To construct a 98% confidence interval for the population mean [tex]\(\mu\)[/tex], we need to follow a series of steps:
1. Identify the sample statistics:
- Sample size ([tex]\(n\)[/tex]) = 187
- Sample mean ([tex]\(\bar{x}\)[/tex]) = 19 ounces
- Sample standard deviation ([tex]\(s\)[/tex]) = 3.3 ounces
- Confidence level = 98%
2. Calculate the standard error of the mean:
The standard error (SE) is given by the formula:
[tex]\[
SE = \frac{s}{\sqrt{n}}
\][/tex]
Plugging in the given values:
[tex]\[
SE = \frac{3.3}{\sqrt{187}} \approx 0.24
\][/tex]
3. Find the critical z-value for the given confidence level:
For a 98% confidence level, the critical z-value (z*) corresponds to the area in the middle 98% of the standard normal distribution, leaving 1% in each tail. The critical z-value is approximately 2.33.
4. Calculate the margin of error (ME):
The margin of error is given by:
[tex]\[
ME = z^* \cdot SE
\][/tex]
Plugging in the values:
[tex]\[
ME = 2.33 \cdot 0.24 \approx 0.56
\][/tex]
5. Determine the confidence interval:
The confidence interval is calculated as:
[tex]\[
(\bar{x} - ME, \bar{x} + ME)
\][/tex]
Substituting the values:
[tex]\[
(19 - 0.56, 19 + 0.56) \approx (18.4, 19.6)
\][/tex]
Therefore, the 98% confidence interval for the population mean [tex]\(\mu\)[/tex] is approximately [tex]\((18.4, 19.6)\)[/tex] ounces.
Comparing this result with the given options:
- [tex]\(17 \text{ oz} < \mu < 19 \text{ oz}\)[/tex]
- [tex]\(19 \text{ oz} < \mu < 21 \text{ oz}\)[/tex]
- [tex]\(18 \text{ oz} < \mu < 21 \text{ oz}\)[/tex]
- [tex]\(18 \text{ oz} < \mu < 20 \text{ oz}\)[/tex]
None of the provided options exactly match the calculated confidence interval of [tex]\((18.4, 19.6)\)[/tex]. However, the confidence interval we found properly answers the question of estimating the population mean [tex]\(\mu\)[/tex].
