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Alex17521 [72]
2 years ago
11

A recent study asked U. S. Adults to name 10 historic events that occurred during their lifetime that have had the greatest impa

ct on the country. The most frequently chosen answer was the September 11, 2001, terrorist attacks, which was included by 77% of the 2,000 randomly selected U. S. Adults.
Construct and interpret a 99% confidence interval for the proportion of all U. S. Adults who would include the 9/11 attacks on their list of 10 historic events. (Be sure to follow the 4-step process)


a. State (1) b. Plan (2) c. Do (2) d. Conclude (2)
Mathematics
1 answer:
Nadusha1986 [10]2 years ago
7 0

Using the z-distribution, as we are working with a proportion, it is found that the 99% confidence interval for the proportion of all U. S. Adults who would include the 9/11 attacks on their list of 10 historic events is (0.7458, 0.7942). It means that we are 99% sure that the true proportion for all U.S. adults is between these two bounds.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, the parameters are:

z = 2.575, n = 2000, \pi = 0.77

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 2.575\sqrt{\frac{0.77(0.23)}{2000}} = 0.7458

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 2.575\sqrt{\frac{0.77(0.23)}{2000}} = 0.7942

The 99% confidence interval for the proportion of all U. S. Adults who would include the 9/11 attacks on their list of 10 historic events is (0.7458, 0.7942). It means that we are 99% sure that the true proportion for all U.S. adults is between these two bounds.

More can be learned about the z-distribution at brainly.com/question/25890103

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