1-Sample inference

Z-test –for Normal Population with Known Population Variance \(\sigma^2\)

Context

• use this test when testing a single sample (size n) against a reference population
• test assumes that the population from which the sample data was drawn from is normal distributed
• test assumes population variance of sample data is same as variance of ref population, \(\sigma^2_1 = \sigma^2_{ref}\)

Steps

0. Specify alpha level, e.g., \(\alpha=0.05\)

1. Formulate null hypothesis, \(H_0: \mu = \mu_{ref}\), this means that the mean of the samples’ population is the same as the mean of the reference population

2. Formulate alternative hypothesis:

   • one-sided, upper-tail. \(H_A: \mu > \mu_{ref}\)
   • one-sided, lower-tail: \(H_A: \mu < \mu_{ref}\)
   • two-sided: \(H_A: \mu \neq \mu_{ref}\)

3. Compute z-statistic: \(z=\frac{\bar{x}-\mu}{\sigma_1 / \sqrt{n}}\)

4. Look up p-value for the z value, corresponding to the alternative hypothesis

   • \(H_A: \mu > \mu_{ref}\): use \(P(Z > z | H_0)\)
   • \(H_A: \mu < \mu_{ref}\): use \(P(Z < z | H_0)\)
   • \(H_A: \mu \neq \mu_{ref}\): use \(2*P(Z > |z| | H_0)\)

Example

• Question: Is Minnesota Secchi depth (a lake clarity measurement) larger than Wisconsin?
• Given information (provided in the question)
  • Wisconsin (population); the reference to test the sample against:
    • population normal distributed, \(X_W \sim N(\mu_W, \sigma^2_W)\)
    • population mean & variance, \(\mu_W = 2.0\), \(\sigma_W^2 = 0.25\)
  • Minnesota; the sample that we want to test against Wisconsin population:
    • sample mean, \(\bar{X}_M = 2.38\),
    • sample size \(n=22\)
• Formulate
  • Null hypothesis, \(H_0: \mu_M = \mu_W\) or \(H_0: \mu_M = 2.0\)
  • Alt. hypothesis (here one sided): \(H_A: \mu_M > 2.0\)
• Assumptions:
  • Minnesota lakes normal distributed: \(X_M \sim N(\mu_M, \sigma^2_M)\)
  • Minnesota sample mean is normal distributed \(\bar{X}_M \sim N(\mu_M, \frac{\sigma_M^2}{n})\)
  • equal variance, \(\sigma_M^2 = \sigma_W^2\); hence \(\sigma_M^2 = 0.25\)

\[P\left(\bar{X} \geq 2.38 \mid H_{0}\right)=P\left(\frac{\bar{X}_M-\mu_W}{\sqrt{\sigma_M^2/n}} \geq \frac{2.38-2.0}{0.1066}\right)=P(Z \geq 3.56)=0.0002\] - assuming \(\alpha=0.05\), we can reject the null hypothesis that the Wisconsin and Minnesota lake depths are the same

# Given (change based on question)
sample_mean <- 2.38
pop_mean <- 2.00
pop_var <- 0.25
n <- 22

# Assumed & calculated (don't change this)
pop_mean_of_sample <- pop_mean # from null hypothesis
pop_var_of_sample <- pop_var # from equal variance assumption
sample_mean_var <- pop_var_of_sample / n
sample_mean_std_dev <- sqrt(sample_mean_var)

# z-value
z <- (sample_mean - pop_mean_of_sample ) / sample_mean_std_dev
cat('z:', z, '\n')

## z: 3.564716

# p-value (change for one-sided upper/lower, two-tailed)
p <- pnorm(z, mean = 0, sd = 1, lower.tail = F)
# lower: p <- pnorm(z, mean = 0, sd = 1, lower.tail = T)
# 2-sided: p <- 2*pnorm(abs(z), mean = 0, sd = 1, lower.tail = F)
cat('p-value, P(Z > z):', p)

## p-value, P(Z > z): 0.0001821252

T-test – for Normal Population with Unknown Population Variance \(\sigma^2\)

• TODO

Approximate Z test – for Non-Normal Population with Known or Unknown Population Variance \(\sigma^2\)

• TODO

Confidence Intervals

• TODO