Math Problem Statement

Given two independent random samples with the following results:

n1=8 xbar1=127 s1=33    n2=12 xbar2=157 s2=27

Use this data to find the 95% confidence interval for the true difference between the population means. Assume that the population variances are not equal and that the two populations are normally distributed.

Step 2 of 3 : Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

Solution

To find the margin of error for the confidence interval for the difference between the population means, we follow these steps:

Formula for the Margin of Error:

ME=ts12n1+s22n2ME = t^* \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} Where:

  • tt^*: Critical value from the tt-distribution for a 95% confidence level, based on the degrees of freedom (dfdf).
  • s1,s2s_1, s_2: Sample standard deviations.
  • n1,n2n_1, n_2: Sample sizes.

Step 1: Compute the degrees of freedom (dfdf):

The degrees of freedom for unequal variances is calculated as: df=(s12n1+s22n2)2(s12n1)2n11+(s22n2)2n21df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}

Plug in the values:

  • s1=33s_1 = 33, s2=27s_2 = 27
  • n1=8n_1 = 8, n2=12n_2 = 12
  1. Compute variances and their ratios: s12n1=3328=10898=136.125\frac{s_1^2}{n_1} = \frac{33^2}{8} = \frac{1089}{8} = 136.125 s22n2=27212=72912=60.75\frac{s_2^2}{n_2} = \frac{27^2}{12} = \frac{729}{12} = 60.75

  2. Sum these terms for the numerator of dfdf: s12n1+s22n2=136.125+60.75=196.875\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} = 136.125 + 60.75 = 196.875

  3. Compute the denominator terms:

    • For s1s_1: (s12n1)2n11=(136.125)281=18518.0156257=2645.430804\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} = \frac{(136.125)^2}{8 - 1} = \frac{18518.015625}{7} = 2645.430804
    • For s2s_2: (s22n2)2n21=(60.75)2121=3690.562511=335.505682\frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1} = \frac{(60.75)^2}{12 - 1} = \frac{3690.5625}{11} = 335.505682
  4. Combine to find dfdf: df=(196.875)22645.430804+335.505682=38753.5156252980.93648613.0df = \frac{(196.875)^2}{2645.430804 + 335.505682} = \frac{38753.515625}{2980.936486} \approx 13.0

Thus, df13df \approx 13.


Step 2: Find tt^*:

Using df=13df = 13 and a 95% confidence level, the critical value tt^* from the tt-distribution table is approximately: t2.160t^* \approx 2.160


Step 3: Calculate the margin of error:

ME=ts12n1+s22n2ME = t^* \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}

  1. Compute the square root term: s12n1+s22n2=136.125+60.75=196.87514.026\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{136.125 + 60.75} = \sqrt{196.875} \approx 14.026

  2. Multiply by tt^*: ME=2.16014.02630.295656ME = 2.160 \cdot 14.026 \approx 30.295656


Final Answer:

The margin of error is: 30.295656\boxed{30.295656}

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Math Problem Analysis

Mathematical Concepts

Confidence Intervals
Margin of Error
Degrees of Freedom
t-Distribution

Formulas

Margin of Error: ME = t* × √((s1²/n1) + (s2²/n2))
Degrees of Freedom: df = [(s1²/n1 + s2²/n2)²] / [(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1)]

Theorems

Central Limit Theorem
t-Distribution Critical Values

Suitable Grade Level

Undergraduate Statistics