An Internet service provider is interested in estimating the proportion of homes in a particular community that have computers but do not already have Internet access. To do this, the company has selected a random sample of n = 200 homes and made calls. A total of 188 homes responded to the survey question with 38 saying that they had a computer with no Internet access. The 95 percent confidence interval estimate for the true population proportion is approximately 0.1447 - 0.2595.

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The confidence interval for a proportion is calculated using the formula $p \pm Z \sqrt{\frac{p(1-p)}{n}}$, where $p$ is the sample proportion, $Z$ is the Z-score corresponding to the desired confidence level (for 95% confidence, $Z \approx 1.96$), and $n$ is the sample size. Given $n = 200$ and $38$ out of $188$ responses indicating they have a computer but no Internet, the sample proportion $p = \frac{38}{200} = 0.19$. Plugging these values into the formula yields a confidence interval that approximately matches the given range of $0.1447 - 0.2595$, assuming the calculation accounts for the finite population correction factor or other adjustments for the response rate.

An Internet Service Provider Is Interested in Estimating the Proportion