Biology Forums - Study Force

The claim is that the proportion of drowning deaths of children attributable to beaches is more than 0.25, and the sample statistics include n=575 drowning deaths of children with 30% of them attributable to beaches.

Null Hypothesis (H_0): P = 0.25
Alternative Hypothesis: (H_A): P > 0.25

\(z=\frac{\overline{p}-p}{\sqrt{\frac{p\cdot q}{n}}},\ where\ is\ q\ =\ 1-p\)

\(z=\frac{0.30-0.25}{\sqrt{\frac{0.25\cdot \left(1-0.25\right)}{575}}}=2.768\)

Rounded to 2 decimal places, 2.77

How much do they want it rounded to?

Some notes to keep in mind:

To test the null hypothesis H₀: p = p₀ against a one- or two-sided alternative hypothesis H_a, replace p with p₀ in the test statistic 
The test statistic follows the standard normal distribution (with mean = 0 and standard deviation = 1).  The test statistic z is used to compute the P-value for the standard normal distribution, the probability that a value at least as extreme as the test statistic would be observed under the null hypothesis.  Given the null hypothesis that the population proportion p is equal to a given value p₀, the P-values for testing H₀ against each of the possible alternative hypotheses are:

P(Z > z) for H_a: p > p₀
P(Z < z) for H_a: p < p₀
2P(Z>|z|) for H_a: p =/ p₀.

Thank you. I got the same answer and I just wanted to know if it was correct. It's just two decimal places so its 2.77.

Welcome, also just finished answering the other.

Let me know if you need anything else.

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I have this other question if that's okay.

Of course, no worries