Demonstrating agreement on a Likert scale question in a single population I have one population of people to whom I am asking one Likert scale question (1 = strongly disagree; 5 = strongly agree).
I want to demonstrate that people mostly agree with the statement being rated.
Specifically, I want to demonstrate that more people answer 'agree' or 'strongly agree' than we would expect if people were choosing answers at random. In other words, there is a significant trend for people to choose 'agree' or 'strongly agree' over the other options.
All the resources I have found talk about comparing two groups. I only have one group.
My first thought was to use a Binomial test, but the possible outcomes are not equally likely. For example, if we consider 'success' to mean 'chose agree or strongly agree,' then there is a 2/5 chance of success and a 3/5 chance of failure... and that's assuming we don't account for central tendency bias. If I understand correctly, this means I can't use a Binomial test in the traditional way... though perhaps someone who understands it better can modify it to account for the unequal probabilities of success and failure.
Can someone please let me know the appropriate test to use?
 A: If you have $n = 100$ subjects choosing among the five categories at
random, then the number of Successes (Likert 4's or 5's) is
$X \sim \mathsf{Binom}(n = 100, p = 2/5 = .4)$ 
Exact binomial test. You want to test $H_0: p = 0.4$ vs. $H_a: p > 0.4.$ You might use an exact test
that rejects $H_0$ in favor of $H_a,$ if $X \ge 49.$
Significance level. That test has significance level 4.23% because
$$P(X \ge 49|p=.4) = 1 - P(X \le 48|p=.4) = 0.0423,$$ as computed in
R statistical software, where pbinom is a binomial CDF.
1 - pbinom(48, 100, .4)
[1] 0.04230142

Power. The power against the specific alternative that there are 60% Successes, is 99%. (The power against alternative $p = 0.55$ is about 90%.)
$$P(X \ge 49|p=.6) = 1 - P(X \le 48|p=.6) = 0.99.$$
 1 - pbinom(48, 100, .6)
 [1] 0.9899949
 1 - pbinom(48, 100, .55)
 [1] 0.9040484

Graphical summary. The upper panel of the figure below shows the null distribution $\mathsf{Binom}(100,.4);$ the significance level is the sum of the
heights to the right of the vertical red line. Similarly the lower
panel shows the alternative distribution $\mathsf{Binom}(100, .6),$
illustrating the power against alternative $p = 0.6.$

Notes: (1) Similar tests can be constructed for different numbers $n$ of subjects. (2) For sufficiently large $n$ (say $n\ge 25)$, you could use a normal approximation to obtain good approximations of the relevant probabilities. (3) Many statistical computer programs have procedures for doing binomial tests, sometimes called 'test of a single proportion'. In R, the test is binom.test.
Minitab's output for such a test is as follows:
Test and CI for One Proportion 

Test of p = 0.4 vs p > 0.4

                                              Exact
Sample   X    N  Sample p  95% Lower Bound  P-Value
1       51  100  0.510000         0.423411    0.017

