Question

In: Statistics and Probability

A poll of 809 adults aged 18 or older asked about purchases that they intended to...

A poll of 809 adults aged 18 or older asked about purchases that they intended to make for the upcoming holiday season. One of the questions asked about what kind of gift they intended to buy for the person on whom they would spend the most. Clothing was the first choice of 498 people. Give a 99% confidence interval for the proportion of people in this population who intend to buy clothing as their first choice.

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Expert Solution

Solution :

Given that,

Point estimate = sample proportion = = x / n = 498 / 809 = 0.616

Z_/2 = 2.576

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

= 2.576 * ( $\sqrt$ ((0.616 * 0.384) / 809)

= 0.044

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.616 - 0.044 < p < 0.616 + 0.044

0.576 < p < 0.660

(0.576 , 0.660)

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orchestra answered 1 year ago

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