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FromTheMoon [43]
3 years ago
11

A researcher wishes to estimate the number of households with two cars. A previous study indicates that the proportion of househ

olds with two cars is 25%. How large a sample is needed in order to be 99% confident that the sample proportion will not differ from the true proportion by more than 3%?
A) 4.
B) 1132.
C) 1842.
D) 1382.
Mathematics
1 answer:
ivanzaharov [21]3 years ago
3 0
The correct answer is D
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Answer:

a) z = 1.645

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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).

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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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Step-by-step explanation:

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A=6+3.5=9.5\ ft^2

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