Use the given degree of confidence and sample data to construct a confidence interval for the population mean $\mu$ . Assume that the population has a normal distribution.
-The amounts (in ounces) of juice in eight randomly selected juice bottles are: $\begin{array} { l l l l } 15.4 & 15.8 & 15.4 & 15.1 \end{array}$
$\begin{array} { l l l l } 15.8 & 15.9 & 15.8 & 15.7 \end{array}$
Construct a $98 \%$ confidence interval for the mean amount of juice in all such bottles.
A) $15.23 \mathrm { oz }

Question

Use the given degree of confidence and sample data to construct a confidence interval for the population mean $\mu$ . Assume that the population has a normal distribution.
-The amounts (in ounces) of juice in eight randomly selected juice bottles are: $\begin{array} { l l l l } 15.4 & 15.8 & 15.4 & 15.1 \end{array}$
$\begin{array} { l l l l } 15.8 & 15.9 & 15.8 & 15.7 \end{array}$
Construct a $98 \%$ confidence interval for the mean amount of juice in all such bottles.
A) $15.23 \mathrm { oz } < \mu < 15.99 \mathrm { oz }$
B) $15.33 \mathrm { oz } < \mu < 15.89 \mathrm { oz }$
C) $15.89 \mathrm { oz } < \mu < 15.33 \mathrm { oz }$
D) $15.99 \mathrm { oz } < \mu < 15.23 \mathrm { oz }$

Quizplus · Accepted Answer

First, we need to calculate the sample mean and standard deviation:
$$\bar{x} = \frac{15.4+15.8+15.4+15.1+15.8+15.9+15.8+15.7}{8} = 15.625$$
$$s = \sqrt{\frac{\sum_{i=1}^n (x_i-\bar{x})^2}{n-1}} = 0.292$$
Using a t-distribution with 7 degrees of freedom (since we have 8 data points), we can find the t-value for a 98% confidence interval:
$$t_{0.01/2,7} = 2.998$$
Now we can construct the confidence interval:
$$\bar{x} - t_{0.01/2,7} \frac{s}{\sqrt{n}} < \mu < \bar{x} + t_{0.01/2,7} \frac{s}{\sqrt{n}}$$
$$15.625 - 2.998 \cdot \frac{0.292}{\sqrt{8}} < \mu < 15.625 + 2.998 \cdot \frac{0.292}{\sqrt{8}}$$
$$15.33 < \mu < 15.89$$
Therefore, the best choice is B.

Use the Given Degree of Confidence and Sample Data to Construct