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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 400 patients with a disease (cases) and 790 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 = 165b = 174c = 235d = 616
OR =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 5% for a disease. A test with sensitivity 68% and specificity 61% comes back positive. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 5%Se = 68%Sp = 61%positive
p′ = %
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 38, 40, 51, 32, 51, 29.
n = 6
s =
4.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 12. Find the z-score z of a score of 88.
μ = 70σ = 12x = 88
z =
5.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 60 phones, mean 41.5 h, known standard deviation 16 h. Model 2: 50 phones, mean 45 h, known standard deviation 17 h. Find the upper limit of the 95% confidence interval for μ₁ − μ₂.
The diameters of 20 machined shafts have a sample standard deviation of 1.6 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 19, 2.5% of the χ² distribution lies below 8.907 and 2.5% lies above 32.852.
n = 20s = 1.6 µmχ² = 8.907, 32.852
σ = µm
7.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 102.3 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 102.3s = 4n = 25
t =
8.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 115. The population standard deviation is 10. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 115σ = 10n = 2595%
L =
3Solutions
1.Evidence-based practice statistics
Answer: OR = 2.49
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: 165/235 = 0.7021
controls: 174/616 = 0.2825
Odds ratio
OR = (165·616)/(174·235) = 2.486 ≈ 2.49
Answer
OR = 2.49
OR above 1: exposure is more common among the cases.
2.Evidence-based practice statistics
Answer: p′ = 8.4 %
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.
Likelihood ratio
LR+ = 0.68/(1 − 0.61) = 1.744
Pre-test odds
0.05/(1 − 0.05) = 0.05263
Post-test odds
0.05263·1.744 = 0.09177
Post-test probability
p′ = 0.09177/(1 + 0.09177) = 0.08405 ≈ 8.4%
Answer
p′ = 8.4%
3.Descriptive statistics
Answer: s = 9.283
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 = (88 − 70)/12 = 1.5
Answer
z = 1.5
5.Confidence intervals
Answer: upper limit = 2.713 h
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.
Point estimate
x̄₁ − x̄₂ = 41.5 − 45 = −3.5
Standard error
SE = √(16²/60 + 17²/50) = 3.17
Margin of error
E = 1.96·3.17 = 6.213
Interval
−3.5 ± 6.213 → [−9.713, 2.713]
Answer
upper limit = 2.713 h
The interval contains 0, so the data do not show a significant difference.
6.Confidence intervals
Answer: σ = 2.337 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 = 19·1.6² = 48.64
Limits for σ²
48.64/32.852 = 1.481
48.64/8.907 = 5.461
Limit for σ
σ = √5.461 = 2.337 µm
Answer
σ = 2.337 µm
95% interval for σ: [1.217, 2.337] µm.
7.Hypothesis tests & confidence intervals
Answer: t = 2.875
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
SE = 4/√25 = 4/5 = 0.8
Test statistic
t = (102.3 − 100)/0.8 = 2.875
Answer
t = 2.875
t0.025; df = 24 = 2.064
|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).
8.Hypothesis tests & confidence intervals
Answer: L = 111.1
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).