# Testing hypotheses proportions

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.

## 1 Answer

Both parts look good to me. I'm not thinking rounding makes much sense because it changes the confidence level.