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3 years ago

9

from 1956 to 1966 the number of automobiles in the country increased at an annual rate of 6.65 percent. what is the doubling tim

e for growth in automobiles in the US?

1 answer:

NeTakaya3 years ago

4 0

A= P(1+i)ⁿ ,

where A is the value generated after a time " n" at an interest rate of i (in %).

When A = 2.P ?

Replace in the formula: 2.P = P(1+0.0665)ⁿ , after simplification by P, we get

2= (1+0.0665)ⁿ

To calculate n, let' use ln (Neperian logarithm)
ln(2) = n.ln(1.0665) and

n=(ln2)/(ln(1.0665) and n=10.76 years ,

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Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

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In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

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Item a:

The sample size is <u>n for which M = 0.04.</u>

There is no estimate, hence $\pi = 0.5$

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$0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}$

$\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2$

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Rounding up:

A sample of 601 is needed.

Item b:

The estimate is $\pi = 0.96$ , hence:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

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$0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}$

$\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2$

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Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to $\pi = 0.5$ , the larger the sample size needed, hence, the correct option is A.

For more on the z-distribution, you can check brainly.com/question/25404151

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