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adoni [48]
3 years ago
9

An entomologist writes an article in a scientific journal which claims that fewer than 16 in ten thousand male fireflies are una

ble to produce light due to a genetic mutation. Use the parameter​ p, the true proportion of fireflies unable to produce light.
Express the null hypothesis and the alternative hypothesis in symbolic form for a test to support this claim.
A. H0: p > 0.0016
H1: p <= 0.0016
B. H0: p < 0.0016
H1: p >= 0.0016
C. H0: p = 0.0016
H1: p < 0.0016
D: H0: p = 0.0016
H1: p > 0.0016
Mathematics
1 answer:
bazaltina [42]3 years ago
6 0

Answer:

a) H0: p > 0.0016

H1: p ≤ 0.0016

Step-by-step explanation:

A null hypothesis states that there is no relation between two variables, it's what the researcher wants to discredit.

On the other hand, the alternative hypothesis is what the researcher tries to prove.

<u>The claim of this entomologists is that fewer than 16 in ten thousand male fireflies are unable to produce light due to a genetic mutation. Therefore, this claim is what he wants to prove and it would be the alternative hypothesis.</u>

H1: p ≤ 0.0016

The null hypothesis would be the opposite of the alternative hypothesis and therefore it would be:

H0: p > 0.0016

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melisa1 [442]

Answer:

P(\hat p>0.14)

And using the z score given by:

z = \frac{\hat p -\mu_p}{\sigma_p}

Where:

\mu_{\hat p} = 0.12

\sigma_{\hat p}= \sqrt{\frac{0.12*(1-0.12)}{474}}= 0.0149

If we find the z score for \hat p =0.14 we got:

z = \frac{0.14-0.12}{0.0149}= 1.340

So we want to find this probability:

P(z>1.340)

And using the complement rule and the normal standard distribution and excel we got:

P(Z>1.340) = 1-P(Z

Step-by-step explanation:

For this case we have the proportion of interest given p =0.12. And we have a sample size selected n = 474

The distribution of \hat p is given by:

\hat p \sim N (p , \sqrt{\frac{p(1-p)}{n}})

We want to find this probability:

P(\hat p>0.14)

And using the z score given by:

z = \frac{\hat p -\mu_p}{\sigma_p}

Where:

\mu_{\hat p} = 0.12

\sigma_{\hat p}= \sqrt{\frac{0.12*(1-0.12)}{474}}= 0.0149

If we find the z score for \hat p =0.14 we got:

z = \frac{0.14-0.12}{0.0149}= 1.340

So we want to find this probability:

P(z>1.340)

And using the complement rule and the normal standard distribution and excel we got:

P(Z>1.340) = 1-P(Z

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Step-by-step explanation:

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