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It is thought that not as many Americans buy presents to celebrate Valentine's Day anymore. A random sample of 40 Americans yielded 22 who bought their significant other a present and celebrated Valentine's Day. Estimate the true proportion of all Americans who celebrate Valentine's Day using a 98% confidence interval. Express the answer in the form the point estimates /- the margin of error.

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Answer:

98% confidence interval of the true proportion of all Americans who celebrate Valentine's Day

 (0.3672 , 0.7328)

Step-by-step explanation:

Explanation:-

Given Random sample size n =40

Sample proportion

                            [tex]p = \frac{x}{n} = \frac{22}{40} = 0.55[/tex]

98% confidence interval of the true proportion of all Americans who celebrate Valentine's Day

   

          [tex](p-Z_{\alpha } \sqrt{\frac{pq}{n} } , p + Z_{\alpha } \sqrt{\frac{pq}{n} } )[/tex]

         

The Z-value Z₀.₉₈ = Z₀.₀₂ = 2.326

98% confidence interval of the true proportion of all Americans who celebrate Valentine's Day

    [tex](0.55-2.326\sqrt{\frac{0.55 X0.45}{40} } , 0.55 + 2.326\sqrt{\frac{0.55 X0.45}{40} } )[/tex]

  ( 0.55 - 0.1828 , 0.55 + 0.1828)

  (0.3672 , 0.7328)

         

alexandriaw460

Answer:

Step-by-step explanation:

98% confidence interval of the true proportion of all Americans who celebrate Valentine's Day

 

         

       

The Z-value Z₀.₉₈ = Z₀.₀₂ = 2.326

98% confidence interval of the true proportion of all Americans who celebrate Valentine's Day

   

 ( 0.55 - 0.1828 , 0.55 + 0.1828)

 (0.3672 , 0.7328)

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