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11 Comparing Two Samples (Comparing Paired Samples (A Nonparametric Method…

- - - - estimate \(\bar{X}-\bar{Y}\)
      - \[\bar{X}-\bar{Y}\sim N\left[\mu_{X}-\mu_{Y},\sigma^{2}\left(\frac{1}{n}+\frac{1}{m}\right)\right]\]
      - If \(\sigma^2\) were known, a confidence interval for \(\mu_X-\mu_Y\) could be based on\[Z=\frac{(\bar{X}-\bar{Y})-(\mu_{X}-\mu_{Y})}{\sigma\sqrt{\frac{1}{n}+\frac{1}{m}}}\] which follows a standard normal distribution. The confidence interval would be\[(\bar{X}-\bar{Y})\pm z_{\alpha/2}\sigma\sqrt{\frac{1}{n}+\frac{1}{m}}\]
      - \(\sigma^2\) estimated by the pooled sample variance
        
        \[s_{p}^{2}=\frac{(n-1)s_{X}^{2}+(m-1)s_{Y}^{2}}{m+n-2}=\frac{\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}+\sum_{i=1}^{m}(Y_{i}-\bar{Y})^{2}}{m+n-2}\]
        
        where
        \(s_{X}^{2}=\frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}\)
        \(s_{Y}^{2}=\frac{1}{m-1}\sum_{i=1}^{m}(Y_{i}-\bar{Y})^{2}\)
        
        Thus the estimated standard deviation of \(\bar{X}-\bar{Y}\) is\[\color{orangered}{s_{\bar{X}-\bar{Y}}=s_{p}\sqrt{\frac{1}{n}+\frac{1}{m}}}\]
    - - Test statistic:\[t=\frac{\bar{X}-\bar{Y}}{\color{orangered}{s_{\bar{X}-\bar{Y}}}}\]
      - \(H_{0}:\,\mu_{X}=\mu_{Y}\)
        
        meaning there is no difference
        between the treatment and control group
      - För \(H_1:\,\mu\neq\mu_Y\) så är rejection-regionen\[|t|>t_{n+m-2}(\alpha/2)\]
      - Man kan visa att likelihood ratio test är ekvivalent med t-testet (s.426 i boken)
    - - If (the square root of) # this estimate is used in the denominator of #, the distribution of the statistic is no longer the \(t\) distribution.
        But it has been shown that its distribution can be closely approximated by the \(t\) distribution with degrees of freedom calculated in the following way and rounded to the nearest integer:\[\textrm{df}=\frac{\left((s_{X}^{2}/n)+(s_{Y}^{2}/m)\right)^{2}}{\frac{(s_{X}^{2}/n)^{2}}{n-1}+\frac{(s_{Y}^{2}/m)^{2}}{m-1}}\]
      - If the underlying distributions are not normal and the sample sizes are large, the use of the \(t\) distribution is justified by the central limit theorem => CI and HT approx. valid.
        In such a case, however, there is little difference between the t and normal distributions.
        If the sample sizes are small, however, and the distributions are not normal, conclusions based on the assumption of normality cannot be tested effectively unless the deviation is quite gross as we saw in Chapter 9.
    - - log
      - sqrt
  - - - Eller för stora samples (\(n\geq10,m\geq10\), använd normalapproximation för nulldistr. av \(R_1\) och \(R_2\).
        
        \[E(R_{1})=\frac{n(n+m+1)}{2}\]
        
        \[\textrm{Var}(R_{1})=\textrm{Var}(R_{2})=\frac{mn(n+m+1)}{12}\]
        
        \[E(R_{2})=\frac{m(n+m+1)}{2}\]
- - - - How to calculate the null distr.
        
        There are \(2^n\) such assignments, and for each we can calculate \(W_+\). We obtain a list of \(2^n\) values (not all distinct) of \(W_+\), each of which occurs with probability \(1/2^n\). The probability of each distinct value of \(W_+\) may thus be calculated, giving the desired null distribution.
        
        The kth largest value of D is thus equally likely to be positive or negative, and any particular assignment of signs to the integers \(1,...,n\) (the ranks) is equally likely.
        
        If \(H_0\) is true, it makes no difference which member of the pair corresponds to treatment and which to control. The difference \(X_i-Y_i=D_i\) has the same distribution as the difference \(Y_i-X_i=-D_i\), so the distribution of \(D_i\) is symmetric about zero.
      - This will be true if the members of pairs of experimental units are assigned randomly to treatment and control conditions, and the treatment has no effect at all.
    - - Small values of n: tables
      - For \(n\geq 20\): use the normal approx. of the null distr. with
        \(\mu_{W}=\frac{n(n+1)}{4}\)
        \(\sigma_{W}^{2}=\frac{n(n+1)(2n+1)}{24}\)