Answer: 4.79 and 11.96
Step-by-step explanation:
The outlier in the data is clearly a mistake, as it is not possible for someone to sleep for 24 hours in a day. We will remove this value (24) from the sample and calculate the mean and standard deviation of the remaining values.
The remaining sample is: 8.5, 9, 8, 6.5, 7.5, 6, 8.5, and 9.5.
The sample mean is:
$\bar{x}=\frac{\sum_{i=1}^{n}x_i}{n}=\frac{8.5+9+8+6.5+7.5+6+8.5+9.5}{8}=7.875$
The sample standard deviation is:
$s=\sqrt{\frac{\sum_{i=1}^{n}(x_i-\bar{x})^2}{n-1}}=\sqrt{\frac{(8.5-7.875)^2+(9-7.875)^2+(8-7.875)^2+(6.5-7.875)^2+(7.5-7.875)^2+(6-7.875)^2+(8.5-7.875)^2+(9.5-7.875)^2}{7}}\approx 1.355$
With a sample size of 8, the degrees of freedom for the t-distribution will be 7. Using a t-distribution table or calculator, we can find the t-value for a 99.5% confidence interval with 7 degrees of freedom to be 4.785.
The margin of error for the mean is:
$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}=4.785\frac{1.355}{\sqrt{8}}\approx 3.08$
Therefore, the 99.5% confidence interval for the mean amount of sleep is:
$\bar{x}\pm ME=7.875\pm 3.08=(4.79,11.96)$
Rounded to at least two decimal places, the confidence interval is (4.79, 11.96).