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Example: Legal Advice
According to a 2002 survey by FundLaw, 20% of Americans needed legal advice during the past
year to resolve such thorny issues as trusts and landlord disputes. Suppose a recent sample of
1000 adult Americans showed that 20% of them needed legal advice in the past year to resolve
such family-related issues. Find a 99% confidence interval for the percentage of American
adults who needed such legal advice.
In our sample of 1000 American adults, there were 20% or 200 successes and 800 failures. We
can use 1-PropZInt with x = 200, n = 1000, and our C-Level set at 0.99.
Press the STAT key.
Press the ► key twice to get to TESTS.
Press the ALPHA key and then the MATH key to get
to the letter A.
Type in 200 for x.
Type in 1000 for n.
Type in .99 for C-Level.
Highlight Calculate and press the ENTER key.
The 1-PropZInt output shows the 99% confidence
interval, the sample proportion, and the sample size.
The 99% confidence interval is from .167 to .233 which
means that we believe with 99% confidence that the
true population proportion is between 16.7% and 23.3%
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