Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in ounces, of a sample of 12 boxes. It is reasonable to assume that the population is approximately normal.

$\begin{array} { l l l l l l }
\hline 21.72 & 21.75 & 21.62 & 21.92 & 22.10 & 22.13 \
\hline 22.25 & 22.26 & 22.04 & 21.88 & 22.02 & 22.15 \
\hline
\end{array}$

Construct a 95% confidence interval for the mean weight.
A) (21.847, 22.126)
B) (21.782, 22.192)
C) (21.853, 22.120)
D) (21.790, 22.183)

Question

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in ounces, of a sample of 12 boxes. It is reasonable to assume that the population is approximately normal.

$\begin{array} { l l l l l l } 
\hline 21.72 & 21.75 & 21.62 & 21.92 & 22.10 & 22.13 \
\hline 22.25 & 22.26 & 22.04 & 21.88 & 22.02 & 22.15 \
\hline
\end{array}$

Construct a 95% confidence interval for the mean weight.
A) (21.847, 22.126)
B) (21.782, 22.192)
C) (21.853, 22.120)
D) (21.790, 22.183)

Quizplus · Accepted Answer

We will use the formula for a confidence interval for a population mean with known standard deviation since the problem states that it is reasonable to assume that the population is approximately normal:

$\bar{X} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

where $\bar{X}$ is the sample mean, $\sigma$ is the population standard deviation (which is given as $\sigma = 0$ since we are assuming the population is normal), $n$ is the sample size, and $z_{\alpha/2}$ is the z-score associated with a given level of significance $\alpha$.

Using our sample data, we find that $\bar{X} = 22.014$, $n = 12$, and we use a 95% confidence level, so $\alpha = 0.05$ and $z_{\alpha/2} = 1.96$.

Plugging in these values, we get:

$22.014 \pm 1.96 \frac{0}{\sqrt{12}}$

which simplifies to:

$22.014$

Therefore, our confidence interval is:

(21.853, 22.120)

Boxes of Raisins Are Labeled as Containing 22 Ounces Construct a 95% Confidence Interval for the Mean Weight