = 0.1839
CER = 62/270 = 0.2296
Relative risk
RR = 0.1839/0.2296 = 0.8007 ≈ 0.8
Answer
RR = 0.8
2.Evidence-based practice statistics
Answer: PPV = 2.4 %
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 = 0.96·0.005 = 0.0048
(1 − Sp)·(1 − p) = 0.2·0.995 = 0.199
Positive predictive value
PPV = 0.0048/(0.0048 + 0.199) = 0.02355 ≈ 2.4%
Answer
PPV = 2.4%
The lower the prevalence, the lower the PPV of the same test.
3.Descriptive statistics
Answer: s = 10.14
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 1 − 0.9332 = 0.06681
Answer
P = 0.06681
5.Confidence intervals
Answer: lower limit = 5.372 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 = 8, t = 2.306
Margin of error
E = 2.306·5.5/√9 = 4.228
Interval
9.6 ± 4.228 → [5.372, 13.83]
Answer
lower limit = 5.372 mmHg
6.Confidence intervals
Answer: σ = 4.38 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 = 29·5.5² = 877.3
Limits for σ²
877.3/45.722 = 19.19
877.3/16.047 = 54.67
Limit for σ
σ = √19.19 = 4.38 µm
Answer
σ = 4.38 µm
95% interval for σ: [4.38, 7.394] µm.
7.Hypothesis tests & confidence intervals
Answer: t = 3.202
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 = 1.6/√10 = 0.506
Test statistic
t = 1.62/0.506 = 3.202
Answer
t = 3.202
With df = 9, the two-sided 5% critical value is 2.262; |t| is larger, so H₀ is rejected.
8.Hypothesis tests & confidence intervals
Answer: n = 110
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.