Inference for two means

Sampling distribution of the difference in sample means
- sampling independently from two populations
- has a mean of $μ_{1} - μ_{2}$
  - $E (\bar{X} - \bar{X}) = E (\bar{X}) - E (\bar{X}) = μ_{1} - μ_{2}$
- has a standard deviation of $σ_{\bar{X_{1}} - \bar{X_{2}}} = \sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{2}^{2}}{n_{2}}}$
  - $V a r ({\bar{X}}_{1} - {\bar{X}}_{2}) = V a r ({\bar{X}}_{1}) + V a r ({\bar{X}}_{2}) = \frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{2}^{2}}{n_{2}}$
- is normally distributed if we are sampling from normally distributed populations
  - if the distributions are not normal but the sample sizes are large:
    - ${\bar{X}}_{1}, {\bar{X}}_{2} and {\bar{X}}_{1} - {\bar{X}}_{2}$ will be approximately normal
Hypothesis Testing and Confidence Intervals for $μ_{1} - μ_{2}$ when $σ_{1}$ and $σ_{2}$ are unknown
- if we are drawing two independent samples from two normally distributed populations
  - (needed here as both procedures assume normality)
  - then
    - $\frac{({\bar{X}}_{1} - {\bar{X}}_{2}) - (μ_{1} - μ_{2})}{\sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{2}^{2}}{n_{2}}}} \sim N (0, 1)$
    - $\frac{(X_{1} - X_{2}) - (μ_{1} - μ_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}}$ has the t-distribution
- two methods
  - both assume
    - the two samples are simple random samples from their respective populations
      - or are the results of a randomized experiment
    - populations are normally distributed
      - not very important for large sample sizes (Central Limit Theorem)
        
        two sample t procedures are more robust to violation of normality than the one sample t procedure as well
      - can be t
    - the two samples are independent
  - pooled-variance two-sample t
    - an exact procedure that requires the additional assumption that $σ_{1}^{2} = σ_{2}^{2}$
      - means that the distributions look the same but for a shift
      - $σ_{2} < σ_{1}$
    - math looks clean, but if assumption is false...
    - assumes normality
    - $S E_{p} ({\bar{X}}_{1} - {\bar{X}}_{2}) = \sqrt{\frac{s_{p}^{2}}{n_{1}} + \frac{s_{p}^{2}}{n_{2}}}$
      - $= s_{p} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}$
    - $d f$ for the t: $n_{1} + n_{2} - 2$
  - welch procedure
    - approximate procedure that does not assume $σ_{1}^{2} = σ_{2}^{2}$
    - assumes normality as well
    - $S E_{W} ({\bar{X}}_{1} - {\bar{X}}_{2}) = \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}$
    - $d f$ for the t: $\frac{(\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}})^{2}}{\frac{1}{n_{1} - 1} (\frac{s_{1}^{2}}{n_{1}})^{2} + \frac{1}{n_{2} - 1} (\frac{s_{2}^{2}}{n_{2}})^{2}}$
      - ${(\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}})}^{2} / \frac{1}{n_{1} - 1} (\frac{s_{1}^{2}}{n_{1}}) + \frac{1}{n_{2} - 1} {(\frac{s_{2}^{2}}{n_{2}})}^{2}$
  - to estimate the common population variance, we pool the sample variances together
    - $s_{p}^{2} = \frac{(n_{1} - 1) s_{1}^{2} + (n_{2} - 1) s_{2}^{2}}{n_{1} + n_{2} - 2}$
- the endpoints of a $(1 - α) 100 %$ confidence interval for $μ_{1} - μ_{2}$ are given by
  - ${\bar{X}}_{1} - {\bar{X}}_{2} \pm t_{α / 2} S E ({\bar{X}}_{1} - {\bar{X}}_{2})$
- t test statistic
  - $t = \frac{{\bar{X}}_{1} - {\bar{X}}_{2}}{S E ({\bar{X}}_{1} - {\bar{X}}_{2})}$ (no mean because we're assuming they're equal for the $H_{0}$ )
    - t test is always done assuming $H_{0}$ is true
  - used to test the null hypothesis that the population means are equal $(H_{0} : μ_{1} = μ_{2})$
    - to choose between the 2 procedures consider
      - pooled-variance t is exact if the assumptions are true, while Welch is only an approximation
      - pooled-variance t is a special case of other common statistical methods (ANOVA/regression), while Welch is its own thing
      - pooled-variance t can perform horribly if the population variances are very different, while Welch does well
        
        even worse if the sample sizes are also very different
      - when $σ_{1}^{2} = σ_{2}^{2}$ Welch performs only a little worse than the pooled-variance t
- example
  - psychotic x nonpsychotic
    - ratio of $\frac{s_{1}}{s_{2}} < 2$ , in this case its 1.10
      - they're pretty close, and sample sd's are never gonna be equal, so its reasonable to say that $σ_{1} = σ_{2}$
      - both pooled-variance t and welch are reasonable
        
        using pooled variance
        
        $s_{p}^{2} = \frac{(n_{1} - 1) s_{1}^{2} + (n_{2} - 1) s_{2}^{2}}{n_{1} + n_{2} - 2} = \frac{(10 - 1) {5.27}^{2} + (15 - 1) {4.77}^{2}}{10 + 15 - 2} = 24.73$
        
        $S E_{p} ({\bar{X}}_{1} - {\bar{X}}_{2}) = s_{p} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}$
        
        $= \sqrt{24.73} \sqrt{\frac{1}{10} + \frac{1}{15}} = 2.03$
        
        $t = \frac{{\bar{X}}_{1} - {\bar{X}}_{2}}{S E_{p} ({\bar{X}}_{1} - {\bar{X}}_{2})}$
        
        $= \frac{24.25 - 16.47}{2.03} = 3.83$
        
        degrees of freedom are $n_{1} + n_{2} - 2 = 23$
        
        95% confidence interval for $μ_{1} - μ_{2}$
        
        ${\bar{X}}_{1} - {\bar{X}}_{2} \pm t_{.025} S E_{p} ({\bar{X}}_{1} - {\bar{X}}_{2})$
        
        $24.25 - 16.47 \pm 2.069 \times 2.03$
        
        $= .78 \pm 4.20$
        
        for a 95% interval of $(3.58, 11, 98)$
        
        using welch
        
        $S E_{W} ({\bar{X}}_{1} - {\bar{X}}_{2}) = \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}$
        
        $t = \frac{{\bar{X}}_{1} - {\bar{X}}_{2}}{S E_{W} ({\bar{X}}_{1} - {\bar{X}}_{2})}$
        
        $= \frac{24.25 - 16.47}{2.07} = 3.75$
        
        $d f = \frac{(\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}})^{2}}{\frac{1}{n_{1} - 1} (\frac{s_{1}^{2}}{n_{1}})^{2} + \frac{1}{n_{2} - 1} (\frac{s_{2}^{2}}{n_{2}})^{2}} = 18.053 .$
        
        95% confidence interval for $μ_{1} - μ_{2}$
        
        ${\bar{X}}_{1} - {\bar{X}}_{2} \pm t_{.025} S E_{W} ({\bar{X}}_{1} - {\bar{X}}_{2})$
        
        $24.25 - 16.47 \pm 2.100 \times 2.07$
        
        $= 7.78 \pm 4.35$
        
        for a 95% interval of $(3.43, 12.13)$
    - just use R
    - we can be $95 %$ confident that the difference in true mean dopamine levels $(μ_{p} - μ_{n})$ lies between $3.6$ and $12.0$
    - p-value = $0.0008569$
      - the mean of the group that became nonpshychotic was significantly lower than that of the group that remained psychotic (two-tailed t-test of independent means: $t (23) = 3.83; P < .001) .$
  - blood calcium level
    - group that drink orange juice vs one who didnt
    - confidence interval
      - 95% CI for $μ_{d} - μ_{c}$
        
        $μ_{d}$ = true mean change in calcium for the vitamin d group
        
        $μ_{d} - μ_{c} =$ treatment effect
    - hypothesis
      - $H_{0} : μ_{d} - μ_{c} = 0 \to H_{0} : μ_{d} = μ_{c}$
      - $H_{a} : μ_{d} \neq μ_{c}$
    - plots look symmetric, we dont know if its normal tho
      - the t-procedure stull works well with normality violations so should be fine
      - sd's are close, either pooled-variance or welch are reasonable
    - R
      - pooled
        
        t.test(Fortified,Control,var.equal=TRUE)
      - welch
        
        t.test(Fortified,Control,var.equal=FALSE)
      - $95 %$ CI for $μ_{d} - μ_{c} = (- 3.3, 2.6)$ mg/dl
        
        we can be 95% confident that the difference in true mean of the change in calcium ( $μ_{d} - μ_{c}$ ) lies between -3.3 and 2.6 ml/dl
      - t(23) = -0.24647
        
        p-value is double the area to the left $\approx 0.81$
        
        large p $\to$ no evidence against $H_{0}$
        
        there is no evidence of a difference in the true mean change in calcium between the vitamin d and the control group
paired-difference procedures (dependent samples)
- both measurements are on the same sample, with different conditions
- designs
  - pretest-posttest design
    - sketchy
    - measurements are taken on the same individuals before and after a treatment
      - normally we have a control group that doesnt get the treatment
  - crossover design
    - participants are randomly assigned to a time order of the treatments
      - treatment 1 $\to$ some time $\to$ treatment 2
      - treatment 2 $\to$ some time $\to$ treatment 1
  - matched pairs design
    - individuals are matched based on relevant variables
      - one will take treatment 1 the other treatment 2
- paired-difference t procedure
  - treat the observed differences as a single sample from a population of differences
    - one data using the difference between the 2
  - $\bar{X} \pm t_{\frac{α}{2}} S E (\bar{X})$
    - $(1 - α) 100 %$ confidence interval for the population mean difference $μ$
    - $\bar{X}$ = sample mean of the differences
    - $d f = n - 1$
      - $n$ is the number of differences
    - $S E (\bar{X}) = \frac{s}{\sqrt{n}}$
  - hypothesis
    - $H_{0} : μ = μ_{0}$ ; almost always $H_{0} : μ = 0$
      - difference between the 2 groups = 0 (theyre equal)
    - $t = \frac{\bar{X} - μ_{0}}{S E (\bar{x})}$