Mathematics, 03.12.2020 19:50 vanessa051266

How large a sample is needed if we wish to be 96% confident that our sample proportion in Exercise 9.53 will be within 0.02 of the true fraction of the voting population? A. A random sample of 200 voters in a town is selected, and 114 are found to support an annexation suit. Find the 96% conﬁdence interval for the fraction of the voting population favoring the suit.
B. What can we assert with 96% conﬁdence about the possible size of our error if we estimate the fraction of voters favoring the annexation suit to be 0.57?

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Answer from: cteal7507

Complete Question

How large a sample is needed if we wish to be 96% confident that our sample proportion in Exercise 9.53 will be within 0.02 of the true fraction of the voting population?

Exercise 9.53

A. A random sample of 200 voters in a town is selected, and 114 are found to support an annexation suit. Find the 96% conﬁdence interval for the fraction of the voting population favoring the suit.

B. What can we assert with 96% conﬁdence about the possible size of our error if we estimate the fraction of voters favoring the annexation suit to be 0.57?

The First question

n = 1879

The 96% confidence interval is

0.51 < p < 0.63

The possible size of our error if we estimate the fraction of voters favoring the annexation suit to be 0.57 is

E = 0.06

Step-by-step explanation:

From the question we are told that

The margin of error is E = 0.02

The sample size is n = 200

The number that supported the annexation suit is k = 114

Considering the first question

Generally our sample in Exercise 9.53 is $\^ p = 0.57$

From the question we are told the confidence level is 96% , hence the level of significance is

$\alpha = (100 - 96 ) \%$

=> $\alpha = 0.04$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.751$

Generally the sample size is mathematically represented as

$n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )$

=> $n = [\frac{1.751}{0.02 } ]^2 * \^ p (1 - \^ p )$

=> $n = [\frac{1.751}{0.02 } ]^2 * 0.57 (1 -0.57 )$

=> n = 1879

Considering question B

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.751 * \sqrt{\frac{ 0.57 (1- 0.57 )}{200} }$

=> E = 0.06

Considering question A

Generally the sample proportion of the number that supported the annexation suit is mathematically represented as

$\^ p = \frac{114}{200}$

=> $\^ p = 0.57$

Generally 96% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> 0.57 - 0.06 < p < 0.57 + 0.06

=> 0.51 < p < 0.63

Answer from: Quest

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Answer from: Quest

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