The 98% confidence interval of the proportion of all Americans who think that living on their own makes someone feels like an adult is (29.0%, 31.0%)
The given parameters are:
The number of success is then calculated as;
[tex]\bar x = np[/tex]
This gives
[tex]\bar x = 2000 * 28\%[/tex]
Evaluate
[tex]\bar x = 560[/tex]
Hence, 560 successes would be found in the sample of 2000 Americans
The proportion that lives on their own is:
p = 30%
The z-score for 98% confidence interval is
z = 2.326
The confidence interval is calculated as:
[tex]CI = \bar p \pm z * \sqrt{\frac{\bar p(1 - \bar p)}{ n}[/tex]
So, we have:
[tex]CI = 30\% \pm 2.326 * \sqrt{\frac{30\%(1 - 30\%)}{2000}[/tex]
Evaluate
[tex]CI = 30\% \pm 1\%[/tex]
Expand
CI = (30% - 1%, 30% + 1%)
Evaluate
CI = (29%, 31%)
Hence, the 98% confidence interval is (29.0%, 31.0%)
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