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out of 300 people sampled, 261 had children. Based on this, construct a 95% confidence interval for the true population proportion of people with children. Preliminary: Is it safe to assume that n < 5% of all people with children?

User Erkmene
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Final answer:

To construct a 95% confidence interval for the true population proportion of people with children, use the formula CI = p-hat ± z * √(p-hat * (1-p-hat) / n). Substituting the given values, the 95% confidence interval for the true population proportion is approximately 0.87 ± 0.0368.

Step-by-step explanation:

To construct a 95% confidence interval for the true population proportion of people with children, we can use the formula:
CI = p-hat ± z * √(p-hat * (1-p-hat) / n)
Where p-hat is the sample proportion (number of people with children divided by the sample size), z is the z-score corresponding to the desired confidence level (in this case, 95% confidence corresponds to a z-score of 1.96) and n is the sample size.

In this case, out of 300 people sampled, 261 had children, so the sample proportion is 261/300 = 0.87. Substituting into the formula:
CI = 0.87 ± 1.96 * √(0.87 * (1-0.87) / 300)

Calculating the values:
CI = 0.87 ± 1.96 * √(0.87 * 0.13 / 300)
CI = 0.87 ± 1.96 * √(0.1131 / 300)
CI = 0.87 ± 1.96 * 0.0188

Simplifying further, the 95% confidence interval for the true population proportion of people with children is approximately 0.87 ± 0.0368.

User RedZ
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