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Sagot :
Certainly! Let's go through the problem step-by-step and provide a detailed solution.
Problem Statement:
A coin-operated drink machine is designed to discharge a mean of 7 fluid ounces of coffee per cup. In a test of the machine, the discharge amounts in 10 randomly chosen cups of coffee were recorded. The sample mean is 6.8 fluid ounces, and the sample standard deviation is 0.23 fluid ounces. We need to determine if there is enough evidence to conclude that the population mean discharge, [tex]\(\mu\)[/tex], differs from 7 fluid ounces, using a 0.05 level of significance.
Step-by-Step Solution:
### (a) State the Null and Alternative Hypotheses:
1. Null Hypothesis [tex]\((H_0)\)[/tex]:
- The null hypothesis states that there is no difference in the population mean discharge from the specified mean.
- [tex]\( H_0: \mu = 7 \)[/tex]
2. Alternative Hypothesis [tex]\((H_1)\)[/tex]:
- The alternative hypothesis states that the population mean discharge is different from the specified mean.
- [tex]\( H_1: \mu \neq 7 \)[/tex]
### (b) Test Statistic Calculation:
To determine if we have enough evidence to reject the null hypothesis, we will use a t-test for the sample mean.
1. Given Data:
- Sample mean, [tex]\(\bar{x} = 6.8\)[/tex] fluid ounces
- Population mean, [tex]\(\mu = 7\)[/tex] fluid ounces
- Sample standard deviation, [tex]\(s = 0.23\)[/tex] fluid ounces
- Sample size, [tex]\(n = 10\)[/tex]
- Significance level, [tex]\(\alpha = 0.05\)[/tex]
2. Formula for the Test Statistic:
[tex]\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \][/tex]
3. Substituting the values:
[tex]\[ t = \frac{6.8 - 7}{0.23 / \sqrt{10}} = \frac{-0.2}{0.0727} = -2.750 \][/tex]
### (c) Calculating the p-value:
A two-tailed test is performed. The p-value tells us the probability of observing the test statistic as extreme as the one computed, under the null hypothesis.
1. Degrees of Freedom (df):
[tex]\[ df = n - 1 = 10 - 1 = 9 \][/tex]
2. Determine the p-value:
From the t-distribution table or using statistical software, we can find the p-value corresponding to the computed t-statistic.
Given:
[tex]\[ t \approx -2.750 \][/tex]
The p-value for a two-tailed test is:
[tex]\[ p = 2 \times P(T > |t|) \][/tex]
Where [tex]\(T\)[/tex] follows a t-distribution with 9 degrees of freedom.
From the calculations:
[tex]\[ p \approx 0.022 \][/tex]
### (d) Decision Rule:
1. Significance Level:
[tex]\(\alpha = 0.05\)[/tex]
2. Decision Rule:
- If [tex]\( p < \alpha \)[/tex], reject the null hypothesis.
- If [tex]\( p \geq \alpha \)[/tex], do not reject the null hypothesis.
### (e) Conclusion:
1. p-value:
[tex]\[ p \approx 0.022 \][/tex]
2. Since [tex]\( p = 0.022 \)[/tex] is less than the significance level [tex]\(\alpha = 0.05\)[/tex], we reject the null hypothesis.
### Final Answer:
- Null Hypothesis [tex]\((H_0)\)[/tex]: [tex]\( \mu = 7 \)[/tex]
- Alternative Hypothesis [tex]\((H_1)\)[/tex]: [tex]\( \mu \neq 7 \)[/tex]
- Test Statistic: [tex]\(-2.750\)[/tex]
- p-value: [tex]\(0.0225\)[/tex]
- Decision: Reject [tex]\(H_0\)[/tex]
Conclusion: There is sufficient evidence at the [tex]\(0.05\)[/tex] significance level to conclude that the population mean discharge of the machine differs from [tex]\(7\)[/tex] fluid ounces.
Problem Statement:
A coin-operated drink machine is designed to discharge a mean of 7 fluid ounces of coffee per cup. In a test of the machine, the discharge amounts in 10 randomly chosen cups of coffee were recorded. The sample mean is 6.8 fluid ounces, and the sample standard deviation is 0.23 fluid ounces. We need to determine if there is enough evidence to conclude that the population mean discharge, [tex]\(\mu\)[/tex], differs from 7 fluid ounces, using a 0.05 level of significance.
Step-by-Step Solution:
### (a) State the Null and Alternative Hypotheses:
1. Null Hypothesis [tex]\((H_0)\)[/tex]:
- The null hypothesis states that there is no difference in the population mean discharge from the specified mean.
- [tex]\( H_0: \mu = 7 \)[/tex]
2. Alternative Hypothesis [tex]\((H_1)\)[/tex]:
- The alternative hypothesis states that the population mean discharge is different from the specified mean.
- [tex]\( H_1: \mu \neq 7 \)[/tex]
### (b) Test Statistic Calculation:
To determine if we have enough evidence to reject the null hypothesis, we will use a t-test for the sample mean.
1. Given Data:
- Sample mean, [tex]\(\bar{x} = 6.8\)[/tex] fluid ounces
- Population mean, [tex]\(\mu = 7\)[/tex] fluid ounces
- Sample standard deviation, [tex]\(s = 0.23\)[/tex] fluid ounces
- Sample size, [tex]\(n = 10\)[/tex]
- Significance level, [tex]\(\alpha = 0.05\)[/tex]
2. Formula for the Test Statistic:
[tex]\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \][/tex]
3. Substituting the values:
[tex]\[ t = \frac{6.8 - 7}{0.23 / \sqrt{10}} = \frac{-0.2}{0.0727} = -2.750 \][/tex]
### (c) Calculating the p-value:
A two-tailed test is performed. The p-value tells us the probability of observing the test statistic as extreme as the one computed, under the null hypothesis.
1. Degrees of Freedom (df):
[tex]\[ df = n - 1 = 10 - 1 = 9 \][/tex]
2. Determine the p-value:
From the t-distribution table or using statistical software, we can find the p-value corresponding to the computed t-statistic.
Given:
[tex]\[ t \approx -2.750 \][/tex]
The p-value for a two-tailed test is:
[tex]\[ p = 2 \times P(T > |t|) \][/tex]
Where [tex]\(T\)[/tex] follows a t-distribution with 9 degrees of freedom.
From the calculations:
[tex]\[ p \approx 0.022 \][/tex]
### (d) Decision Rule:
1. Significance Level:
[tex]\(\alpha = 0.05\)[/tex]
2. Decision Rule:
- If [tex]\( p < \alpha \)[/tex], reject the null hypothesis.
- If [tex]\( p \geq \alpha \)[/tex], do not reject the null hypothesis.
### (e) Conclusion:
1. p-value:
[tex]\[ p \approx 0.022 \][/tex]
2. Since [tex]\( p = 0.022 \)[/tex] is less than the significance level [tex]\(\alpha = 0.05\)[/tex], we reject the null hypothesis.
### Final Answer:
- Null Hypothesis [tex]\((H_0)\)[/tex]: [tex]\( \mu = 7 \)[/tex]
- Alternative Hypothesis [tex]\((H_1)\)[/tex]: [tex]\( \mu \neq 7 \)[/tex]
- Test Statistic: [tex]\(-2.750\)[/tex]
- p-value: [tex]\(0.0225\)[/tex]
- Decision: Reject [tex]\(H_0\)[/tex]
Conclusion: There is sufficient evidence at the [tex]\(0.05\)[/tex] significance level to conclude that the population mean discharge of the machine differs from [tex]\(7\)[/tex] fluid ounces.
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