Questions: HW10: Problem 4
(10 points)
Starting salaries of 64 college graduates who have taken a statistics course have a mean of 42,500 with a standard deviation of 6,800. Find an 90% confidence interval for μ. (NOTE: Do not use commas or dollar signs in your answers. Round each bound to three decimal places.)
Lower-bound: Upper-bound:
Note: You can earn partial credit on this problem.
We are given the following information about the starting salaries of 64 college graduates who have taken a statistics course:
Sample mean (\(\bar{x}\)): \(42500\)
Sample standard deviation (\(s\)): \(6800\)
Sample size (\(n\)): \(64\)
Confidence level: \(90\%\)
Step 2: Determine the Z-Score
For a \(90\%\) confidence level, the significance level (\(\alpha\)) is:
\[
\alpha = 1 - 0.90 = 0.10
\]
Since we are looking for a two-tailed confidence interval, we divide \(\alpha\) by \(2\):
\[
\frac{\alpha}{2} = 0.05
\]
The corresponding Z-score for \(0.05\) in the upper tail is approximately \(1.645\).
Step 3: Calculate the Margin of Error
The margin of error (\(E\)) can be calculated using the formula:
\[
E = z \cdot \frac{s}{\sqrt{n}}
\]
Substituting the values:
\[
E = 1.645 \cdot \frac{6800}{\sqrt{64}} = 1.645 \cdot \frac{6800}{8} = 1.645 \cdot 850 = 1398.25
\]
Step 4: Calculate the Confidence Interval
The confidence interval is given by:
\[
\bar{x} \pm E
\]
Calculating the lower and upper bounds:
\[
\text{Lower bound} = 42500 - 1398.25 = 41101.75
\]
\[
\text{Upper bound} = 42500 + 1398.25 = 43898.25
\]
Final Answer
The \(90\%\) confidence interval for the mean starting salary is:
\[
\boxed{(41101.874, 43898.126)}
\]