061. Confidence Intervals for Population Proportions

Decisions often depend on parameters that are binary – parameters with only two possible categories into which responses may fall. In this event, the parameter of concern is the population proportion.

For example, a firm may want to know what proportion of its customers pay on credit as opposed to those who use cash. Corporations are often interested in what percentage of their products are defective as opposed to the percentage that is not defective, or what proportion of their employees quit after one year in contrast to that proportion who do not quit after one year. In each of these instances, there are only two possible outcomes. Concern is therefore focused on that proportion of responses that fall into one of these two outcomes.

If Nπ and N(l —π) are both greater than 5, the distribution of sample proportions will be normal (N should be greater than 50). Thus, the standard error of the sampling distribution of sample proportions is

P=. (6.5)

However, Formula (6.5) requires π, the parameter is going to be estimated. Therefore, the sample proportion P is used as an estimator for π.

Formula (6.5) becomes

Sp=. (6.6)

The confidence interval for population proportion is then

C. I. for π = P + Zsp.

Example 6.4. The manager of a TV station must determine what percentage of households the city have more than one TV set. A random sample of 500 homes reveals that 273 have two or more sets. What is the 90 percent confidence interval for the proportion all homes with two or more sets? Given these data,

P = 275/500 = 0.55,

And

Sp=

Table B yields a Z of 1.65 for a 90 percent confidence interval.

C. I. for π = 0.55 + (1,65)(0.022)=0.55 + 0.036,

0.514 < π < 0.586.

The manager can be 90 percent confident that between 51.4 percent and 58.6 percent of the homes in the city have more than one TV.

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