To construct a confidence interval for the population mean (\(\mu\)) with a given degree of confidence, we can use the t-distribution since the sample size is small and the population standard deviation is unknown.
First, let's calculate the sample mean (\(\bar{x}\)) and the sample standard deviation (\(s\)):
Sample mean:
\[
\bar{x} = \frac{15.4 + 15.8 + 15.4 + 15.1 + 15.8 + 15.9 + 15.8 + 15.7}{8} = 15.6125 \text{ ounces}
\]
Sample standard deviation:
\[
s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}} = \sqrt{\frac{(15.4-15.6125)^2 + (15.8-15.6125)^2 + \ldots + (15.7-15.6125)^2}{7}} \approx 0.2459 \text{ ounces}
\]
Next, we need to find the t-value for a 98% confidence interval with \(n-1 = 7\) degrees of freedom. Using a t-table or a statistical software, the t-value is approximately 2.3646.
Now, we can construct the confidence interval using the formula:
\[
\text{Confidence Interval} = \bar{x} \pm \left( t \times \frac{s}{\sqrt{n}} \right)
\]
Substituting the values:
\[
\text{Confidence Interval} = 15.6125 \pm \left( 2.3646 \times \frac{0.2459}{\sqrt{8}} \right)
\]
Calculating the interval:
Lower limit: \(15.6125 - (2.3646 \times 0.0871) \approx 15.4071\) ounces
Upper limit: \(15.6125 + (2.3646 \times 0.0871) \approx 15.8179\) ounces
Therefore, the 98% confidence interval for the mean amount of juice in all such bottles is approximately \(15.4071\) ounces to \(15.8179\) ounces.