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 1


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


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