How can I perform a t test in R that tests the mean difference from a target ratio? In my data I have a ratio that aims to be 5:1. To test how close the mean difference is between the ratios in my data (of 10 items) and the ideal 5:1 ratio I carried out the following t test.
t.test(SratioA$ratio, c(5,5,5,5,5,5,5,5,5,5), paired = TRUE)
t.test(SratioB$ratio, c(5,5,5,5,5,5,5,5,5,5), paired = TRUE)

Would a better test be:?
t.test(SratioA$ratio, c(5,5,5,5,5,5,5,5,5,5))
t.test(SratioB$ratio, c(5,5,5,5,5,5,5,5,5,5))

If not one of those how would I go about testing this? 
 A: You should really have posted also your data. Without them we cannot judge if the normality hypothesis needed for t-test is reasonable. But, however, the main problem with your approach have nothing to do with that, and would remain with a nonparametric test. 
Let us first compare your two proposals:
> n <- 10
> set.seed(7*11*13)
> test.dat <- rnorm(n, 5, sd=1.5)
> t.test(test.dat, rep(5,n), paired=TRUE)

    Paired t-test

data:  test.dat and rep(5, n)
t = -0.8295, df = 9, p-value = 0.4283
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -1.9087770  0.8845178
sample estimates:
mean of the differences 
             -0.5121296 

> t.test(test.dat, rep(5,n), paired=FALSE)

    Welch Two Sample t-test

data:  test.dat and rep(5, n)
t = -0.8295, df = 9, p-value = 0.4283
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -1.9087770  0.8845178
sample estimates:
mean of x mean of y 
  4.48787   5.00000 

They give the same result, why? Concentrate on the numerator of the t-stat (we will look at the denominator later). For the paired test we get
$$
   \text{mean}(x_1-5,x_2-5, \dotsc, x_n-5)=\text{mean}(x_1,\dotsc,x_n)-5
$$
and for the nonpaired we get:
$$
\text{mean}(x_1,x_2, \dotsc, x_n)-\text{mean}(5, 5, \dotsc, 5)
$$
and this are equal. But, your error lies elsewhere, you do not really have two groups (whether paired or unpaired), you have one group, the 5's are not observations, they are specifying the hypothesis to test. By treating them as data, your variance estimate is in error. The way to do this is:
> t.test(test.dat, mu=5)

    One Sample t-test

data:  test.dat
t = -0.8295, df = 9, p-value = 0.4283
alternative hypothesis: true mean is not equal to 5
95 percent confidence interval:
 3.091223 5.884518
sample estimates:
mean of x 
  4.48787 

(and then maybe replace the t test with some nonparametric test)
