Question

Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size $12$to check the amount of soda served in the $16$oz. serving size. The sample mean is $13.30$with a sample standard deviation of $1.55$. Assume the underlying population is normally distributed.

Find the $95\%$Confidence Interval for the true population mean for the amount of soda served.

$$\begin{gathered} a.(12.42,14.18) \\ b.(12.32,14.29) \\ c.(12.50,14.10) \end{gathered}$$

d. Impossible to determine

Short answer

b.$(12.32,14.29)$

Step-by-step solution

Given information

A quality control specialist for a restaurant chain takes a random sample of size $12$to check the amount of soda served in the $16$oz. serving size.

The sample mean is $13.30$with a sample standard deviation of $1.55$.

Explanation

The confidence interval will be $\mu +/-\mathrm{t}\sigma /\sqrt{n}$

$\begin{array}{r}\mu =13.30\\ \sigma =1.55\\ n=12\end{array}$

We use a t-distribution rather than $z$because the population standard deviation isn't known and the sample size is $<30$

degrees of freedom $=12-1=11$

two tailed$alpha=0.05(1-0.95)$

so on a $TI-842nd$vars invt $area=0.975$

localid="1650542364085" $df=11;t=2.26$

you can calculate the confidence interval$(12.32,14.29)$