Statistical Inference

Confidence Interval

• It is reported in 2 parts: Confidence level (e.g. 95%) and Interval (e.g. 0.2 $\pm$ 0.01)
• It is used to infer info on the population parameter using a sample of the population.
• 95% CI means out of 100 samples, and confidence intervals constructed for each sample, 95 of them will contain the population parameter. So, we are 95% confident that the population parameter lies within the CI.
• To construct a CI for population proportion (for discrete variables)
  • $p\ast \pm z\ast \sqrt{\frac{p\ast (1-p\ast )}{n}}$
  • $p\ast$ : sample proportion
  • $z\ast$ : z-value from standard normal distribution (depends on confidence level)
  • $n$ : sample size
• To construct a CI for population mean
  • $\bar{x}\pm t\ast \frac{s}{\sqrt{n}}$
  • $\bar{x}$ : sample mean
  • $t\ast$ : t-value from t-distribution (depends on $n$ and confidence level)
  • $s$ : sample standard deviation
  • $n$ : sample size
• Properties
  • When a smaller sample is taken, the CI will be wider
  • At a lower confidence level, the CI will be narrower

Hypothesis Testing

Used to decide if the data from a random sample is sufficient to support a hypothesis about the population

1. State (mutually exclusive):
   1. Null Hypothesis (always lower probability)
   2. Alternate Hypothesis
2. Collect sample data
3. State the level of significance (typically 0.05, 0.01 or 0.1). Lower value: harder to reject null hypothesis
4. Calculate P value (probability that null hypothesis is true)
5. Compare with level of significance
   1. p value < significance level: null hypothesis rejected, accept alternate hypothesis
   2. p value > significance level: not enough information to reject null hypothesis

Example

A coin manufacturer claims they produced a biased coin with $P(H)=0.3$ When testing the coin 8 times, there were 7H and 1T

1. Hypotheses
   1. Null Hypothesis $H_0$: Coin is as claimed i.e. P(H) = 0.3
   2. Alternate Hypothesis $H_1$: P(H) > 0.3 (one tailed test) (two tailed test is P(H) \neq 0.3)
2. Let the level of significance be 0.1
3. P-value = P(observation | null is true) + P(equally extreme outcomes | null is true) + P(outcomes that are even more extreme | null is true) Note: at least as extreme → at least as favourable to the alternate hypothesis
   1. P(HHHHHHHT | P(H) = 0.3) = (0.3)^7 x 0.7
   2. • P(other combinations of 7H 1T | P(H) = 0.3) = 7(0.3)^7 x (0.7)
   3. • P(HHHHHHHH | P(H) = 0.3) = (0.3)
4. Since p-value < level of significance, we can reject the null hypothesis, and conclude the alternate hypothesis

Chi-sq test

Hypothesis testing for whether two variables are associated

Example

Null Hypothesis: Smoking is not associated with heart disease: rate(HD|S) = rate(HD|NS) = rate(HD) Alternate hypothesis: Smoking is associated with heart disease: rate(HD|S) $\ne$ rate(HD|NS)

Assume Null Hypothesis is true, calculate rate(HD).

Construct table for observation:

	Heart Diseas	No Heart Disease	Row Total
Smoker	38	14962	15000
Non Smoker	44	84956	85000
Col Total	82	99918	100000
rate(HD) = 82/100000
Then, draw the table for the expected outcome (null hypothesis) and compare:

	Heart Disease	No Heart Disease	Row Total
Smoker	12.3	14897.7	15000
Non Smoker	69.7	844930.3	85000
Col Total	82	99918	100000
p value is low if there is a big difference between expectation and observation

if p-value < level of significance, we can reject the null hypothesis and conclude alternate hypothesis if p-value > level of significance, we cannot reject the null hypothesis, and therefore cannot conclude the alternate hypothesis (we cannot conclude the null hypothesis)

Note: Don’t need to know how to calculate p value, just use software