Statistical Hypothesis Testing with SAS and R

by Dirk Taeger and Sonja Kuhnt

(c) John Wiley & Sons, Ltd

Test 2.2.5: Paired t-test

Description:

Tests if the difference of two population means \(\mu_d = \mu_1 - \mu_2\) differ from a value \(d_0\) in the case that observations are collected in pairs.

Assumptions:

• Data are measured on an interval or ratio scale and randomly sampled in pairs \((X_1, X_2)\).

• \(X_1\) follows Gaussian distributions with \(\mu_1\) and variance \(\sigma_1^2\). \(X_2\) follows Gaussian distributions with \(\mu_2\) and variance \(\sigma_2^2\). The covariance of \(X_1\) and \(X_2\) is \(\sigma_{12}\).

• The standard deviations are unknown. The standard deviation \(\sigma_d\) of the differences is estimated through the population standard deviation \(S_d\) of the differences.

Hypotheses:

1. \(H_0\) : \(\mu_d = d_0\) vs \(H_1\) : \(\mu_d \neq d_0\).
2. \(H_0\) : \(\mu_d \le d_0\) vs \(H_1\) : \(\mu_d \gt d_0\).
3. \(H_0\) : \(\mu_d \ge d_0\) vs \(H_1\) : \(\mu_d \lt d_0\).

Test statistic:

\[T = \frac{\bar{D} - d_0}{\sigma_d}\sqrt{S_d}\] with \[S_d = \sqrt{\frac{1}{n-1}\sum_{i=1}^n (D_i - \bar{D})^2}\]

\[\bar{D} = \frac{1}{n}\sum_{i=1}^n D_i\] and \(D_i = X_{1i} - X_{2i}\), \(i = 1,...,n\)

Test decision:

Rejection \(H_0\) if for the observed value t of T:

1. \(t \lt t_{\alpha/2, n-1}\) or \(t \gt t_{1-\alpha/2, n-1}\)

2. \(t \gt t_{1-\alpha, n-1}\)

3. \(t \lt t_{\alpha,n-1}\)

P-value:

1. \(\rho = 2P(T \le (-|t|))\)

2. \(\rho = 1 - P(T \le t)\)

3. \(\rho = P(T \le t)\)

Annotation:

• The test statistic \(T\) is t-distribution with \(n-1\) degree of freedom.
• \(t_{\alpha, n_1}\) quantitle of t-distribution with \(n-1\) degree of freedom.
• The assumtion of a Gaussian distribution can be relaxed if the distribution of the difference is symmetric.

Example

To test if the mean intelligence quotient increases by 10 comparing before training (IQ1) and after training (IQ2). Note: Because we are interested in a negative difference of means of \(IQ1 - IQ2\), we must test against \(d_0 = -10\)

#iq dataset
no <- seq(1:20)
IQ1 <- c(127, 98,105,83,133,90,107,98,91,100,88,96,110,87,88,88,105,95,79,106)
IQ2 <- c(137, 108,115,93,143,100,117,108,101,110,98,106,120,97,98,100,115,111,89,116)
iq <-data.frame(no,IQ1, IQ2)

t.test(iq$IQ1,iq$IQ2,mu=-10,alternative="two.sided",paired=TRUE)

## 
##  Paired t-test
## 
## data:  iq$IQ1 and iq$IQ2
## t = -1.2854, df = 19, p-value = 0.2141
## alternative hypothesis: true difference in means is not equal to -10
## 95 percent confidence interval:
##  -11.051338  -9.748662
## sample estimates:
## mean of the differences 
##                   -10.4

Remarks:

• The command \(paired=TRUE\) forces R to calculate the paired t-test.
• \(mu=value\) is optional and defines the value \(mu_0\) to test against. Default is 0.
• \(alternative="value"\) is optional and defines the type of alternative hypothesis: “two.sided”= two sided (A); “greater”=true mean difference is greater (B); “less”=true mean difference is lower (C). Default is “two.sided”.

Noted: Sang Nguyen

Nashville,TN - NOV 2016