A company claims its iced tea beverages average 300 mL with a standard deviation of 3 mL. A consumer protection group sampled 20 beverages and found an average of 298.4 mL. Using the given table, which is the most restrictive level of significance that indicates the company is packaging less than the required average of 300 mL?

| a | 5% | 2.5% | 1% |
|--------|------|------|------|
| z-value| 1.65 | 1.96 | 2.58 |

Options:
A. 1%
B. 2.5%
C. 5%
D. 10%



Answer :

To determine the most restrictive level of significance that indicates the company is packaging an average amount of iced tea beverages less than the required average of [tex]\(300 \, \text{mL}\)[/tex], we need to perform a hypothesis test.

Here’s a step-by-step solution:

### Step 1: Formulate the Hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): The true mean volume is [tex]\(300 \, \text{mL}\)[/tex], i.e., [tex]\( \mu = 300 \, \text{mL}\)[/tex].
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The true mean volume is less than [tex]\(300 \, \text{mL}\)[/tex], i.e., [tex]\( \mu < 300 \, \text{mL}\)[/tex].

### Step 2: Calculate the Standard Error
The standard error of the mean (SE) is computed using the population standard deviation ([tex]\(\sigma\)[/tex]) and the sample size ([tex]\(n\)[/tex]):
[tex]\[ \text{SE} = \frac{\sigma}{\sqrt{n}} \][/tex]

Given:
- Population standard deviation ([tex]\(\sigma\)[/tex]) = [tex]\(3 \, \text{mL}\)[/tex]
- Sample size ([tex]\(n\)[/tex]) = 20

Plugging in these values:
[tex]\[ \text{SE} = \frac{3}{\sqrt{20}} \approx 0.6708 \][/tex]

### Step 3: Calculate the Test Statistic (Z-score)
The Z-score is calculated with the formula:
[tex]\[ Z = \frac{\bar{X} - \mu}{\text{SE}} \][/tex]

Where:
- [tex]\(\bar{X}\)[/tex] is the sample mean = [tex]\(298.4 \, \text{mL}\)[/tex]
- [tex]\(\mu\)[/tex] is the population mean = [tex]\(300 \, \text{mL}\)[/tex]

Plugging in the values:
[tex]\[ Z = \frac{298.4 - 300}{0.6708} \approx -2.385 \][/tex]

### Step 4: Compare the Test Statistic to Critical Z-values
We compare the calculated Z-score to the critical Z-values from the table for different levels of significance.

Given critical Z-values for the upper tail:
- [tex]\(5\%\)[/tex] significance level: [tex]\(Z = 1.65\)[/tex]
- [tex]\(2.5\%\)[/tex] significance level: [tex]\(Z = 1.96\)[/tex]
- [tex]\(1\%\)[/tex] significance level: [tex]\(Z = 2.58\)[/tex]

Since our test is one-tailed (left-tailed), we compare the negative of these critical values with our calculated Z-score:

Critical Z-values (left-tailed):
- [tex]\(5\%\)[/tex]: [tex]\(-1.65\)[/tex]
- [tex]\(2.5\%\)[/tex]: [tex]\(-1.96\)[/tex]
- [tex]\(1\%\)[/tex]: [tex]\(-2.58\)[/tex]

The test statistic [tex]\(Z \approx -2.385\)[/tex] is less than [tex]\(-1.96\)[/tex] but greater than [tex]\(-2.58\)[/tex].

### Conclusion:
The most restrictive level of significance at which we can reject the null hypothesis and conclude that the company is packaging less than the required average [tex]\(300 \, \text{mL}\)[/tex] is [tex]\(2.5\%\)[/tex].

Therefore, the answer is [tex]\( \boxed{2.5\%} \)[/tex].