Question

In Exercises 6.188 to 6.191 , use the t-distribution to find a confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) given the relevant sample results. Give the best estimate for \(\mu_{1}-\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=75.2, s_{1}=10.7, n_{1}=30\) and \(\bar{x}_{2}=69.0, s_{2}=8.3, n_{2}=20 .\)

Short answer

The best estimate for \(\mu_{1}-\mu_{2}\) is 6.2 with a standard error of 2.68. The margin of error is 5.61. Thus, the \(95\%\) confidence interval for \(\mu_{1}-\mu_{2}\) is \((0.59, 11.81)\).

Step-by-step solution

Compute the best estimate.

The best estimate for \(\mu_{1}-\mu_{2}\) is the difference between the sample means. We denote this as \(D\), so our best estimate is calculated as \(D = \bar{x}_{1} - \bar{x}_{2} = 75.2 - 69.0 = 6.2\).

Calculate the standard error of difference.

The standard error of difference is calculated using the formula: \(SE_{D} = sqrt(\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}})\) where \(s_{1}\) and \(s_{2}\) are the sample standard deviations and \(n_{1}\) and \(n_{2}\) are the sample sizes. Plugging in the given values, we get \(SE_{D} = sqrt(\frac{10.7^{2}}{30} + \frac{8.3^{2}}{20}) = 2.68\).

Identify the t-value corresponding to the given level of confidence.

To construct a \(95\%\) confidence interval, we would refer the t-table for degrees of freedom \(df = min(n_{1}, n_{2}) - 1 = min(30,20) - 1 = 19\). The t-value for \(95\%\) confidence and \(19\) degrees of freedom is approximately \(2.093\).

Calculate the margin of error.

The margin of error (MoE) is calculated as: \(MoE = t \times SE_{D}\) where \(t\) is the t-value and \(SE_{D}\) is the standard error of difference. By plugging in the values from steps 2 and 3, we get \(MoE = 2.093 \times 2.68 = 5.61\).

Construct the confidence interval.

The confidence interval is then \(D \pm MoE\) where \(D\) is our best estimate and \(MoE\) is the margin of error. Thus our \(95\%\) confidence interval is \(6.2 \pm 5.61\), or \((0.59, 11.81)\).