Math Problem Statement

A random sample of 14 subjects was asked to perform a given task. The time in seconds it took each of them to complete the task is recorded as follows: 32, 38, 43, 49, 45, 27, 47, 33, 39, 43, 44, 32, 49, 26. Find the 99% confidence interval for the true mean completion time for this task.

Solution

To find the 99% confidence interval for the mean completion time, let's go through the steps required, as the sample data is already provided. Here’s the approach we’ll use:

  1. Calculate the Sample Mean (xˉ\bar{x}): Add up all the completion times and divide by the number of subjects (14).

  2. Calculate the Sample Standard Deviation (s):

    • Find each score's deviation from the mean.
    • Square each deviation.
    • Sum these squared deviations, divide by n1n - 1 (where n=14n = 14), and then take the square root.
  3. Determine the T-Score for 99% Confidence Level:

    • Since n=14n = 14, we have 13 degrees of freedom (n1n - 1).
    • Using a t-table or calculator, find the t-score corresponding to 99% confidence with 13 degrees of freedom.
  4. Calculate the Margin of Error (ME): ME=t×snME = t \times \frac{s}{\sqrt{n}}

  5. Construct the Confidence Interval:

    • Lower limit = xˉME\bar{x} - ME
    • Upper limit = xˉ+ME\bar{x} + ME

Let's go through the calculations in detail:

1. Sample Mean (xˉ\bar{x})

Completion times: 32,38,43,49,45,27,47,33,39,43,44,32,49,2632, 38, 43, 49, 45, 27, 47, 33, 39, 43, 44, 32, 49, 26

xˉ=32+38+43+49+45+27+47+33+39+43+44+32+49+2614\bar{x} = \frac{32 + 38 + 43 + 49 + 45 + 27 + 47 + 33 + 39 + 43 + 44 + 32 + 49 + 26}{14}

2. Sample Standard Deviation (s)

Calculate the deviations from the mean, square each deviation, sum them, and divide by n1n - 1, then take the square root.

3. T-Score for 99% Confidence with 13 Degrees of Freedom

From the t-table, look up the t-value for 99% confidence and 13 degrees of freedom.

4. Margin of Error (ME)

ME=t×s14ME = t \times \frac{s}{\sqrt{14}}

5. Confidence Interval

Once we have the margin of error, we can construct the confidence interval:

Lower limit=xˉME\text{Lower limit} = \bar{x} - ME Upper limit=xˉ+ME\text{Upper limit} = \bar{x} + ME

Let me compute these values for you.The calculations yield the following results for the 99% confidence interval:

  • Sample Mean (xˉ\bar{x}): 39.1 seconds (rounded to one decimal place)
  • Sample Standard Deviation (s): 7.9 seconds
  • T-Score for 99% Confidence with 13 Degrees of Freedom: 3.012
  • Margin of Error (ME): 6.3 seconds

Confidence Interval

  • Lower limit: 32.7 seconds (rounded to one decimal place)
  • Upper limit: 45.4 seconds (rounded to one decimal place)

Therefore, the 99% confidence interval for the true mean completion time is: (32.7,45.4)(32.7, 45.4)

Would you like further details on these calculations or have any questions?

Here are five related questions to deepen your understanding:

  1. What would happen to the confidence interval if the confidence level were changed to 95%?
  2. How does increasing the sample size affect the width of the confidence interval?
  3. What role does the t-score play in the calculation of the confidence interval?
  4. How would the interval change if we used a larger sample standard deviation?
  5. Why do we use the t-distribution instead of the normal distribution in this calculation?

Tip: Always check the degrees of freedom when using the t-distribution, as it directly affects the t-score value.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Normal Distribution
T-Distribution

Formulas

Sample Mean: \(\bar{x} = \frac{\sum x}{n}\)
Sample Standard Deviation: \(s = \sqrt{\frac{\sum (x - \bar{x})^2}{n - 1}}\)
Margin of Error: \(ME = t \times \frac{s}{\sqrt{n}}\)
Confidence Interval: \((\bar{x} - ME, \bar{x} + ME)\)

Theorems

Central Limit Theorem
T-Distribution

Suitable Grade Level

College-level Statistics