Homework: Chapter 7 Homework Question 3, 7.2.3 Choose the correct answer below. HW Score: 0%, 0 of 14 points O Points: 0 of 2 Refer to the technology output given to the right that results from measured hemoglobin levels (g/dL) in 100 randomly selected adult females. The results to the right are based on a 90% confidence level. Write a statement that correctly interprets the confidence level. OA. We have 90% confidence that the limits of 12.787 g/dL and 13.239 g/dL contain the true value of the mean hemoglobin level of the population of all adult females. OB. The true value of the mean hemoglobin level of the population of all adult females is between 12.787 g/dL and 13.239 g/dL. OC. Approximately 90% of the hemoglobin levels in the population of all adult females is between 12.787 g/dL and 13.239 g/dL. OD. There is a 90% probability that the limits of 12.787 g/dL and 13.239 g/dL contain the true value of the mean hemoglobin level of the population of all adult females. Clear all Save Tinterval (12.787,13.239) x=13.013 Sx=1.363 n=100 Final check
Homework: Chapter 7 Homework Question 3, 7.2.3 Choose the correct answer below. HW Score: 0%, 0 of 14 points O Points: 0 of 2 Refer to the technology output given to the right that results from measured hemoglobin levels (g/dL) in 100 randomly selected adult females. The results to the right are based on a 90% confidence level. Write a statement that correctly interprets the confidence level. OA. We have 90% confidence that the limits of 12.787 g/dL and 13.239 g/dL contain the true value of the mean hemoglobin level of the population of all adult females. OB. The true value of the mean hemoglobin level of the population of all adult females is between 12.787 g/dL and 13.239 g/dL. OC. Approximately 90% of the hemoglobin levels in the population of all adult females is between 12.787 g/dL and 13.239 g/dL. OD. There is a 90% probability that the limits of 12.787 g/dL and 13.239 g/dL contain the true value of the mean hemoglobin level of the population of all adult females. Clear all Save Tinterval (12.787,13.239) x=13.013 Sx=1.363 n=100 Final check
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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Transcribed Image Text:### Homework: Chapter 7 Homework
#### Question 3, 7.2.3
Refer to the technology output given to the right that results from measured hemoglobin levels (g/dL) in 100 randomly selected adult females. The results to the right are based on a 90% confidence level. Write a statement that correctly interprets the confidence level.
**Technology Output:**
- TInterval
- (12.787, 13.239)
- x̄ (mean) = 13.013
- Sx (standard deviation) = 1.363
- n (sample size) = 100
#### Choose the correct answer below:
- **A.** We have 90% confidence that the limits of 12.787 g/dL and 13.239 g/dL contain the true value of the mean hemoglobin level of the population of all adult females.
- **B.** The true value of the mean hemoglobin level of the population of all adult females is between 12.787 g/dL and 13.239 g/dL.
- **C.** Approximately 90% of the hemoglobin levels in the population of all adult females is between 12.787 g/dL and 13.239 g/dL.
- **D.** There is a 90% probability that the limits of 12.787 g/dL and 13.239 g/dL contain the true value of the mean hemoglobin level of the population of all adult females.
---
This problem pertains to the interpretation of a 90% confidence interval for the mean hemoglobin levels in adult females, based on a sample of 100 individuals. The provided data includes the mean, standard deviation, and the calculated confidence interval limits.
![**Chapter 7 Homework: Confidence Intervals for the Population Mean**
---
**Question 6**
**Task:** Use the given level of confidence and statistics to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution.
**Scenario:** Thirty randomly selected students took a calculus final. If the sample mean was 95 and the standard deviation was 6.6, construct a 99% confidence interval for the mean score of all students.
---
**Options:**
A. \( 91.69 < \mu < 98.31 \)
B. \( 91.68 < \mu < 98.32 \)
C. \( 92.03 < \mu < 97.97 \)
D. \( 92.95 < \mu < 97.05 \)
---
**Explanation:**
To find the 99% confidence interval for the population mean, we follow these steps:
1. **Identify the sample statistics:**
- Sample Mean (\(\bar{x}\)): 95
- Standard Deviation (\(s\)): 6.6
- Sample Size (\(n\)): 30
2. **Find the critical value for 99% confidence:**
- Use the z-distribution since \( n > 30 \).
- Critical z-value for 99% confidence is approximately 2.576.
3. **Calculate the standard error (SE):**
\[
SE = \frac{s}{\sqrt{n}} = \frac{6.6}{\sqrt{30}} \approx 1.205
\]
4. **Construct the confidence interval:**
\[
\bar{x} \pm (z \times SE) = 95 \pm (2.576 \times 1.205)
\]
\[
95 \pm 3.10
\]
\[
91.90 < \mu < 98.10
\]
Therefore, the closest answer from the choices provided is:
A. \( 91.69 < \mu < 98.31 \)
This interval means we are 99% confident that the true population mean score of all students lies between 91.69 and 98.31.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffc887dfe-8b6d-4ebc-8d81-f0d1356179cf%2F29868e70-7a4e-488f-b827-83a350005a13%2Fvinqkt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chapter 7 Homework: Confidence Intervals for the Population Mean**
---
**Question 6**
**Task:** Use the given level of confidence and statistics to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution.
**Scenario:** Thirty randomly selected students took a calculus final. If the sample mean was 95 and the standard deviation was 6.6, construct a 99% confidence interval for the mean score of all students.
---
**Options:**
A. \( 91.69 < \mu < 98.31 \)
B. \( 91.68 < \mu < 98.32 \)
C. \( 92.03 < \mu < 97.97 \)
D. \( 92.95 < \mu < 97.05 \)
---
**Explanation:**
To find the 99% confidence interval for the population mean, we follow these steps:
1. **Identify the sample statistics:**
- Sample Mean (\(\bar{x}\)): 95
- Standard Deviation (\(s\)): 6.6
- Sample Size (\(n\)): 30
2. **Find the critical value for 99% confidence:**
- Use the z-distribution since \( n > 30 \).
- Critical z-value for 99% confidence is approximately 2.576.
3. **Calculate the standard error (SE):**
\[
SE = \frac{s}{\sqrt{n}} = \frac{6.6}{\sqrt{30}} \approx 1.205
\]
4. **Construct the confidence interval:**
\[
\bar{x} \pm (z \times SE) = 95 \pm (2.576 \times 1.205)
\]
\[
95 \pm 3.10
\]
\[
91.90 < \mu < 98.10
\]
Therefore, the closest answer from the choices provided is:
A. \( 91.69 < \mu < 98.31 \)
This interval means we are 99% confident that the true population mean score of all students lies between 91.69 and 98.31.
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