Question

Out of 300 people sampled, 78 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids.

Give your answers as decimals, to three places

_________ < p <_________

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 78 /300 = 0.260

1 - = 0.740

Z_{/2}
= 2.576

Margin of error = E = Z_{
/ 2} *
((
* (1 -
)) / n)

= 2.576 * (((0.260 * 0.740) / 300)

= 0.065

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.260 - 0.065 < p < 0.260 + 0.065

0.195 < p < 0.325

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Give your answers as decimals, to three places
___________ < p < ____________

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people with kids. As in the reading, in your calculations: --Use z
= 1.645 for a 90% confidence interval --Use z = 2 for a 95%
confidence interval --Use z = 2.576 for a 99% confidence interval.
Give your answers to three decimals Give your answers as decimals,
to three decimal places. [,]

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Give your answers as decimals, to three places.
We can write this answer as
+/-
Or
<p<

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Give your answers as decimals, to three places

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proportion of people with kids. As in the reading, in your
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= 2 for a 95% confidence interval --Use z = 2.576 for a 99%
confidence interval. Give your answers to three decimals Give your
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< p <

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Give your answers as decimals, to three places

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