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Step-By-Step Example of a Confidence Interval for a Proportion (used Ex 8.8)
Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. 500 randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes- they own cell phones.
Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones.
| Guidelines | Example |
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| Plan: State what we need to know. | We are asked to find a 95% confidence interval for the true proportion of city residents who have cell phones. We have a sample of 500 residents. |
| Model: Think about the assumptions and check the conditions. |
Randomization Condition: The sample is a random sample. Independence Assumption: It is reasonable to think that the cell phone ownership of 500 residents are independent. 10% Condition: Since this is a large city the number of adult residents must be over 5000, so 500 subjects is less than 10% of the population. Success/Failure Condition: n = 500(0.842) = 421>10; n(1 - )= 500(0.158) = 79>10 |
| State the parameters and the sampling model | The conditions are satisfied so I can use a Normal model to find a confidence interval for a proportion.
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Mechanics: CL = 0.95, so α = 1-CL = 1-0.95 = 0.05.
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| Conclusion: Interpret your result in the proper context, and relate it to the original question. | I am 95% confident that between 81% and 87.4% of all adult residents of this city have cell phones.95% of all confidence intervals constructed in this way contain the true proportion of adults in this city who have cell phones. |
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