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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 270 patients with a disease (cases) and 190 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 = 105b = 106c = 165d = 84
OR =
2.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 86% and a specificity of 95%. It is used in a population where the prevalence is 5%. Find the negative predictive value NPV as a percentage. Round to one decimal place.
Se = 86%Sp = 95%p = 5%
NPV = %
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 57, 12, 11, 49, 55.
n = 5
s =
4.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is above 84?
μ = 100σ = 20x = 84
P =
5.Confidence intervalsMedium
The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 4.4 and standard deviation 7.9. Find the lower limit of the 99% confidence interval for the mean difference.
n = 15d̄ = 4.4sd = 7.999%t(14) = 2.977
lower limit = mmHg
6.Confidence intervalsHard
The diameters of 21 machined shafts have a sample standard deviation of 11.8 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 = 11.8 µmχ² = 9.591, 34.17
σ = µm
7.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 50; the population standard deviation is known to be 7.5. A random sample of 25 observations has a mean of 51.89. Test H₀: μ = 50 against H₁: μ ≠ 50. Find the p-value p.
μ₀ = 50σ = 7.5n = 25x̄ = 51.89H₁: μ ≠ μ₀
p =
8.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 80 against H₁: μ > 80 is done at α = 0.10 with a sample of 100; the standard deviation is known to be 5. In reality the mean is 80.85. Find the probability of a Type II error β in percent.
μ₀ = 80μ = 80.85σ = 5n = 100α = 0.10
β = %
3Solutions
1.Evidence-based practice statistics
Answer: OR = 0.5
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: 105/165 = 0.6364
controls: 106/84 = 1.262
Odds ratio
OR = (105·84)/(106·165) = 0.5043 ≈ 0.5
Answer
OR = 0.5
OR below 1: exposure is less common among the cases, which suggests a protective effect.
2.Evidence-based practice statistics
Answer: NPV = 99.2 %
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 negatives (per person)
Sp·(1 − p) = 0.95·0.95 = 0.9025
(1 − Se)·p = 0.14·0.05 = 0.007
Negative predictive value
NPV = 0.9025/(0.9025 + 0.007) = 0.9923 ≈ 99.2%
Answer
NPV = 99.2%
3.Descriptive statistics
Answer: s = 23.29
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 = −0.8
Standard normal table
Φ(−0.8) = 0.2119
P = 1 − 0.2119 = 0.7881
Answer
P = 0.7881
5.Confidence intervals
Answer: lower limit = −1.672 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 = 14, t = 2.977
Margin of error
E = 2.977·7.9/√15 = 6.072
Interval
4.4 ± 6.072 → [−1.672, 10.47]
Answer
lower limit = −1.672 mmHg
6.Confidence intervals
Answer: σ = 9.028 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 = 20·11.8² = 2785
Limits for σ²
2785/34.17 = 81.5
2785/9.591 = 290.4
Limit for σ
σ = √81.5 = 9.028 µm
Answer
σ = 9.028 µm
95% interval for σ: [9.028, 17.04] µm.
7.Hypothesis tests & confidence intervals
Answer: p = 0.2077
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 = (51.89 − 50)/1.5 = 1.26
Tail area
Φ(1.26) = 0.8962
1 − Φ(1.26) = 0.1038
p-value
p = 2·0.1038 = 0.2077
Answer
p = 0.2077
p > 0.05, so H₀ is not rejected at the 5% level.
8.Hypothesis tests & confidence intervals
Answer: β = 33.8 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).