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You are told that a 90% confidence intertal for a population proportion is (0.4086, 0.5914). What was the sample proportion that lead to this confidence interval? Also, what was the size of the sample used?​

Sagot :

ajeigbeibraheem

The size of the sample used is 81

What is the sample size (n)?

The sample size is a part of the population sample used in an empirical study to make a statistical judgment.

From the parameters given:

  • The confidence interval level for the population proportion is = 90%

The level of significance ∝ = 1 - C.I

∝ = 1 - 0.90

∝ = 0.10

From the  [tex]\mathbf{Z_{\alpha /2}}[/tex] tables:

[tex]= \mathbf{Z_{0.10/2} = \mathbf{Z_{0.05} = 1.64}}[/tex]

We are given the lower limit and the upper limit of the population proportion to be: (0.4086, 0.5914)

The sample proportion is:

[tex]\mathbf{\hat p = \dfrac{lower \ limit + upper \ limit }{2}}[/tex]

[tex]\mathbf{\hat p = \dfrac{0.4086 + 0.5914 }{2}}[/tex]

[tex]\mathbf{\hat p = \dfrac{1}{2}}[/tex]

[tex]\mathbf{\hat p =0.5}[/tex]

The size of the sample(n) can be determined by using the margin of error formula:

i.e.

[tex]\mathbf{M.O.E = Z_{\alpha/2} \sqrt{\dfrac{\hat p ( 1 - \hat p)}{n}}}}[/tex]

where;

  • Margin of Error M.O.E = 0.5 - 0.4086 = 0.0914

[tex]\mathbf{0.0914 = 1.645 \sqrt{\dfrac{0.5 ( 1 -0.5)}{n}}}}[/tex]

[tex]\mathbf{n= {\dfrac{1.645^2 \times 0.5^2}{0.0914^2}}}}[/tex]

n = 80.98

n ≅ 81

Learn more about sample size (n) here:

https://brainly.com/question/17203075

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