A U.S. Travel Data Center’s survey of 1500 adults found that 46% of respondents stated that...

A U.S. Travel Data Center’s survey of 1500 adults found that 46% of respondents stated that they favor historical sites as vacations. Find the 99% confidence interval of the true proportion of all adults who favor visiting historical sites as vacations.

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Expert Solution

Solution :

Given that,

n = 1500

Point estimate = sample proportion = = 46%=0.46

1 - = 1-0.46 =0.54

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

Margin of error = E = Z/2 * ( $\sqrt$ ((( * (1 - )) / n)

= 2.576* ( $\sqrt$ ((0.46*0.54) /1500 )

E = 0.033

A 99% confidence interval for proportion p is ,

- E < p < + E

0.46-0.033 < p < 0.46+0.033

0.427< p < 0.493

The 99% confidence interval for the proportion p is : (0.427 , 0.493)

orchestra answered 1 year ago

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