In a 60-day period in Ithaca 12 days were rainy

Question

I hope that someone could tell me if my answer is correct and if this is not the case, could you tell me what I do wrong.

 In a 60-day period in Ithaca 12 days were rainy. Is this observation consistent
with the belief that the true proportion of rainy days is 1/3?

For this exercise I used the Z single proportion formula: $$ z = \frac{(p1 - P0)}{σ} $$ $$ σ = \sqrt{\frac{P * ( 1 - P )}{n}} $$

$$ h0 = P0 = \frac{1}{3} $$ $$ h1 \ne \frac{1}{3}$$ $$ p1 = \frac{12}{60} $$ $$ n = 60 $$ $$ z = \frac{(\frac{12}{60} - \frac{1}{3})}{\sqrt{\frac{\frac{1}{3} * ( \frac{2}{3} )}{n}}} = -0.822 $$

Then I filled "-0.822" in a Z calculator. This gave me 0.2055. So my answer would be "No".

The answer of the textbook is also "No", but I think the 0.2055 is to high.

Answer 1

$\begin{align} z &= \frac{\tilde p - \mu}{σ} \\ σ &= \sqrt{\frac{P \times ( 1 - P )}{n}} \\ H_0 &: \mu = \frac{1}{3} \\ H_1 &:\mu \ne \frac{1}{3}\\ \tilde p &= \frac{12}{60} \\ n &= 60 \\ z &= \frac{\frac{12}{60} - \frac{1}{3}}{\sqrt{\frac{\frac{1}{3}\frac{2}{3} }{60}}} =-2\sqrt{\frac{6}{5}}\approx -2.19\\ P(Z>|-2.19|)&=2P(Z<-2.19)\approx 0.0286 \end{align}$

With a significance level of $0.01$ we would reject it.

Answer 2

There are a number of errors with this solution, all of them in the last few lines.

I think you made an error when calculating your z value. It should be $z \approx 2.19$.

You have also calculated the p value for a one-sided test - but you are doing a two sided test.

A p-value of 0.2055 would not be reason to reject the null hypothesis. Concluding that the observations are inconsistent from a p-value of 0.2055 is wrong. Usually you require your p-value to be less than a 'significance level' (usually 0.05) for you to reject the null hypothesis.