Math Problem Statement

Given two dependent random samples with the following results:

Population 1 30 15 28 44 26 47 32 Population 2 44 28 36 35 18 35 22 Use this data to find the 95% confidence interval for the true difference between the population means. Assume that both populations are normally distributed.

Step 3 of 4 : Calculate the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

Solution

To calculate the margin of error (ME) for the 95% confidence interval, follow these steps:

1. Calculate the Differences

Compute the differences between the paired data points for Population 1 and Population 2:

di=X1iX2id_i = X_{1i} - X_{2i}

The differences are: d=[3044,1528,2836,4435,2618,4735,3222]d = [30 - 44, 15 - 28, 28 - 36, 44 - 35, 26 - 18, 47 - 35, 32 - 22]

d=[14,13,8,9,8,12,10]d = [-14, -13, -8, 9, 8, 12, 10]

2. Compute the Mean (dˉ\bar{d}) and Standard Deviation (sds_d) of the Differences

  • Mean (dˉ\bar{d}):

dˉ=din\bar{d} = \frac{\sum d_i}{n}

dˉ=14+(13)+(8)+9+8+12+107=470.571429\bar{d} = \frac{-14 + (-13) + (-8) + 9 + 8 + 12 + 10}{7} = \frac{4}{7} \approx 0.571429

  • Standard Deviation (sds_d):

sd=(didˉ)2n1s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}}

  1. Differences from mean: (140.571429)2,(130.571429)2,...,(100.571429)2(-14 - 0.571429)^2, (-13 - 0.571429)^2, ..., (10 - 0.571429)^2

    Squared differences: [211.5918,181.7347,73.4694,71.0204,55.1837,130.6122,88.1633][211.5918, 181.7347, 73.4694, 71.0204, 55.1837, 130.6122, 88.1633]

  2. Sum of squared differences: =811.7755\sum = 811.7755

  3. Compute sds_d: sd=811.775571=811.77556=135.295911.6308s_d = \sqrt{\frac{811.7755}{7-1}} = \sqrt{\frac{811.7755}{6}} = \sqrt{135.2959} \approx 11.6308

3. Determine the Critical tt-Value

For a 95% confidence level and n1=71=6n-1 = 7-1 = 6 degrees of freedom, the tt-value is approximately t0.025,6=2.447t_{0.025, 6} = 2.447.

4. Calculate the Margin of Error (ME)

The formula for the margin of error is:

ME=tsdnME = t \cdot \frac{s_d}{\sqrt{n}}

Substitute the values: ME=2.44711.63087ME = 2.447 \cdot \frac{11.6308}{\sqrt{7}}

  1. Compute 7\sqrt{7}: 72.6458\sqrt{7} \approx 2.6458

  2. Compute MEME: ME=2.44711.63082.6458=2.4474.392010.7526ME = 2.447 \cdot \frac{11.6308}{2.6458} = 2.447 \cdot 4.3920 \approx 10.7526

Final Answer:

The margin of error is approximately: 10.752646\boxed{10.752646}

Let me know if you want the remaining steps explained or have additional questions!


Here are 5 related questions:

  1. How do you interpret a confidence interval in this context?
  2. How does the assumption of normality affect the calculations?
  3. What happens to the margin of error if the confidence level is increased?
  4. How do you calculate the tt-value for a different sample size?
  5. Why do we use the standard error in the margin of error formula?

Tip: Always ensure the assumptions of the test (e.g., normality and dependency of samples) are valid before interpreting the results.

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Math Problem Analysis

Mathematical Concepts

Confidence Interval
Margin of Error
Dependent Samples
Normal Distribution
t-Distribution

Formulas

d_i = X_{1i} - X_{2i}
Mean of differences: \bar{d} = \frac{\sum d_i}{n}
Standard deviation of differences: s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}}
Margin of Error: ME = t \cdot \frac{s_d}{\sqrt{n}}

Theorems

Central Limit Theorem
t-Distribution for Confidence Intervals

Suitable Grade Level

Undergraduate (Statistics)