Calculating the p-value for two events can anyone help me with that problem?
The school newspaper claims that 35% of students have credit card debt. You think it’s different. We simple random sample “M” students and “H” say that they have credit card debt. M=104 H= 30
Is there sufficient evidence at the 5% significance level (α = .05) to indicate that the percent of students with credit card debt is different than 35%
 A: Exact binomial test. You want to test $H_0: p = .35$ against $H_a: p\ne .36$ at 5% level. Your data are $x = 30$ successes in $n = 104$ trials.
If $H_0$ is true then $X \sim \mathsf{Binom}(n=104, p=.35).$
In R statistical software, this test gives P-value
about $0.2 > 0.05 = 5\%,$ so the null hypothesis is not rejected.
One might say that the sample proportion $\hat p = 0.2885$ of successes is not significantly different from $0.35 = 35\%.$
binom.test(30, 104, p=.35)

        Exact binomial test

data:  30 and 104
number of successes = 30, number of trials = 104, p-value = 0.2173
alternative hypothesis: true probability of success is not equal to 0.35
95 percent confidence interval:
 0.2038192 0.3855334
sample estimates:
probability of success 
             0.2884615 

Because you have shown no work of your own, there is no way for me to know
whether this answer matches what you are studying in class. Maybe you are expected to use an approximate normal tests based
on $Z = \frac{X - \mu}{\sigma},$ where $\mu = np_0 = 104(.35);$
$\sigma = \sqrt{np_0(1-p_0)},$ using standard normal CDF
tables to provide an approximate P_value.
Minitab statistical software shows results of such an approximate test:
Test and CI for One Proportion 

Test of p = 0.35 vs p ≠ 0.35


Sample   X    N  Sample p         95% CI         Z-Value  P-Value
1       30  104  0.288462  (0.201390, 0.375533)    -1.32    0.188

Using the normal approximation.

From R: Here is a plot of the $\mathsf{Binom}(104, .35)$ along
with its approximating normal density curve.

If this response gives you some cue how to work the problem
using what you have studied, perhaps you can edit your Question
to show what you have been able to do.
Note: The sample size as well as the observed proportion of
successes is important for this kind of test. If you had $90$
successes is $312$ trials, then there would be sufficient
evidence to rejest $H_0.$
binom.test(90, 312, p=.35)

        Exact binomial test

data:  90 and 312
number of successes = 90, number of trials = 312, p-value = 0.02398
alternative hypothesis: true probability of success is not equal to 0.35
95 percent confidence interval:
  0.2388094 0.3421823
sample estimates:
probability of success 
             0.2884615 

