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].
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].
