A consumer protection group randomly checks the volume of different beverages to ensure that companies are packaging the stated amount. Each individual volume is not exact, but a volume of iced tea beverages is supposed to average [tex]$300 \text{ mL}$[/tex] with a standard deviation of [tex]$3 \text{ mL}$[/tex]. The consumer protection group sampled 20 beverages and found the average to be [tex]$298.4 \text{ mL}$[/tex]. Using the given table, which of the following is the most restrictive level of significance for a hypothesis test that would indicate the company is packaging less than the required average of [tex]$300 \text{ mL}$[/tex]?

\begin{tabular}{|c|c|c|c|}
\hline \multicolumn{4}{|c|}{ Upper-Tail Values } \\
\hline a & [tex]$5\%$[/tex] & [tex]$2.5\%$[/tex] & [tex]$1\%$[/tex] \\
\hline \begin{tabular}{r}
Critical \\
[tex]$z$[/tex]-values
\end{tabular} & 1.65 & 1.96 & 2.58 \\
\hline
\end{tabular}

A. [tex]$1\%$[/tex]

B. [tex]$2.5\%$[/tex]

C. [tex]$5\%$[/tex]

D. [tex]$10\%$[/tex]



Answer :

In this problem, we are asked to determine the most restrictive level of significance on a hypothesis test that would indicate the company is packaging less than the required average of 300 mL of iced tea beverages.

Let's solve the problem step-by-step:

1. State the hypotheses:
- Null hypothesis (\(H_0\)): \(\mu = 300\) mL (The average volume is 300 mL)
- Alternative hypothesis (\(H_a\)): \(\mu < 300\) mL (The average volume is less than 300 mL)

2. Given Values:
- Population mean (\(\mu\)) = 300 mL
- Sample mean (\(\bar{x}\)) = 298.4 mL
- Population standard deviation (\(\sigma\)) = 3 mL
- Sample size (\(n\)) = 20

3. Calculate the standard error of the mean (SEM):
The standard error is given by the formula:
[tex]\[ SEM = \frac{\sigma}{\sqrt{n}} \][/tex]
Substituting the values:
[tex]\[ SEM = \frac{3}{\sqrt{20}} \approx 0.6708 \][/tex]

4. Calculate the z-score:
The z-score is calculated using the formula:
[tex]\[ Z = \frac{\bar{x} - \mu}{SEM} \][/tex]
Substituting the values:
[tex]\[ Z = \frac{298.4 - 300}{0.6708} \approx -2.385 \][/tex]

5. Compare the computed z-score with the critical z-values:
We refer to the given critical z-values for different significance levels:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline \multicolumn{4}{|c|}{ Upper-Tail Values } \\ \hline a & 5\% & 2.5\% & 1\% \\ \hline Critical \(z\)-values & 1.65 & 1.96 & 2.58 \\ \hline \end{tabular} \][/tex]

Since this is a one-tailed test (we are only interested if the average is less than 300 mL), we will consider the negative values of the critical z-scores.

- For \(5\%\) significance level, \(Z_{critical} = -1.65\)
- For \(2.5\%\) significance level, \(Z_{critical} = -1.96\)
- For \(1\%\) significance level, \(Z_{critical} = -2.58\)

6. Determine the most restrictive level of significance:
- Our computed z-score \(Z = -2.385\) is less than \(-1.96\) but greater than \(-2.58\).
- This implies that the \(Z\)-score falls in between the \(2.5\%\) and \(1\%\) significance levels.

Since the z-score is more extreme than the critical value for \(2.5\%\) but not as extreme as for \(1\%\), the most restrictive level of significance at which we can reject the null hypothesis is \(2.5\%\).

Thus, the most restrictive level of significance that would indicate the company is packaging less than the required average of 300 mL is [tex]\(\boxed{2.5\%}\)[/tex].