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Biala Biala
wrote...
Posts: 67
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4 years ago
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.
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wrote...
Educator
4 years ago Edited: 4 years ago, bio_man
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?
wrote...
Educator
4 years ago
Some notes to keep in mind:

To test the null hypothesis H0: p = p0 against a one- or two-sided alternative hypothesis Ha, 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 Ha: p > p0
P(Z < z) for Ha: p < p0
2P(Z>|z|) for Ha: p =/ p0.
Biala Author
wrote...
4 years ago
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.
wrote...
Educator
4 years ago
Welcome, also just finished answering the other.

Let me know if you need anything else.

Oh, and if you get a chance, use the link in my signature below to review us :-)
Biala Author
wrote...
4 years ago
I have this other question if that's okay.
wrote...
Educator
4 years ago Edited: 4 years ago, bio_man
I have this other question if that's okay.

Of course, no worries
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