Hypothesis testing for binomial probability of rain--too low/high? Here is a problem
The probability there is heavy rain at least once in a week in a region is thought to be 0.7 for any given week.
Over a period of 52 consecutive weeks, there is heavy rain in the following thirteen weeks:
1, 2, 5, 6, 7, 10, 12, 13, 15, 16, 17, 18, 19

a - Assuming this probability to be true, find the probability that there is heavy rain in more than one week out of four consecutive weeks.
b - By considering the weeks independently, perform a hypothesis test at the 5% significance level to test if 0.7 is too low an estimate for the probability of heavy rain in any week. Make sure to state your hypotheses clearly.
c - By considering the weeks in thirteen groups of four consecutive weeks, perform a hypothesis test at the 5% significance level to test if your answer to part a) is too high an estimate for the probability there is heavy rain in more than one week in any four-week period. Make sure to state your hypotheses clearly.
Here is what I have tried
For (a), $0.7^4=0.2401$.
For (b),
$X \sim B(52,0.7)$, 
$$ H_0: p=0.7, \quad H_1: p<0.7 (????) $$
Then
$$
P(X \leq 13) = 2.89*10^{-11} \quad < \quad 0.05,
$$
which is significantly less than 5%. Hence there is significant evidence to reject $H_0$. But is this the correct understanding of the phrase too low.
Now for (c), how do I undestand this too high? How does that compare to answer in (a)? I am a bit confused.
Start with $X \sim B(13,0.7^4)$, 
$$ H_0: p=0.7^4, \quad H_1: p>0.7^4 \quad (\text{????  not sure}) $$
$$
P(X \geq 1) = 1 - P( X=0 ) = 0.971826... 
$$
or
$$
P(X<1) = P( X=0 ) = 0.0238173...
$$
How do I answer this question here, compared to (a), too high and the 5% level ect?? I am fully confused by these words here and trying to understand it.
Thanks.
 A: Some comments and some computations in R to get you started. In R pbinom denotes a binomial CDF and dbinom denotes a PDF.
(a) $Y \sim \mathsf{Binom}(n = 4, p = .7).$ You seek 
$P(Y > 1) = 1 - P(X \le 1) = 0.9163.$
1 - pbinom(1, 4, .7)
[1] 0.9163

y = 0:4;  pdf = dbinom(y, 4, .7)
plot(y, pdf, type="h", ylim=c(0,max(pdf)), lwd=2,
      ylab="PDF", main = "PDF of BINOM(4,.7)")
  abline(h=0, col="green2")


(b) If there is a 70% chance of rain in any one week then we'd expect something near 36 or 37
rainy weeks out of 52. But we observed only 13 rainy weeks. So the 70% probability seems
too high.
Let $X$ be the number of rainy days in $n = 52$ weeks.
Test null hypothesis $H_0: p = .7;$ alternative hypothesis
$H_a: p > .7.$ 
The 'null distribution' is $X \sim \mathsf{Binom}(n=52,p=0.7).$
Observed $X = 13.$ Intuitively, we have no evidence the 70% probability
is too low. So we do not reject $H_0$ against the alternative that
the probability should be even higher.
P-value is the probability under $H_0$ of an
outcome as extreme or more extreme (in the direction of $H_a)$ than observed.
Formally, $P(X \ge 13\,|\,p=.7) = 1 -P(X \le 12) \approx 1,$ which is not smaller than 5%,
so no rejection of the null hypothesis.
1 - pbinom(12, 52, .7)
[1] 1


Unless your instructor or text has put some emphasis on using technology to compute
binomial probabilities, I suppose you are expected to compute the P-value in (b) by using the normal approximation to binomial distributions.  The probability computation in (a)
is easy enough to do on a calculator using the formula for the binomial PDF.
