Reducing confidence intervals and alpha into a single confidence measure for layperson? Is there a statistical approach or other practical convention for converting the statistical outputs into a single measure of 'confidence' in an estimate that a layperson can easily interpret?
Example
Suppose we observe a binomial event occur with outcomes: 0, 1, 0, 1, 0, 1, 0, 1, 0. A good example could be whether a customer read a newsletter sent to them (we may try to estimate whether they will read a future newsletter sent to them) based on their history of reading / not reading emails
We can say (in R code)

# Sample of whether customer read email in the past (9 previous emails sent)
logical_sample <- c(0,1,0,1,0,1,0,1,0) # 0 = did not read; 1 = did read

sample_mean <- mean(logical_sample) # 0.4444444
sample_st_deviation <- sd(logical_sample) # 0.5270463
n <- length(logical_sample) # 9
alpha <- 0.95 

confidence_interval <- function(alpha, sample_mean, sample_st_deviation) {
  qnorm(1-(1-alpha)/2) * (sample_st_deviation / sqrt(n)) # note: qnorm(0.975)# [1] 1.959964
}

confidence_interval(alpha, sample_mean, sample_st_deviation) # 0.3443306


Now suppose a non-technical person does not want to know these statistics, but instead wants to know only two things, 1. the expected outcome, and 2. "how confident we are as a percent". E.g. 10% might be a fairly wild guess, 50% might be a little confident, 90% would be quite confident etc. 
The first is straight forward 0.444, but for the second..
We can say the event that we witnessed 9 (n) times, has an expected outcome of 0.444 and that we are 95% confident that the mean is within 0.444 +/- 0.344 (i.e. between 0.10 and 0.789)
How can we map these stats to a 'confidence' percentage (that a layperson could use to gauge how confident we are of the estimate)
Question
How can we map these statistics into a single figure that the layperson describes as 'confidence'? 
 A: I'm not sure there is a direct answer to this, so this is certainly not an answer. But here is something to think about: 
CI as interval of acceptable hypothetical values. Consider the one-sample t test of $H_0: \mu = \mu_0$ against $H_a: \mu \ne \mu_0.$ One fails to reject (let's just say "Accepts") $H_0$ at the 5% level if $T = \frac{\bar X - \mu_0}{S/\sqrt{n}}$ lies in the interval between $\pm t^*,$ where $t^*$ cuts probability 0.025 from the upper tail of Student's t dist'n with $df = n-1.$
"Inverting the test," one has the 95% confidence interval (CI) $\bar X \pm t^*\frac{S}{\sqrt{n}}.$ Thus, in effect, the 95% CI is the set of "Acceptable" values $\mu_0.$
Approximate binomial test. In your binomial example, the test of $H_0: p = p_0$ against $H_a: p \ne p_0,$ is often Accepted at the 5% level if 
$Z = \frac{\hat p - p_0}{\sqrt{p_0(1-p_0)/n}}$ has $|Z| < 1.96,$ where $\hat p = x/n$ (the proportion of Successes in $n$ trials) and 1.96 cuts probability 0.025 from the upper tail of the standard normal distribution.  This is not an exact test because it relies on a normal approximation to the binomial.
Wald interval is asymptotic, not for small samples. Furthermore, the Wald CI for $p$ does not even try to "invert" the approximate test. The CI is of the form $\hat p \pm 1.96\sqrt{{\hat p(1-\hat p)}/{n}}.$ This introduces a second approximation, estimating
the standard error $\sqrt{p_0(1-p_0)/n}$ by $\sqrt{\hat p(1-\hat p)/n}.$
The normal approximation together with the estimation of the standard
error can go badly wrong for small $n$ and $p_0$ far from 1/2. The result
can be a so-called "95%" confidence interval with much less than 95% probability
of including the true value of $p.$ (Perhaps see graphs here.)
Wilson CI leads to Agresti CI It is possible, but messy to find and solve an appropriate quadratic equation to invert the normal test. The messy result is the Wilson CI shown on Wikipedia. The Agresti-Coull CI ignores small terms in the Wilson formula, conflates
1.96 with 2, and simplifies the algebra.
Approximate duality of binomial test and CI. Thus, up to normal approximation and Agresti-Coull simplification, one can say that a 95% A-C CI consists of "Acceptable" values of $p_0$ in the test of $H_0.$ This is "pretty much" true if the normal approximation is valid.
(Some authors give the rule of thumb that $\min(np_0, n(1-p_0) > 5$ for that.)
