Math Problem Statement
Solution
Thank you for uploading the images. I see they relate to confidence interval calculations. Let's go step by step to find the correct values.
(a) Calculations
Given:
- (sample mean)
- (sample standard deviation)
- (sample size)
- Confidence level = 95%
-
Calculate the standard error (SE):
-
Determine the critical -value: For degrees of freedom and 95% confidence level, the -value is approximately 2.056 (from -table or software).
-
Margin of Error (ME):
-
Confidence interval:
(b) Calculations
Given:
- Confidence level = 99%
-
Calculate the standard error (SE):
-
Determine the critical -value: For degrees of freedom and 99% confidence level, the -value is approximately 2.977.
-
Margin of Error (ME):
-
Confidence interval:
Final Results
- (a) Confidence interval:
- (b) Confidence interval:
Let me know if you'd like further clarifications or step-by-step explanations!
Follow-Up Questions
- How does changing the confidence level affect the interval width?
- Why is the -distribution used instead of the normal distribution in these cases?
- What happens to the confidence interval when the sample size increases?
- Can you explain the role of the critical -value in the calculation?
- How would the calculations differ if the population standard deviation were known?
Tip: Always ensure to check the degrees of freedom when finding the critical -value for smaller sample sizes!
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Math Problem Analysis
Mathematical Concepts
Statistics
Confidence Intervals
t-distribution
Formulas
Standard Error: SE = s / sqrt(n)
Margin of Error: ME = t * SE
Confidence Interval: [x̄ - ME, x̄ + ME]
Theorems
t-distribution theorem
Suitable Grade Level
Grades 11-12 or introductory college statistics
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