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 the volume of iced tea beverages is supposed to average 300 mL with a standard deviation of 3 mL. The consumer protection group sampled 20 beverages and found the average to be 298.4 mL. Using the given table, which of the following is 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?

\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 \\
z-values
\end{tabular} & 1.65 & 1.96 & 2.58 \\
\hline \hline
\end{tabular}

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

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

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

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



Answer :

To determine whether the company is packaging less than the required average of 300 mL, we need to perform a hypothesis test. The hypothesis tested will be:

- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu = 300\)[/tex] mL (the population mean is 300 mL).
- Alternative hypothesis ([tex]\(H_1\)[/tex]): [tex]\(\mu < 300\)[/tex] mL (the population mean is less than 300 mL).

To test this hypothesis, we'll use the sample data provided:
- Population mean ([tex]\(\mu\)[/tex]) = 300 mL
- Sample mean ([tex]\(\bar{x}\)[/tex]) = 298.4 mL
- Sample size ([tex]\(n\)[/tex]) = 20
- Population standard deviation ([tex]\(\sigma\)[/tex]) = 3 mL

We employ the z-test for the sample mean. The z-score formula for a sample mean is given by:

[tex]\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \][/tex]

Substituting the given values:

[tex]\[ z = \frac{298.4 - 300}{\frac{3}{\sqrt{20}}} \][/tex]

Simplifying the denominator:

[tex]\[ \frac{3}{\sqrt{20}} = \frac{3}{4.4721} \approx 0.6708 \][/tex]

Now compute the z-score:

[tex]\[ z = \frac{298.4 - 300}{0.6708} = \frac{-1.6}{0.6708} \approx -2.385 \][/tex]

Next, we need to check this z-score against the critical z-values for different significance levels to determine the most restrictive level of significance. From the table:
[tex]\[ \begin{array}{|c|c|c|c|} \hline \text{Upper-Tail Values} & 5\% & 2.5\% & 1\% \\ \hline \text{Critical z-values} & 1.65 & 1.96 & 2.58 \\ \hline \end{array} \][/tex]

Given this, we compare the computed z-score of [tex]\(-2.385\)[/tex] against the negative of these critical values because we are looking at a left-tailed test (since [tex]\(\bar{x}\)[/tex] < [tex]\(\mu\)[/tex]):

- For 5% significance level: [tex]\( \text{Critical } z = -1.65 \)[/tex]
- For 2.5% significance level: [tex]\( \text{Critical } z = -1.96 \)[/tex]
- For 1% significance level: [tex]\( \text{Critical } z = -2.58 \)[/tex]

Given our calculated z-score of -2.385:

- It is less than -1.65 (thus reject [tex]\(H_0\)[/tex] at 5% level).
- It is less than -1.96 (thus reject [tex]\(H_0\)[/tex] at 2.5% level).
- It is not less than -2.58 (thus fail to reject [tex]\(H_0\)[/tex] at 1% level).

The most restrictive level at which we can reject the null hypothesis is at the 5% significance level.

Thus, the most restrictive level of significance that indicates the company is packaging less than the required average of 300 mL is:

[tex]\[ \boxed{5\%} \][/tex]