Question

For Problem Set 4, you worked with data on the number of siblings a respondent had. In this sample of 15,508 adults, the average number of siblings was 3.44 (s=2.68). 1. Calculate a 95% confidence int

Answer 1

For the sample of 15,508 adults with an average number of siblings of
`\(3.44\)` (standard deviation,
`\(s=2.68\))`, the 95% confidence interval for the mean number of siblings is approximately
`\(3.42 \leq \mu \leq 3.46\).`

Step-by-step explanation:

To calculate the 95% confidence interval for the mean
`(\(\mu\))`, the formula
`\(\bar{X} \pm Z * (s)/(√(n))\)` is used. Here,
`\(\bar{X}=3.44\), \(s=2.68\)`, and
`\(n=15,508\).` The Z-score for a 95% confidence interval is approximately
`\(1.96\).`

The calculation involves substituting these values into the formula:

`\[3.44 \pm 1.96 * (2.68)/(√(15,508))\]`

Now, compute the margin of error:

`\[1.96 * (2.68)/(√(15,508)) \approx 0.02\]`

The confidence interval is then:

`\[3.44 \pm 0.02\]`

This results in the interval
`\(3.42 \leq \mu \leq 3.46\)`. Therefore, we can be 95% confident that the true mean number of siblings for the entire population falls within this range.

In conclusion, the confidence interval is a statistical tool that provides a range within which the true population mean is likely to lie. In this context, the interval
`\(3.42 \leq \mu \leq 3.46\)` gives a measure of the precision of the sample estimate, offering insights into the average number of siblings among adults based on the given data.