Question

A sample of size n=11 has a sample mean x=15.6 and sample standard deviation s=2.4. Construct a 95% confidence interval for the population mean μ.

Answer 1

The 95% confidence interval for the population mean μ is (13.658, 17.542).

Step-by-step explanation:

To construct a confidence interval for the population mean μ, we use the formula:
`\[ \bar{x} \pm \left( (t \cdot s)/(√(n)) \right) \]` where
`\(\bar{x}\)` is the sample mean,
`\(s\)` is the sample standard deviation,
`\(n\)` is the sample size, and
`\(t\)` is the critical t-value. For a 95% confidence interval with
`\(n-1\)` degrees of freedom, the critical t-value is approximately 2.201. Plugging in the given values, we get:
`\[ 15.6 \pm \left( (2.201 \cdot 2.4)/(√(11)) \right) \]`.

Calculating this expression yields the confidence interval (13.658, 17.542). This means we are 95% confident that the true population mean μ falls within this range. The width of the interval is influenced by the sample standard deviation
`\(s\)` and the square root of the sample size
`\(√(n)\)`. As the sample size increases or the standard deviation decreases, the interval becomes narrower, reflecting increased precision in our estimate of the population mean.

Interpreting the confidence interval, we can say, with 95% confidence, that the true average of the population lies between 13.658 and 17.542 based on the information gathered from our sample of size
`\(n=11\)`. This statistical approach provides a range of plausible values for the population mean, giving a measure of the uncertainty associated with our estimation.