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3 years ago

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Si la época prehistórica sucedió desde hace 2.5 millones de años a.C., hasta hace 2.500 años a.C.; el período Paleolítico (2.5 m

illones de años a.C. Hasta 10.000 a.C.; el Mesolítico 10.000 a.C. Hasta 5.000 a.C. Y, el Neolítico sucedió desde el 5.000 a.C. Hasta el 2.500 a.C. ¿A qué fracción de la época prehistórica corresponde el período paleolítico? *

1 answer:

tia_tia [17]3 years ago

5 0

Answer:

El período paleolítico corresponde a $\frac{332}{333}$ de la época prehistórica.

Step-by-step explanation:

En primer lugar, se determina la extensión de tiempo, medida en años, tanto de la época prehistórica como el período paleolítico: (Nótese que el signo negativo significa a.C.)

Época Prehistórica

$\Delta T = -2500-(-2500000)$

$\Delta T = 2497500\,years$

Período Paleolítico

$\Delta t = -10000 - (-2500000)$

$\Delta t = 2490000\,years$

La fracción de la época prehistórica que corresponde al período paleolítico es:

$r = \frac{\Delta t}{\Delta T}$

$r = \frac{2490000\,years}{2497500\,years}$

$r = \frac{332}{333}$

El período paleolítico corresponde a $\frac{332}{333}$ de la época prehistórica.

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Answer:

$z=\frac{0.333 -0.3}{\sqrt{\frac{0.3(1-0.3)}{195}}}=1.01$

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Step-by-step explanation:

Data given and notation

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$\hat p=\frac{65}{195}=0.333$ estimated proportion of women who complain of nausea between the 24th and 28th week of pregnancy

$p_o=0.3$ is the value that we want to test

$\alpha$ represent the significance level

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$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportion is higher than 0.3.:

Null hypothesis: $p\leq 0.3$

Alternative hypothesis: $p > 0.3$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

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Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.333 -0.3}{\sqrt{\frac{0.3(1-0.3)}{195}}}=1.01$

Statistical decision

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Since is a right tailed test the p value would be:

$p_v =P(z>1.01)=0.156$

So the p value obtained was a very low value and using the significance level asumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%

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