To determine the outlier and construct a 98% confidence interval for the mean amount of sleep, we first eliminate the outlier from the dataset. The given data is:
- 8.5
- 24 (this appears to be an outlier)
- 9
- 7.5
- 8
- 6.5
- 8.5
- 6
After removing the outlier (24), the remaining data is:
- 8.5
- 9
- 7.5
- 8
- 6.5
- 8.5
- 6
Now we'll calculate the mean and standard deviation of the remaining values:
-
Calculate the Mean (\(\bar{x}\)): \[ \bar{x} = \frac{\sum x}{n} \]
Where \( n \) is the number of values, and \( x \) is each individual value.
\[ \bar{x} = \frac{8.5 + 9 + 7.5 + 8 + 6.5 + 8.5 + 6}{7} = \frac{54.0}{7} \approx 7.7143 \]
-
Calculate the Standard Deviation (s): \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]
First, we'll calculate each squared difference:
- \( (8.5 - 7.7143)^2 \approx 0.6102 \)
- \( (9 - 7.7143)^2 \approx 1.6494 \)
- \( (7.5 - 7.7143)^2 \approx 0.0457 \)
- \( (8 - 7.7143)^2 \approx 0.0802 \)
- \( (6.5 - 7.7143)^2 \approx 1.4784 \)
- \( (8.5 - 7.7143)^2 \approx 0.6102 \)
- \( (6 - 7.7143)^2 \approx 2.9347 \)
Now we sum these squared differences: \[ \sum (x_i - \bar{x})^2 \approx 0.6102 + 1.6494 + 0.0457 + 0.0802 + 1.4784 + 0.6102 + 2.9347 \approx 7.5088 \]
Plugging this into the standard deviation formula: \[ s = \sqrt{\frac{7.5088}{6}} \approx \sqrt{1.2515} \approx 1.1187 \]
-
Calculate the 98% Confidence Interval:
The formula for the confidence interval is: \[ \bar{x} \pm t^* \times \frac{s}{\sqrt{n}} \]
We need to find the t-value for a 98% confidence interval with \( n-1 = 6 \) degrees of freedom. From the t-table, the critical t-value (\( t^* \)) for 98% confidence and 6 degrees of freedom is approximately 2.447.
Now, we calculate the margin of error: \[ \text{Margin of Error} = t^* \times \frac{s}{\sqrt{n}} = 2.447 \times \frac{1.1187}{\sqrt{7}} \approx 2.447 \times 0.4226 \approx 1.0353 \]
Finally, the confidence interval is: \[ \left( \bar{x} - \text{Margin of Error}, \bar{x} + \text{Margin of Error} \right) = \left( 7.7143 - 1.0353, 7.7143 + 1.0353 \right) \approx (6.68, 8.75) \]
Thus, the 98% confidence interval for the mean amount of sleep is approximately:
\[ \boxed{(6.68, 8.75)} \]