Correlation and Sample Size Calculations

Correlation

• Correlation assesses the association between two variables.

• Pearson’s Correlation: Parametric method.

• Spearman’s Correlation: Non-parametric method.

• Values range from -1 to 1.

  • +1: Perfect positive linear association.

  • -1: Perfect negative linear association.

  • 0: No linear relation.

• Correlation is valid only within the sample range.

• Strength of Correlation:

  • 0 – 0.19: Very weak

  • 0.2 – 0.39: Weak

  • 0.4 – 0.59: Moderate

  • 0.6 – 0.79: Strong

  • 0.8 – 1.0: Very strong

• Pearson’s Correlation Coefficient:

  • Measures linear association.

  • Does not account for curved relationships.

  • r > 0: Positive correlation (both variables increase).

  • r < 0: Negative correlation (one variable increases, the other decreases).

  • r = +1 or -1: Perfect straight line.

  • r = 0: No linear association.

• Assumptions for Pearson’s Correlation:

  • At least one variable is normally distributed for P-value validity.

  • Both variables should ideally be normally distributed in practice for confidence interval validity.

  • All observations must be independent.

• Hypothesis Test:

  • Null hypothesis: $\rho = 0$

  • Alternative hypothesis: $\rho \neq 0$

  • Test statistic: T-test

• Sample Size:

  • The value at which the correlation coefficient becomes significant at the 5% level.

  • N=10, correlation=0.63

  • N=20, correlation=0.44

  • N=50, correlation=0.28

  • N=100, correlation=0.20

  • N=150, correlation=0.16

• Spearman’s Correlation:

  • Measures general association, not just linear.

  • Non-parametric.

  • Applicable to ordinal data.

  • Outliers have less effect.

• Presentation of Correlation:

  • Include graph, correlation coefficient (to 2 decimal places), P-value, and number of observations.

  • Example: N=20, R=0.58 (P<0.01)

• Problems with Correlation:

  • Correlation does not imply causation.

  • It is useful for generating hypotheses.

  • Correlation does not quantify agreement.

  • Multiple testing issues arise when calculating numerous correlations.

Sample Size Calculations

• Understanding hypothesis testing is crucial.

• Understanding errors related to hypothesis testing is necessary.

• Need to comprehend the power of a test.

Importance of Sample Size:

• Larger studies increase power to detect differences.

• Estimates population parameters with precision.

• Many clinical trials are too small.

• Increasing sample size increases test power.

Ethical Considerations:

• Too small or too large sample sizes can be unethical.

• Impacts time, manpower, risk, and cost.

• Aim for the minimum sample size to achieve a high probability (power) of detecting a clinically worthwhile effect.

Objectives of Study:

• Should be quantified.

• Relate to a single primary outcome.

• Predetermined in grant applications and study protocols.

Factors Influencing Sample Size:

• Outcome type.

• Number of groups for comparison.

• Sampling variability.

• Effect size.

• Response rate.

• Clinical outcome should be expressed in a patient-relevant way.

Components for Sample Size Calculation:

• Effect size: The magnitude of the effect we wish to detect (clinically relevant difference).

• Variability: Standard Deviation (SD) to calculate effect size.

• Chosen Level of Significance: Type I error (e.g., 0.05).

• Chosen Power: 1 – Type II error (e.g., 80%).

Type I and Type II Errors:

• Type I Error (False Positive):

  • Rejecting the null hypothesis when it is true.

  • Risk = $\alpha$

• Type II Error (False Negative):

  • Failing to reject the null hypothesis when it is false.

  • Risk = $\beta$

• Power:

  • The probability of rejecting a false null hypothesis.

  • Power = $1 - \beta$

Power of a Test:

• Used to compare two tests with the same $\alpha$

• Helps decide appropriate sample size.

• Larger sample sizes generally lead to more powerful tests.

Significance Level ($\alpha$):

• The probability of a Type I error.

• Controlled by the location of the rejection or critical region.

• Common values: 0.05 or 0.01.

• Choosing a small $\alpha$ may increase $\beta$ (Type II error).

Sample Size Calculation - Basics:

• Continuous outcome scale (blood pressure, age, QoL score): compare two group means.

• Binary outcome scale (yes/no, disease present/absent): compare proportions of 'yes' between two groups.

Sample Size Calculation - Advanced:

• Longitudinal Studies: Account for dropouts and correlation between repeated measures.

• Neuroimaging Studies: Consider multiple comparisons problem.

• Genetic Studies: Sample size must be large to detect small effect sizes.

• Clinical Trials: Account for the expected effect size and population variance.

• Animal Studies: Balance ethical considerations with the need for statistical power.

• Consult a statistician for complex scenarios.

Examples of comparing variables:

• Comparing continuous variables

• Comparing proportions

Sample size calculations- comparing two group means for a continuous outcome

• Outcome should be on continuous scale (quality of life score, blood pressure, cholesterol level etc.)

• In a trial if considered two sample t-test that means comparing two group means in continuous scale

• Formula: $N = \frac{2s^2 (z<em>\alpha + z</em>\beta)^2}{T^2}$

  • Where:

    • $N$ is the number of patients in each group.

    • $s$ is the standard deviation of the outcome.

    • $T$ is the smallest clinically important difference in the mean outcome between the groups.

    • $z<em>\alpha$ is the quantile related to the hypothesis test significance level. For a 5% significance level, $z</em>\alpha = 1.96$.

    • $z<em>\beta$ is the quantile related to the hypothesis test of power. For 90% power, $z</em>\beta = 1.28$.

Example Problem:

• A double-blind, placebo-controlled, randomized trial investigates a statin drug's effect on LDL cholesterol levels.

• Adults aged 55-70 at low stroke risk are followed for six months.

• The primary outcome is LDL cholesterol level (mmol/L) at six months.

• A clinically important reduction is 1.8 mmol/L.

• Previous study SD is approximately 3.0 mmol/L.

• Significance level of 5% and power of 90% are desired.

Solution:

• s = 3

• T = 1.8

• $z_\alpha = 1.96$

• $z_\beta = 1.28$

• $N = \frac{2 \times 3^2 (1.96 + 1.28)^2}{1.8^2} = 59$

• Approximately 59 patients per group.

Using STATA:

• power twomeans 4.5 6.3 , sd(3) power(0.9)

• power twomeans 14.5 16.3 , sd(3) power(0.9)

• power twomeans 16.3 14.5 , sd(3) power(0.9)

• power twomeans 5.0, diff(1.8) sd(3) power(0.9)

• power twomeans 15.0, diff(1.8) sd(3) power(0.9)

• All yield approximately n = 60 per group.

Accounting for Drop-Out Rate:

• $N^* = \frac{N}{ (1 - d) }$

  • Where $d$ is the expected drop-out rate.

  • $N$ is the original sample size.

  • Then $N^*$ is the adjusted sample size.

• If expected drop out rate is 10% in LDL trial (d=0.10):
  $N^* = \frac{ 59 }{1-0.1} = 66$

• So, calculate revised sample size after considering expected drop out rate.

Sample size calculations-comparing two proportions

• Outcome should be on proportion, rate, prevalence…

• In a trial if considers comparing two rates or prevalence or proportions..

• Assume equal sample per group / arm of the trial

• Formula:
  $N = \frac{[z<em>\alpha \times \sqrt{2 \times p(1-p)} + z</em>\beta \times \sqrt{p<em>1(1-p</em>1)+p<em>2(1-p</em>2)}]^2}{(p<em>1-p</em>2)^2}$

  • Where

    • $p_1$ = Estimated current prevalence rate.

    • $p_2$ = Maximum prevalence rate that would signify a clinically important reduction (if interest lies in a prevalence reduction)

    • $p$ = $\frac{p<em>1+p</em>2}{2}$, mean of $p<em>1$ and $p</em>2$

    • $z<em>\alpha$ = Quantile related to hypothesis test significance level. For a 5% significance level, $z</em>\alpha$= 1.96.

    • $z<em>\beta$ = Quantile related to hypothesis test of power. For 80% power, $z</em>\beta$ = 0.84.

Example:

• Multi-center randomized trial to investigate the effect of a beta-blocker on mortality as a treatment for severe heart failure.

• The trial will be double-blind, placebo controlled and will include the same number of patients in each group.

• Patients will be followed for two years and the primary outcome is all-cause mortality during this two-year period.

• A previous study suggests that there is a two-year mortality rate of approximately 36% amongst the population of patients to be considered.

• The investigators consider that it would be clinically important to detect a reduction to 24% in two-year mortality when treated with the beta-blocker.

• The investigators wish to use a significance level of 5% and a power of 80%.

• $p_1 = 0.36$

• $p_2 = 0.24$

• $p = \frac{0.36 + 0.24}{2} = 0.30$

• $z_\alpha = 1.96$

• $z_\beta = 0.84$

• $N = \frac{[1.96 \times \sqrt{2 \times 0.30 (1-0.30)} + 0.84 \times \sqrt{0.36(1-0.36)+0.24(1-0.24)}]^2}{(0.36-0.24)^2} = 228$

• Therefore, we should recruit 228 patients in each group.

Using STATA:

• power twoproportions 0.24 0.36 , test(chi2) power(0.8)

• n= 228 / group

• power twoproportions 0.24 0.36 , test(chi2) power(0.9)

• n= 305/ group

The more power you want to achieve the more sample you need.