The amounts (in ounces) of juice in eight randomly selected juice bottles are:
$\begin{array} { l l l l l } 15.3 & 15.3 & 15.7 & 15.7\15.3&15 .9&15 .3&15 .9\ \end{array}$
Construct a $98 \%$ confidence interval for the mean amount of juice in all such bottle
A) $15.83 \mathrm { oz }

Question

The amounts (in ounces) of juice in eight randomly selected juice bottles are:
$\begin{array} { l l l l l } 15.3 & 15.3 & 15.7 & 15.7\15.3&15 .9&15 .3&15 .9\ \end{array}$
Construct a $98 \%$ confidence interval for the mean amount of juice in all such bottle
A) $15.83 \mathrm { oz } < \mu < 15.27 \mathrm { oz }$
B) $15.93 \mathrm { oz } < \mu < 15.17 \mathrm { oz }$
C) $15.27 \mathrm { oz } < \mu < 15.83 \mathrm { oz }$
D) $15.17 \mathrm { oz } < \mu < 15.93 \mathrm { oz }$

Quizplus · Accepted Answer

The correct answer is C because the confidence interval is calculated as:$\overline{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$where $\overline{x}$ is the sample mean, $z_{\alpha/2}$ is the critical value from the standard normal distribution, $\sigma$ is the population standard deviation, and $n$ is the sample size. In this case, we have:$\overline{x} = 15.56$$z_{\alpha/2} = 2.326$$\sigma = 0.28$$n = 8$Plugging these values into the formula, we get:$15.56 \pm 2.326 \frac{0.28}{\sqrt{8}}$$15.56 \pm 0.27$$15.29 \mathrm { oz } < \mu < 15.83 \mathrm { oz }$

The Amounts (In Ounces) of Juice in Eight Randomly Selected 15.315.315.715.715.315.915.315.9\begin{array} { l l l l l } 15.3 & 15.3 & 15.7 & 15.7\\15.3&15 .9&15 .3&15 .9\\ \end{array}15.315.3​15.315.9​15.715.3​15.715.9​

The Amounts (In Ounces) of Juice in Eight Randomly Selected $\begin{array} { l l l l l } 15.3 & 15.3 & 15.7 & 15.7\\15.3&15 .9&15 .3&15 .9\\ \end{array}$