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Phantasy [73]
3 years ago
12

It is thought that not as many Americans buy presents to celebrate Valentine's Day anymore. A random sample of 40 Americans yiel

ded 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.
Mathematics
2 answers:
shtirl [24]3 years ago
7 0

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)

evablogger [386]3 years ago
6 0

Answer:

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

 (0.3672 , 0.7328)

Step-by-step explanation:

<u><em>Explanation:</em></u>-

Given Random sample size n =40

Sample proportion

                            p = \frac{x}{n} = \frac{22}{40} = 0.55

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

   

          (p-Z_{\alpha } \sqrt{\frac{pq}{n} } , p + Z_{\alpha } \sqrt{\frac{pq}{n} } )

         

The Z-value Z₀.₉₈ = Z₀.₀₂ = 2.326

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

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

  ( 0.55 - 0.1828 , 0.55 + 0.1828)

  (0.3672 , 0.7328)

         

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