At a large university in Conway, South Carolina, the Dean wants to know what proportion of students enjoyed their introductory statistics course. In order to answer this question, 200 randomly selected students who have taken an introductory statistics course were contacted and asked whether they enjoyed this course. 145 out of the 200 answered that they enjoyed the course. Using this data, construct a 99% confidence interval for the true proportion of students who enjoyed their introductory statistics course. (0.6631,0.7869) (0.7206,0.7294) (0.6437,0.8063) (0.6731,0.7769)
At a large university in Conway, South Carolina, the Dean wants to know what proportion of students enjoyed their introductory statistics course. In order to answer this question, 200 randomly selected students who have taken an introductory statistics course were contacted and asked whether they enjoyed this course. 145 out of the 200 answered that they enjoyed the course. Using this data, construct a 99% confidence interval for the true proportion of students who enjoyed their introductory statistics course. (0.6631,0.7869) (0.7206,0.7294) (0.6437,0.8063) (0.6731,0.7769)
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![### Confidence Interval for Proportion of Students Enjoying Statistics Course
At a large university in Conway, South Carolina, the Dean wants to know what proportion of students enjoyed their introductory statistics course. In order to answer this question, 200 randomly selected students who had taken an introductory statistics course were contacted and asked whether they enjoyed the course. Out of the 200 students, 145 answered that they enjoyed the course.
Using this data, we can construct a 99% confidence interval for the true proportion of students who enjoyed their introductory statistics course. The following intervals represent possible values for this proportion:
1. (0.6631, 0.7869)
2. (0.7206, 0.7294)
3. (0.6437, 0.8063)
4. (0.6731, 0.7769)
To understand how to calculate these intervals, consider the following steps for constructing a confidence interval for a population proportion:
1. **Determine the sample proportion (p-hat)**:
\[
\hat{p} = \frac{\text{Number of students who enjoyed the course}}{\text{Total number of students surveyed}} = \frac{145}{200} = 0.725
\]
2. **Calculate the standard error (SE)**:
\[
SE = \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}} = \sqrt{\frac{0.725 \times (1 - 0.725)}{200}} \approx 0.032
\]
3. **Find the critical value (z*) for a 99% confidence interval** (which is approximately 2.576 for a two-tailed test).
4. **Construct the confidence interval**:
\[
\hat{p} \pm z^* \times SE = 0.725 \pm 2.576 \times 0.032 \implies (0.74 \pm 0.08)
\]
Therefore, the confidence interval is approximately (0.6437, 0.8063), which matches one of the provided intervals (third option).
Selecting the right confidence interval helps in understanding the range within which the true proportion of students who enjoyed the course is likely to fall. This can aid the Dean in making informed decisions about the course's effectiveness and areas for improvement.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa7e444e2-a441-4a28-bca8-26e04764874a%2F4c37fe4a-ff31-43c6-beb0-b07f5a118c3b%2Fofaabno_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Confidence Interval for Proportion of Students Enjoying Statistics Course
At a large university in Conway, South Carolina, the Dean wants to know what proportion of students enjoyed their introductory statistics course. In order to answer this question, 200 randomly selected students who had taken an introductory statistics course were contacted and asked whether they enjoyed the course. Out of the 200 students, 145 answered that they enjoyed the course.
Using this data, we can construct a 99% confidence interval for the true proportion of students who enjoyed their introductory statistics course. The following intervals represent possible values for this proportion:
1. (0.6631, 0.7869)
2. (0.7206, 0.7294)
3. (0.6437, 0.8063)
4. (0.6731, 0.7769)
To understand how to calculate these intervals, consider the following steps for constructing a confidence interval for a population proportion:
1. **Determine the sample proportion (p-hat)**:
\[
\hat{p} = \frac{\text{Number of students who enjoyed the course}}{\text{Total number of students surveyed}} = \frac{145}{200} = 0.725
\]
2. **Calculate the standard error (SE)**:
\[
SE = \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}} = \sqrt{\frac{0.725 \times (1 - 0.725)}{200}} \approx 0.032
\]
3. **Find the critical value (z*) for a 99% confidence interval** (which is approximately 2.576 for a two-tailed test).
4. **Construct the confidence interval**:
\[
\hat{p} \pm z^* \times SE = 0.725 \pm 2.576 \times 0.032 \implies (0.74 \pm 0.08)
\]
Therefore, the confidence interval is approximately (0.6437, 0.8063), which matches one of the provided intervals (third option).
Selecting the right confidence interval helps in understanding the range within which the true proportion of students who enjoyed the course is likely to fall. This can aid the Dean in making informed decisions about the course's effectiveness and areas for improvement.
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