Math Problem Statement

Find the confidence intervals for the population mean based on the given sample data and confidence levels.

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:

  • xˉ=211.2\bar{x} = 211.2 (sample mean)
  • s=44.37s = 44.37 (sample standard deviation)
  • n=27n = 27 (sample size)
  • Confidence level = 95%
  1. Calculate the standard error (SE): SE=sn=44.3727=44.375.1968.54SE = \frac{s}{\sqrt{n}} = \frac{44.37}{\sqrt{27}} = \frac{44.37}{5.196} \approx 8.54

  2. Determine the critical tt-value: For n1=26n-1 = 26 degrees of freedom and 95% confidence level, the tt-value is approximately 2.056 (from tt-table or software).

  3. Margin of Error (ME): ME=t×SE=2.056×8.5417.55ME = t \times SE = 2.056 \times 8.54 \approx 17.55

  4. Confidence interval: Lower limit=xˉME=211.217.55193.65\text{Lower limit} = \bar{x} - ME = 211.2 - 17.55 \approx 193.65 Upper limit=xˉ+ME=211.2+17.55228.75\text{Upper limit} = \bar{x} + ME = 211.2 + 17.55 \approx 228.75


(b) Calculations

Given:

  • xˉ=74.42\bar{x} = 74.42
  • s=31.8s = 31.8
  • n=15n = 15
  • Confidence level = 99%
  1. Calculate the standard error (SE): SE=sn=31.815=31.83.8738.21SE = \frac{s}{\sqrt{n}} = \frac{31.8}{\sqrt{15}} = \frac{31.8}{3.873} \approx 8.21

  2. Determine the critical tt-value: For n1=14n-1 = 14 degrees of freedom and 99% confidence level, the tt-value is approximately 2.977.

  3. Margin of Error (ME): ME=t×SE=2.977×8.2124.44ME = t \times SE = 2.977 \times 8.21 \approx 24.44

  4. Confidence interval: Lower limit=xˉME=74.4224.4449.98\text{Lower limit} = \bar{x} - ME = 74.42 - 24.44 \approx 49.98 Upper limit=xˉ+ME=74.42+24.4498.86\text{Upper limit} = \bar{x} + ME = 74.42 + 24.44 \approx 98.86


Final Results

  • (a) Confidence interval: [193.65,228.75][193.65, 228.75]
  • (b) Confidence interval: [49.98,98.86][49.98, 98.86]

Let me know if you'd like further clarifications or step-by-step explanations!


Follow-Up Questions

  1. How does changing the confidence level affect the interval width?
  2. Why is the tt-distribution used instead of the normal distribution in these cases?
  3. What happens to the confidence interval when the sample size increases?
  4. Can you explain the role of the critical tt-value in the calculation?
  5. 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 tt-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