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sergeinik [125]

2 years ago

8

A 2011 Gallup Poll found that 76% of Americans believe that high achieving high school students should be recruited to become te

achers. This poll was based on a random sample of 1002 Americans. Find a 90% confidence interval for the proportion of Americans who would agree with this. Interpret your interval in this context. Explain what “90% confidence” means. Do these data refute a pundit’s claim that 2 / 3 of Americans believe this statement? Explain.

1 answer:

Anna007 [38]2 years ago

8 0

Answer:

Step-by-step explanation:

Let p be the proportion of Americans who believe that high achieving high school students should be recruited to become teachers

Given that p = 0.76 and q = 1-p =0.24

Std error for proportion= $\sqrt{\frac{pq}{n} } \\=0.0135$

For 90% confidence interval we have

0.76±Margin of error

= 0.76±1.645(0.0135)

=0.76±0.0222

thus confidence interval

= (0.7378, 0.7822)

This means we are 90% confidence for randomly selected samples of large size, proportion will fall within this interval

Since proportion lower bound is 73.78% we can say that pundit claim of 2/3 is correct.

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Step-by-step explanation:

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This gives:

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$P(0) =\frac{1200}{1+11*1}$

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$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

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$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

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$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

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