John is an avid soccer player. Last season, he scored a goal $50$ times out of $200$ attempted shots. This season, he will have $200$ attempts.
(i) Using John's previous statistics, determine the minimum and maximum number of hits we expect from him this season with $90$% confidence.
(ii) John claims to have improved his skills since last season. Let $w$ be the probability that he scores a goal. He claims: $H_0: w = \frac{1}{4}$ should be rejected in favor of $H_1: w > \frac{1}{4}$. You observe John's next $300$ attempts and he scores $90$ goals. Find the $p$-value of this test. Do you agree that he has improved?
My attempt:
(i) The proportion, $p*$ from the last season is $p* = \frac{50}{200} = \frac{1}{4}$.
The confidence is $90$% so the $z_{1 - \alpha/2}$ value is $z_{0.95} = 1.645$
The error is approx.:
$$z_{1 - \alpha/2}\cdot \sqrt{\frac{p*(1-p*)}{n}}$$
$$= 1.645\cdot \sqrt{\frac{0.25\cdot 0.75}{200}}$$
$$= 0.05$$
So, the interval would be $(\frac{1}{4} - 0.05, \frac{1}{4} + 0.05) = (0.199637, 0.300363)$.
We expect the minimum this season to be $0.199637 \cdot 400 = 79.85$ and the maximum to be $0.300363 \cdot 400 = 120.15$.
We round so that the minimum is $80$ and the maximum is $121$.
(ii) $\alpha = 0.1$ so the one-tailed $z_{1 - \alpha}$ is $z_{0.9} = 1.28$.
The test statistic is
$$s = \frac{\frac{y}{n} - w}{\sqrt{\frac{w(1-w)}{n}}} = \frac{\frac{90}{300} - 0.25}{\sqrt{\frac{0.25\cdot 0.75}{300}}} = 2$$
The test statistic, $2$ is larger than $1.28$ so we reject the null hypothesis.
The $p-value$ is the probability that the standard normal $Z > 2$:
$$P(Z > 2) = 0.023$$
John improved.
Is this correct? I sometimes misinterpret the formulas I use especially for determining the error and the test statistic. Any assistance is much appreciated.
Both parts look good to me. I'm not thinking rounding makes much sense because it changes the confidence level.