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In a survey of 284 smokers, 197 smokers reported that they wanted to quit smoking. Compute...

In a survey of 284 smokers, 197 smokers reported that they wanted to quit smoking. Compute a 90% confidence interval for the true proportion of smokers that want to quit smoking. (Adapted from the American Lung Association) What is the mean and standard deviation?

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Answer #1

We need to construct the 90% confidence interval for the population proportion. We have been provided with the following information about the number of favorable cases:

Favorable Cases X = 197
Sample Size N = 284

The sample proportion is computed as follows, based on the sample size N=284 and the number of favorable cases X = 197

The critical value for α=0.1 is . The corresponding confidence interval is computed as shown below:

Therefore, based on the data provided, the 90% confidence interval for the population proportion is 0.649<p<0.739, which indicates that we are 90% confident that the true population proportion p is contained by the interval (0.649, 0.739).

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