Using binomial distribution in hypothesis testing

Question

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

$$H_1 : p \neq 0.7$$

Answer 1

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 $5\%$.

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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.

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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 $5\%$