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Neporo4naja [7]
3 years ago
9

A chemist examines 12 sedimentary samples for bromide concentraction. The mean bromide concentration for the sample date is 0.43

7 cc/cubic meter with a standard deviation of 0.0325. Determine the 90% confidence interval for the population mean bromide consentraction. Assume the population is approximately normal.
a. Find the critical vaule that should be used in constructing the confidence interval. Round your answer to three decimal places. Answer:_________

b. Construct the 90% confidence interval. Round your answer to tree decimal places.Lower endpoint:_____________ Upper endpoint:________________
Mathematics
1 answer:
Umnica [9.8K]3 years ago
4 0

Answer:

a) z = 1.645

b) Lower endpoint: 0.422cc/m³

Upper endpoint: 0.452 cc/m³

Step-by-step explanation:

Population is approximately normal, so we can find the normal confidence interval.

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645. This is the critical value, the answer for a).

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.645*\frac{0.0325}{\sqrt{12}} = 0.015

The lower end of the interval is the sample mean subtracted by M. So it is 0.437 - 0.015 = 0.422cc/m³.

The upper end of the interval is the sample mean added to M. So it is 0.437 + 0.015 = 0.452 cc/m³.

b)

Lower endpoint: 0.422cc/m³

Upper endpoint: 0.452 cc/m³

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Una manera de resolver este problema es la siguiente:

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\\ 2km^{2} = 2km * 1km = 2000m * 1000m = 2000000m^{2} = 2 * 10^{6}m^{2}

En palabras, \\ 2km^{2} = 2000000m^{2}, o <em>dos kilómetros cuadrados</em> son iguales a <em>2 millones de metros cuadrados</em>.

Sabemos que:

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De esta manera, <em>la parte que ocupan los pinos es el total del bosque menos el área ocupada por las hayas</em>. Por lo tanto, el área ocupada por los pinos es:

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