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Sagot :
### Step-by-Step Solution:
1. State the hypotheses for the test:
- The null hypothesis [tex]\(H_0\)[/tex] is the assumption that the passing rate has not changed.
- The alternative hypothesis [tex]\(H_a\)[/tex] is the assumption that the passing rate has improved.
Given the problem's context, we set up the hypotheses as follows:
[tex]\[ H_0: p = 0.65 \quad \text{and} \quad H_a: p > 0.65 \][/tex]
Thus, the correct answer is:
- B. [tex]\( H_0: p=0.65 \)[/tex] and [tex]\( H_a: p>0.65 \)[/tex]
2. Explain the meaning of the [tex]\( p \)[/tex]-value in this context:
- The [tex]\( p \)[/tex]-value tells us the probability of observing a sample proportion as extreme as, or more extreme than, the one observed (147 out of 200 students passing), assuming that the null hypothesis [tex]\(H_0\)[/tex] is true.
Specifically, it is the probability of 147 or more students passing given that the population proportion [tex]\( p \)[/tex] is actually 0.65.
Given the [tex]\( p \)[/tex]-value of 0.0059, the correct interpretation in the context of the problem is:
- C. The probability of 147 or more introductory chemistry students passing out of a random sample of 200 students is 0.0059, assuming the population proportion is 0.65.
3. Determine if the instructor should believe the success rate has improved:
- To make this decision, we compare the [tex]\( p \)[/tex]-value to the significance level α. Common practice is to use α = 0.05.
- If the [tex]\( p \)[/tex]-value is less than α, we reject the null hypothesis [tex]\(H_0\)[/tex] in favor of the alternative hypothesis [tex]\(H_a\)[/tex].
In this case:
- The [tex]\( p \)[/tex]-value is 0.0059, which is less than 0.05.
- Therefore, we reject [tex]\( H_0 \)[/tex] and accept [tex]\( H_a \)[/tex], meaning we have enough evidence to support the claim that the success rate has improved.
The answer is:
- Based on the [tex]\( p \)[/tex]-value of 0.0059, the instructor should believe that the success rate has improved.
### Summary:
1. Hypotheses: B. [tex]\( H_0: p=0.65 \)[/tex] and [tex]\( H_a: p>0.65 \)[/tex]
2. [tex]\( p \)[/tex]-value interpretation: C. The probability of 147 or more introductory chemistry students passing out of a random sample of 200 students is 0.0059, assuming the population proportion is 0.65.
3. Conclusion: The instructor should believe that the success rate has improved.
1. State the hypotheses for the test:
- The null hypothesis [tex]\(H_0\)[/tex] is the assumption that the passing rate has not changed.
- The alternative hypothesis [tex]\(H_a\)[/tex] is the assumption that the passing rate has improved.
Given the problem's context, we set up the hypotheses as follows:
[tex]\[ H_0: p = 0.65 \quad \text{and} \quad H_a: p > 0.65 \][/tex]
Thus, the correct answer is:
- B. [tex]\( H_0: p=0.65 \)[/tex] and [tex]\( H_a: p>0.65 \)[/tex]
2. Explain the meaning of the [tex]\( p \)[/tex]-value in this context:
- The [tex]\( p \)[/tex]-value tells us the probability of observing a sample proportion as extreme as, or more extreme than, the one observed (147 out of 200 students passing), assuming that the null hypothesis [tex]\(H_0\)[/tex] is true.
Specifically, it is the probability of 147 or more students passing given that the population proportion [tex]\( p \)[/tex] is actually 0.65.
Given the [tex]\( p \)[/tex]-value of 0.0059, the correct interpretation in the context of the problem is:
- C. The probability of 147 or more introductory chemistry students passing out of a random sample of 200 students is 0.0059, assuming the population proportion is 0.65.
3. Determine if the instructor should believe the success rate has improved:
- To make this decision, we compare the [tex]\( p \)[/tex]-value to the significance level α. Common practice is to use α = 0.05.
- If the [tex]\( p \)[/tex]-value is less than α, we reject the null hypothesis [tex]\(H_0\)[/tex] in favor of the alternative hypothesis [tex]\(H_a\)[/tex].
In this case:
- The [tex]\( p \)[/tex]-value is 0.0059, which is less than 0.05.
- Therefore, we reject [tex]\( H_0 \)[/tex] and accept [tex]\( H_a \)[/tex], meaning we have enough evidence to support the claim that the success rate has improved.
The answer is:
- Based on the [tex]\( p \)[/tex]-value of 0.0059, the instructor should believe that the success rate has improved.
### Summary:
1. Hypotheses: B. [tex]\( H_0: p=0.65 \)[/tex] and [tex]\( H_a: p>0.65 \)[/tex]
2. [tex]\( p \)[/tex]-value interpretation: C. The probability of 147 or more introductory chemistry students passing out of a random sample of 200 students is 0.0059, assuming the population proportion is 0.65.
3. Conclusion: The instructor should believe that the success rate has improved.
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