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 frac

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?
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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Thành Đạt 6 months 2021-08-23T10:30:27+00:00 1 Answers 0 views 0

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    2021-08-23T10:31:59+00:00

    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?

    Answer:

    The First question

                         n = 1879

    A

        The 96% confidence interval is  

                          0.51 <  p <  0.63

    B

       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

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