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

parameter statistic

mean x

standard deviation ?Formulas:

p̂p

ˆ

1p

p p

n

proportion p p̂

Confidence Intervals for proportions

x zn

*

Formula:

p z *

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.

p 46821

.056

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.

821 .056 10np

1 821 1 .056 10n p

Sample size is large enoughto use a normal distribution.

1 proportion z-intervalproportion of crashesp

Safe to infer population is at least 8210 crashes.

We assume the sample is a random sample.

p 46821

.056

We are 95% confident that the true proportion of crashes lies between .0403 and .0718.

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.

.07176 .0403.01573

2m

2 576. 2 576.

..

05290061

n2

2

.0529 .00003729

1n

1418.76 n

We need a sample sizeof 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?