Introduction to Inference Confidence Intervals for Proportions.

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Presentation transcript:

Introduction to Inference Confidence Intervals for Proportions

Example problem In a study of air-bag effectiveness, it was found that in 821 crashes of midsize cars equipped with air bags, 46 of the crashes resulted in hospitalization of the drivers. –Give a 95% confidence interval for the percent of crashes resulting in hospitalization. Interpret the confidence interval.

Sample means to sample proportions parameterstatistic mean standard deviation Formulas: proportion

Confidence Intervals for proportions Formula: Draw a random sample of size n from a large population with unknown proportion p of successes. Z-interval One-proportion Z-interval

Conditions for proportions The data are a random sample from the population of interest. Issue of normality: –np > 10 and n(1 – p) > 10 The population is at least 10 times as large as the sample.

Give a 95% confidence interval for the percent of crashes resulting in hospitalization. In a study of air-bag effectiveness, it was found that in 821 crashes of midsize cars equipped with air bags, 46 of the crashes resulted in hospitalization of the drivers. Sample size is large enough to use a normal distribution. 1 proportion z-interval Safe to infer population is at least 8210 crashes. We assume the sample is a random sample.

We are 95% confident that the true proportion of crashes lies between.0403 and Give a 95% confidence interval for the percent of crashes resulting in hospitalization. Since we had to assume the crashes were a random sample, we have doubts about the accuracy.

We need a sample size of at least 1419 crashes. How large a sample would be needed to obtain the same margin of error in part “a” for a 99% confidence interval?