$t$ test for paired samples in R

Question

I have two vector samples $x,y$:

x = c(107.32362,  89.15552,  90.75148, 109.51189,  98.43394,  97.25716 ,115.39168,
     95.68554,  93.60000, 103.00000 ,112.70000, 138.40000,  93.30000 , 87.70000,
     83.30000, 105.00000 ,103.40000 ,105.70000, 107.80000,  96.10000, 102.00000,
     105.40000, 123.90000 ,141.10000)
y =  c(100.1,  82.0,  39.0,  18.0,  83.8 ,109.1 ,114.0 , 95.4,  93.1, 116.8  ,56.9 , 56.6,  86.3
      ,62.0,  38.9,  85.2 ,124.4, 124.3, 139.2, 105.6, 117.4 ,125.2, 111.5, 131.7)

I want to calculate the t.test for equality of means (or the differences).I am assuming the values are paired.

Because the samples are paired (matched) we must use the differences between the paired values as : $$D = X_{1i} - X_{2i}$$ The point estimate for the paired difference population mean $\mu_{D}=\mu_{x} - \mu_{y}=0$ is denoted as $\bar{D} = \frac{\sum_{i=1}^{n}D_{i}}{n}$ and the standard deviation of the differences $$S_{D} = \sqrt{ \frac{\sum_{i=1}^{n}(D_{i}-\bar{D})^2 }{n-1}}$$ The test statistic for the equality of means is :

$$t =\frac{\bar{D}-\mu_{D}}{ \frac{S_D}{\sqrt{n}}} = 1.835492$$

d = x-y
t = (mean(d)-0)/ (sd(d)/sqrt(length(d)));t
[1] 1.835492

For a two-tail test with a given level of significance $\alpha$, we reject the null hypothesis if the calculated test statistic is greater than the upper-tail critical value $t_{n-1}$ from the $t$ distribution, or if the calculated test statistic is less than the lower-tail critical value $-t_{n-1}$ from the $t$ distribution. That is, the decision rule is: Reject $H_{0}$ if $t > t_{n-1}$ or if $t< -t_{n-1}$ otherwise, do not reject $H_{0}$. The degrees of freedom in our example and $24-1=23$ and for $\alpha =0.10$ the inverse cumulative distribution for $1-\alpha =0.90$ with 23 degrees of freedom of $t$ distribution is $1.31946$

Therefore $t>t_{n-1}$ or $1.83>1.31$ and I reject the null hypothesis.

t>qt(0.9,23)
[1] TRUE

But when I run the t.test() function in R it reports me a p.value greater than 0.05 (0.0794) and I must not reject the null hypothesis.

> t.test(x,y, 
        paired = TRUE,
        alternative = "two.sided",
        conf.level = 0.90)

    Paired t-test

data:  x and y
t = 1.8355, df = 23, p-value = 0.0794
alternative hypothesis: true difference in means is not equal to 0
90 percent confidence interval:
  0.7990197 23.3185494
sample estimates:
mean of the differences 
               12.05878 

What is wrong here ?

Answer 1

For a two-sided test the relevant critical point is $t_{n-1,\alpha/2}$ so you would find this as follows:

#Find critical point
n     <- length(d)
alpha <- 0.1
CRIT  <- qt(1-alpha/2, df = n-1)

#Show critical point
CRIT
[1] 1.713872

Since $|t| = 1.835492 > 1.713872 = t_{n-1,\alpha/2}$ you reject the null hypothesis at this significance level. This matches the decision from the t.test output --- since $p = 0.0794 < 0.1 = \alpha$ you likewise reject the null.