To calculate the 80% confidence interval for the population mean (\(\mu\)), we will use the formula for the confidence interval when the sample size is large (in this case, \(n = 949\)), which uses the z-distribution.
The formula for the confidence interval is:
\[ \bar{x} \pm z_{c} \left(\frac{\sigma}{\sqrt{n}}\right) \]
where:
- \(\bar{x}\) = sample mean = 30.3
- \(z_{c}\) = critical z-value for the desired confidence level
- \(\sigma\) = sample standard deviation = 18.1
- \(n\) = sample size = 949
Step 1: Find the critical value \(z_{c}\)
For an 80% confidence interval, we are interested in the critical value that leaves 10% in each tail (100% - 80% = 20%, and half of that is 10%).
Using the standard normal distribution table, the z-value that corresponds to an area of 0.90 (which is the 80% coverage) is approximately \(z_{c} \approx 1.281\).
Step 2: Calculate the standard error (SE)
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{18.1}{\sqrt{949}} \]
Calculating \(\sqrt{949}\):
\[ \sqrt{949} \approx 30.8 \]
Now, calculate the standard error (SE):
\[ SE = \frac{18.1}{30.8} \approx 0.587 \]
Step 3: Calculate the margin of error (ME)
\[ ME = z_{c} \times SE = 1.281 \times 0.587 \approx 0.752 \]
Step 4: Calculate the confidence interval
Now we can calculate the lower and upper limits of the confidence interval:
Lower limit:
\[ \bar{x} - ME = 30.3 - 0.752 = 29.548 \approx 29.5 \]
Upper limit:
\[ \bar{x} + ME = 30.3 + 0.752 = 31.052 \approx 31.1 \]
Final Answer
Hence, the 80% confidence interval for the population mean \(\mu\) is:
\[ 29.5 < \mu < 31.1 \]
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