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mafiozo [28]
2 years ago
10

You recently sent out a survey to determine if the percentage of adults who use social media has changed from 66%, which was the

percentage of adults who used social media five years ago. Of the 2809 people who responded to survey, 1634 stated that they currently use social media. Use the data from this survey to construct a 98% confidence interval estimate of the proportion of adults who use social media.
Mathematics
1 answer:
il63 [147K]2 years ago
6 0

Answer:

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

Of the 2809 people who responded to survey, 1634 stated that they currently use social media.

This means that n = 2809, \pi = \frac{1634}{2809} = 0.5817

98% confidence level

So \alpha = 0.02, z is the value of Z that has a pvalue of 1 - \frac{0.02}{2} = 0.99, so Z = 2.327.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 - 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.56

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 + 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.6034

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

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Answer:

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Step-by-step explanation:

see the attached figure to better understand the problem

step 1

In the right triangle ACD

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step 2

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