To find the 98% confidence interval for the population mean μ, we can use the following formula for the confidence interval:
\[ \text{CI} = \bar{x} \pm z^* \left( \frac{s}{\sqrt{n}} \right) \]
Where:
- \(\bar{x}\) is the sample mean,
- \(z^*\) is the critical value for a 98% confidence level,
- \(s\) is the sample standard deviation,
- \(n\) is the sample size.
Step 1: Identify the Sample Statistics
- Sample size \(n = 401\)
- Sample mean \(\bar{x} = 78.1\)
- Sample standard deviation \(s = 14.4\)
Step 2: Find the Critical Value \(z^*\)
For a 98% confidence interval, we need the critical value that captures the central 98% of the standard normal distribution. This leaves 1% in each tail.
Using a standard normal distribution table or calculator:
\[ z^* \approx 2.33 \]
Step 3: Calculate the Standard Error (SE)
The standard error (SE) of the sample mean is calculated as follows:
\[ SE = \frac{s}{\sqrt{n}} = \frac{14.4}{\sqrt{401}} \]
Calculating \(\sqrt{401}\):
\[ \sqrt{401} \approx 20.0249843945 \]
Now, calculate SE:
\[ SE = \frac{14.4}{20.0249843945} \approx 0.719 \]
Step 4: Compute the Margin of Error (ME)
The margin of error (ME) is given by:
\[ ME = z^* \cdot SE = 2.33 \cdot 0.719 \]
Calculating that:
\[ ME \approx 1.674 \]
Step 5: Calculate the Confidence Interval
Now we can calculate the lower and upper bounds of the confidence interval:
\[ \text{Lower bound} = \bar{x} - ME = 78.1 - 1.674 \approx 76.426 \]
\[ \text{Upper bound} = \bar{x} + ME = 78.1 + 1.674 \approx 79.774 \]
Step 6: Round the Results
To one decimal place for the final values:
- Lower bound: \( 76.4 \)
- Upper bound: \( 79.8 \)
Final Result
Thus, the 98% confidence interval for μ can be expressed as:
\[ 76.4 < μ < 79.8 \]
This gives us the final answer in a tri-linear inequality format.
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