Math Problem Statement

Construct a confidence interval for the mean of the paired differences for the two populations with the following information. Round the endpoints of the interval to three decimal places, if necessary.

n=31, d¯=2.538, α=0.05, sd=0.556

Solution

To construct a confidence interval for the mean of the paired differences, we can use the formula for the confidence interval of a mean with a t-distribution, since the population standard deviation is unknown and the sample size is small.

The formula for the confidence interval is:

dˉ±tα/2,n1(sdn)\bar{d} \pm t_{\alpha/2, \, n-1} \left( \frac{s_d}{\sqrt{n}} \right)

Where:

  • dˉ\bar{d} is the mean of the paired differences (2.538)
  • sds_d is the standard deviation of the paired differences (0.556)
  • nn is the sample size (31)
  • tα/2,n1t_{\alpha/2, \, n-1} is the critical value from the t-distribution for a confidence level of 1α1 - \alpha with n1n-1 degrees of freedom.

Step 1: Determine the tt-critical value

For a 95% confidence level (α=0.05\alpha = 0.05), the tt-critical value can be found using a t-distribution table or calculator with n1=30n-1 = 30 degrees of freedom. The critical value for t0.025,30t_{0.025, \, 30} is approximately 2.042.

Step 2: Compute the margin of error

Calculate the margin of error using:

Margin of Error=t0.025,30(sdn)=2.042(0.55631)\text{Margin of Error} = t_{0.025, \, 30} \left( \frac{s_d}{\sqrt{n}} \right) = 2.042 \left( \frac{0.556}{\sqrt{31}} \right)

=2.042(0.5565.568)2.042×0.09980.204= 2.042 \left( \frac{0.556}{5.568} \right) \approx 2.042 \times 0.0998 \approx 0.204

Step 3: Construct the confidence interval

Now, calculate the confidence interval as:

2.538±0.2042.538 \pm 0.204

This gives the endpoints:

  • Lower endpoint: 2.5380.204=2.3342.538 - 0.204 = 2.334
  • Upper endpoint: 2.538+0.204=2.7422.538 + 0.204 = 2.742

Final Confidence Interval

The 95% confidence interval for the mean of the paired differences is:

(2.334,2.742)(2.334, \, 2.742)

Would you like further details on the calculations, or do you have any questions?

Here are five related questions to expand your understanding:

  1. How would the confidence interval change if the sample size were larger?
  2. What happens to the margin of error if the standard deviation increases?
  3. How does changing the confidence level (e.g., to 90% or 99%) affect the interval?
  4. Why do we use the t-distribution instead of the z-distribution here?
  5. What does the confidence interval suggest about the true mean of the paired differences?

Tip: The width of a confidence interval depends on the sample size, variability in the data, and the confidence level chosen. Increasing the sample size reduces the interval width.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Paired Differences
T-Distribution

Formulas

Confidence Interval Formula: d̄ ± t(α/2, n-1) * (sd / √n)
Margin of Error: t(α/2, n-1) * (sd / √n)

Theorems

T-Distribution

Suitable Grade Level

College-level Statistics