Evidence-Based Practice: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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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

  1. Draw the 2 × 2 table and label a, b, c and d before you calculate.
  2. Read sensitivity and specificity down the columns, predictive values along the rows.
  3. For another prevalence, imagine 10 000 people and fill the table with Se, Sp and the prevalence.
  4. For a likelihood ratio, convert the probability to odds, multiply, and convert back.
  5. 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

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. 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

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. 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
eventno eventtotaltreatment117303420abcontrol344586930cd

Practice exercise (made-up values, not clinical guidance). In a randomised trial, 420 patients received a new treatment and 930 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: 117/420control: 344/930
RR =
2.Evidence-based practice statisticsHard
p (%)NPV (%)1%NPV = ?100

Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 76% and a specificity of 92%. It is used in a population where the prevalence is 1%. Find the negative predictive value NPV as a percentage. Round to one decimal place.

Se = 76%Sp = 92%p = 1%
NPV = %
3.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 29, 24, 41, 14, 51.

n = 5
s =
4.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 100 and standard deviation 12. Find the z-score z of a score of 115.

μ = 100σ = 12x = 115
z =
5.Confidence intervalsMedium
n = 20, s = 5.5LU = ?x̄ = 84.5E90%

A sample of 20 items has mean 84.5 and standard deviation s = 5.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 20x̄ = 84.5s = 5.590%t(19) = 1.729
upper limit =
6.Confidence intervalsHard
n = 30Lσ = ?s = 9.795%

The diameters of 30 machined shafts have a sample standard deviation of 9.7 µm. Assuming the diameters are normally distributed, find the upper 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 = 9.7 µmχ² = 16.047, 45.722
σ = µm
7.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 50.88 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 50.88s = 5n = 16
t =
8.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.05βσ = 19, power 90%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 90%. The standard deviation is 19. Find the minimum sample size n.

α = 0.05power = 90%δ = 4σ = 19zα = 1.645zβ = 1.282
n =

3Solutions

1.Evidence-based practice statistics

Answer: RR = 0.75

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 = 117/420 = 0.2786
CER = 344/930 = 0.3699

Relative risk

RR = 0.2786/0.3699 = 0.7531 ≈ 0.75

Answer

RR = 0.75
2.Evidence-based practice statistics

Answer: NPV = 99.7 %

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) = 0.92·0.99 = 0.9108
(1 − Se)·p = 0.24·0.01 = 0.0024

Negative predictive value

NPV = 0.9108/(0.9108 + 0.0024) = 0.9974 ≈ 99.7%

Answer

NPV = 99.7%
3.Descriptive statistics

Answer: s = 14.48

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.

Mean

x̄ = 159/5 = 31.8

Squared deviations

Σ(x − x̄)² = (−2.8)² + (−7.8)² + 9.2² + (−17.8)² + 19.2² = 838.8

Standard deviation

s = √(838.8/4) = 14.48

Answer

s = 14.48
4.Descriptive statistics

Answer: z = 1.25

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 = (115 − 100)/12 = 1.25

Answer

z = 1.25
5.Confidence intervals

Answer: upper limit = 86.63

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Critical value

df = n − 1 = 19, t = 1.729

Margin of error

E = 1.729·5.5/√20 = 2.126

Interval

84.5 ± 2.126 → [82.37, 86.63]

Answer

upper limit = 86.63
6.Confidence intervals

Answer: σ = 13.04 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 = 29·9.7² = 2729

Limits for σ²

2729/45.722 = 59.68
2729/16.047 = 170

Limit for σ

σ = √170 = 13.04 µm

Answer

σ = 13.04 µm

95% interval for σ: [7.725, 13.04] µm.

7.Hypothesis tests & confidence intervals

Answer: t = 0.704

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 = 5/√16 = 5/4 = 1.25

Test statistic

t = (50.88 − 50)/1.25 = 0.704

Answer

t = 0.704
t0.025; df = 15 = 2.131

|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).

8.Hypothesis tests & confidence intervals

Answer: n = 194

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.

Solve for n

n = ((1.645 + 1.282)·19/4)² = 193.3

Round up

n = 194

Answer

n = 194

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