Final answer:
The 99% confidence interval for the average hours slept following the administration of the anesthetic for the sampled population is (5.19, 9.81).
Step-by-step explanation:
To find the 99% confidence interval for the average hours slept, we can use the formula:
![\[ \bar{x} \pm z(a/2) * (s)/(√(n)) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/p553ytc3peeo3qmiuof8bg9evyfabxxrtp.png)
where:
is the sample mean,
is the z-score corresponding to the desired confidence level,
is the sample standard deviation, and
is the sample size.
In this case, the sample mean
is calculated as (4 + 8 + 11 + 11 + 5 + 11 + 4 + 8 + 8) / 9 = 8.
The sample standard deviation
is approximately 2.36.
The z-score for a 99% confidence interval is 2.92 for a two-tailed test.
Substituting these values into the formula:
![\[ 8 \pm 2.92 * (2.36)/(√(9)) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/ndw60ayrfz6w15r94n6goq6l483d1n6m27.png)
Calculating this expression results in the 99% confidence interval of (5.19, 9.81).
This means that we are 99% confident that the true average hours slept following the administration of the anesthetic for the sampled population falls within the interval of 5.19 to 9.81 hours. The population does not need to be normal for the confidence interval to be valid, as we are relying on the Central Limit Theorem, which states that the distribution of the sample mean becomes approximately normal as the sample size increases.