Questions: Question 8 1 pt 1 97 Details A sample of 50 ravens taken from a park in Richmond has a mean mass of 520.1 g with a standard deviation of 92.1 g. Find the 93% confidence interval of the mean mass of all the ravens in the park. Round your endpoints to 4 decimal places. E=

Solution Steps

We have a sample of 50 ravens with the following statistics:

• Sample mean (\(\bar{x}\)): \(520.1 \, \text{g}\)
• Sample standard deviation (\(s\)): \(92.1 \, \text{g}\)
• Sample size (\(n\)): \(50\)
• Confidence level: \(93\%\)

For a \(93\%\) confidence level, the significance level (\(\alpha\)) is: \[ \alpha = 1 - 0.93 = 0.07 \] Since this is a two-tailed test, we divide \(\alpha\) by \(2\): \[ \frac{\alpha}{2} = 0.035 \] Using a Z-table or calculator, the Z-score corresponding to \(0.035\) in the upper tail is approximately \(1.8119\).

The margin of error (\(E\)) is calculated using the formula: \[ E = z \cdot \frac{s}{\sqrt{n}} \] Substituting the values: \[ E = 1.8119 \cdot \frac{92.1}{\sqrt{50}} \approx 1.8119 \cdot 13.007 \approx 23.54 \]

The confidence interval is given by: \[ \bar{x} \pm E \] Calculating the endpoints: \[ \text{Lower limit} = 520.1 - 23.54 \approx 496.56 \] \[ \text{Upper limit} = 520.1 + 23.54 \approx 543.64 \]

Rounding the endpoints to four decimal places:

• Lower limit: \(496.5\)
• Upper limit: \(543.7\)

Final Answer