To calculate the 99% confidence interval for the population mean using a sample, we'll need the following components:
- Sample size \( n = 45 \)
- Sample mean \( \bar{x} = 78.2 \)
- Sample standard deviation \( s = 10.7 \)
Since the sample size is larger than 30, we can apply the Central Limit Theorem and use the z-distribution for the confidence interval.
Step 1: Find the critical value
For a 99% confidence level, the critical z-value (denoted as \( z^* \)) can be found using the standard normal distribution. The area in each tail at 99% confidence level is \( \frac{1 - 0.99}{2} = 0.005 \).
Using the standard normal distribution table, we find:
- \( z^* \approx 2.576 \) (you may round to three decimal places)
Step 2: Calculate the standard error (SE)
The standard error of the sample mean is calculated with the formula: \[ SE = \frac{s}{\sqrt{n}} = \frac{10.7}{\sqrt{45}} \approx \frac{10.7}{6.7082} \approx 1.5912 \]
Step 3: Calculate the margin of error (ME)
The margin of error is given by: \[ ME = z^* \times SE = 2.576 \times 1.5912 \approx 4.0995 \]
Step 4: Calculate the confidence interval
The confidence interval is given by: \[ (\bar{x} - ME, \bar{x} + ME) \]
So we calculate:
- Lower limit: \( \bar{x} - ME = 78.2 - 4.0995 \approx 74.1005 \)
- Upper limit: \( \bar{x} + ME = 78.2 + 4.0995 \approx 82.2995 \)
Final Step: Write the confidence interval
Now rounding to one decimal place, we have:
- Lower limit: \( 74.1 \)
- Upper limit: \( 82.3 \)
Thus, the 99% confidence interval is: \[ \text{99% C.I.} = (74.1, 82.3) \]