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Sagot :
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]
- 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]
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