How to calculate the standard deviation and t-test statistic of matched pairs

Question

So, I have two variables (matched pairs) that I believe are means, with a collection of 10 samples a piece (although if I'm reading it right, they're the same 10 individuals undergoing different trials). It looks a bit like this:

Situation 1: 125, 156, 140, 175, 153, 148, 180, 135, 168, 157
Situation 2: 120, 145, 142, 150, 160, 148, 160, 142, 162, 150

Now the problem is asking for a t test statistic and a p-value, the test statistic is $\frac{\bar x_d}{s_d/\sqrt{n_d}}$, where:

• $\bar x_d$ should be the difference between all of the means summed, so Avg(situation1) - Avg(situation2), which comes out to 5.8.

• $s_d$ is the standard deviation (that's all my textbook is giving me).

And $n_d$ is just 10.

Any help would be appreciated, my past two attempts at this problem have been complete bummers - the standard deviation I started trying was 17.512 because that's just what my technology spits out when I put all the values in.

Answer 1

@chi has given a verbal description. Here is how it looks in the R t.test procedure.

x1 =  c(125, 156, 140, 175, 153, 148, 180, 135, 168, 157)
x2 =  c(120, 145, 142, 150, 160, 148, 160, 142, 162, 150)
d = x1-x2

summary(d); length(d); sd(d)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   -7.0    -1.5     5.5     5.8    10.0    25.0 
[1] 10        # number of differences
[1] 10.65416  # standard deviation of 10 differences

stripchart(d, meth="stack", pch=20)

Notice that the t statistic for the t tests below is as follows:

(mean(d) - 0)/(sd(d)/sqrt(10))
[1] 1.721507

First, doing a one-sample test on the differences d to test the null hypothesis that the population mean of the paired differences is $0.$ The sample mean of the differences is greater $\bar D = 5.8 > 0,$ but compared with the standard deviation $S_d = 10.65$ we cannot say that $\bar D$ is significantly larger than $0.$ The P-value exceeds $0.05 = 5\%,$ so the null hypothesis is not rejected. [The population difference might be anywhere between $-1.822$ and $13.422.]$

t.test(d)

        One Sample t-test

data:  d
t = 1.7215, df = 9, p-value = 0.1193
alternative hypothesis: 
  true mean is not equal to 0
95 percent confidence interval:
 -1.821526 13.421526
sample estimates:
mean of x 
      5.8 

Second, if you use the parameter pair=T for t.test, then the program find the differences and does the same thing as above; the t-statistic, degrees of freedom, and P-value are all the same as before.

t.test(x1, x2, pair=T)

        Paired t-test

data:  x1 and x2
t = 1.7215, df = 9, p-value = 0.1193
alternative hypothesis: 
 true difference in means is not equal to 0
95 percent confidence interval:
  -1.821526 13.421526
sample estimates:
mean of the differences 
                   5.8 

Notes: (1) In both cases the null hypothesis is that the population of pairs has mean $0$ and the alternative hypothesis is two-sided. (Additional parameters in t.test are required if you want to change either.) (2) Because each pair is for a different individual it is reasonable to assume that there is a positive correlation between x1 and x2. [Because x1 and x2 are not independent samples, it would be incorrect to do a two-sample t.test.]

cor(x1,x2)
[1] 0.799743

Also, t tests are exact only if the data (differences here) are from a normal population. With only $n=10$ differences it is difficult to assess normality of the differences. However, there is no evidence of strong skewness in the stripchart, and there are no extreme outliers. Moreover, a Shapiro-Wilk test shows that differences are consistent with a normal sample.

shapiro.test(d)

        Shapiro-Wilk normality test

data:  d
W = 0.93627, p-value = 0.5123