Question

In a sample of 390 adults, 254 had children. Construct a 99% confidence interval for the true population proportion of adults with children.

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Answer 1

The 99% confidence interval for the true population proportion of adults with children is approximately
`\( \left(0.602, 0.748\right) \).`

Step-by-step explanation:

To construct a confidence interval for the population proportion, we can use the formula:

`\[ \text{Confidence Interval} = \hat{p} \pm Z * \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]`

where:

`- \(\hat{p}\) is the sample proportion of adults with children (\( (254)/(390) \)),`

-
`\(Z\)`is the Z-score corresponding to the desired confidence level (99% corresponds to a Z-score of 2.576 for a two-tailed test),

-
`\(n\)` is the sample size (390).

Plugging in the values, we get:

`\[ \text{Confidence Interval} = (254)/(390) \pm 2.576 * \sqrt{((254)/(390) * \left(1 - (254)/(390)\right))/(390)} \]`

After calculating, the confidence interval is
`\( \left(0.602, 0.748\right) \)`.

This means we are 99% confident that the true population proportion of adults with children lies between 60.2% and 74.8%. The width of the interval reflects our level of confidence; a wider interval indicates greater certainty. This result suggests that the majority of adults in the population likely have children, based on the sample data.

Answer 2

The 99% confidence interval for the true population proportion of adults with children is approximately
`\( \underline{0.591} \) to \( \underline{0.701} \).`

Step-by-step explanation:

To construct a confidence interval for a population proportion, we use the formula:

Confidence Interval =
`\hat{p} \pm Z * \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

-
`\(\hat{p}\)` is the sample proportion (254/390 in this case).

-Z is the Z-score corresponding to the desired confidence level (99% in this case).

-n is the sample size (390 in this case).

The Z-score for a 99% confidence interval is approximately 2.576 (you can find this value from a standard normal distribution table). Plugging in the values, we get:

Confidence Interval =
`0.65 \pm 2.576 * \sqrt{(0.65 * 0.35)/(390)}`

Calculating this gives the interval [0.591, 0.701], so we can say with 99% confidence that the true population proportion of adults with children is between 59.1% and 70.1%.

In this context, a 99% confidence level means that if we were to take many samples and construct a confidence interval from each, about 99% of those intervals would contain the true population proportion.

Therefore, we are highly confident that the proportion of adults with children in the population falls within the calculated interval. This information is useful for decision-making and understanding the variability associated with estimating population parameters based on a sample. Keep in mind that the confidence interval provides a range, and the true proportion is expected to lie within this range with a 99% probability.