Using binomial distribution in hypothesis testing I stumbled along this question, however I haven't found any solutions, the question is:
"A social media app was rated 'excellent' by 70% of its users based upon a large sample. They wondered if the percentage of users rating the app as 'excellent has changed' A random sample of 28 customers were asked to rate the app and 15 rated it as 'excellent' 
Use binomial distribution to investigate at 5% significance level whether there is evidence to support the claim that the percentage of users who rate the app as 'excellent' has changed."
What I have done:
I let x be the R.V. denoting the number of excellent ratings so
$$X \sim\ B(N,P) $$ where n = 28 and p = 0.7 
My expected value is $ E(X) = np = 28$ x $0.7 = 19.6 $ but we have $15$ rating it as excellent.
Next I created a null hypothesis and alternative hypothesis which is,
$$H_0 : p = 0.7$$
$$H_1 : p > 0.7$$ 
where p is the proportion of people who voted excellent.
Assuming the previous is correct I am trying to solve this,
$$P(X \geq 15) = 1 - P(X\leq 14)$$
However, I'm not sure if this is correct or how I solve this using binomial since $n = 28$ isn't in my binomial table. Also the question asks whether "percentage of users rating the app as 'excellent has changed" does this mean my alternative hypothesis should be 
$$H_1 : p \neq 0.7$$ 
 A: The alternative hypothesis should be $p\neq 0.7$, cause the number of users who rate "excellent" could both grow and decrease, and the problem asks for "changed".
You should compute the threshold values of the binomial distribution $Bin(28,0.7)$ such that the total weight of the upper and lower tails of the distribution (that is, parts of the distribution lying out of threshold values) is $5\%$, and check if the number $15$ is lying within the threshold values or not; if not, you reject null hypothesis at significance level $5\%$.

PS: Using R, the thresholds are
> qbinom(0.025,28,0.7)
[1] 15
> qbinom(0.975,28,0.7)
[1] 24
> 

$15$ is still within the thresholds, so we fail to reject.

PPS: p-value of observing $15$ is
> pbinom(15, 28, 0.7)+1-pbinom(24,28,0.7)
[1] 0.06475742

which also shows that we fail to reject at significance level $5\%$
A: Here is a way using the $\bf{p}$-value of your test statistic $x = 15$.
Since the test asks only about a possible change of the proportion $p = 0.7$, you have $H_1: p\neq 0.7$.
So, this is a two-tailed test based on the distribution $X\sim B(28,0.7)$.
The $\bf{p}$-value for $x=15$ is the probability - under the assumption that $p=0.7$ - of getting a result with the same or even larger absolute deviation from the expected mean $E(X) = np=19.6 \Rightarrow 19.6-15 = 4.6$:
$${\bf{p}} = P(|X-np|\geq 4.6\,|\, H_0)$$
If ${\bf{p}}\leq 5\%$ (the significance level), $H_0$ is rejected.
Using Wolfram|Alpha you get
$${\bf{p}} \approx 0.065 > 5\% \Rightarrow H_0 \text{ is not rejected }$$
