Question

Let two independent random samples, each of size 10 , from two normal distributions \(N\left(\mu_{1}, \sigma^{2}\right)\) and \(N\left(\mu_{2}, \sigma^{2}\right)\) yield \(\bar{x}=4.8, s_{1}^{2}=8.64, \bar{y}=5.6, s_{2}^{2}=7.88\). Find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\).

Short answer

The 95% confidence interval for the difference between the means \(\mu_{1}-\mu_{2}\) is (-3.53, 1.93).

Step-by-step solution

Compute sample standard error

The first step is to compute the standard error (SE) of the difference in sample means, which is given by the formula: \(SE = \sqrt{\frac{s_{1}^{2}}{n_1} + \frac{s_{2}^{2}}{n_2}}\). Substituting the provided values, we get: \(SE = \sqrt{\frac{8.64}{10} + \frac{7.88}{10}} = \sqrt{0.864 + 0.788} = \sqrt{1.672} \approx 1.29\).

Calculate degrees of freedom

Next, we compute the degrees of freedom (df) for the t-distribution. For two independent samples, it's given by \(df = n_1 + n_2 - 2 = 10 + 10 - 2 = 18\).

Find critical value from t-distribution

We're seeking a 95% confidence interval, hence this corresponds to a 5% significance level (alpha = 0.05). With a two-tailed test, we distribute this alpha evenly on both sides, getting alpha/2 = 0.025. Therefore, we have to find the t-value for df = 18 and alpha/2 = 0.025. Consulting the t-distribution table, we find the critical t-value is approximately 2.101.

Construct the confidence interval

The 95% confidence interval for the difference between the means \(\mu_{1} - \mu_{2}\) is given by: \(\bar{x}-\bar{y} \pm t_{\alpha/2,df}\times SE\). Substituting the obtained values, we get: \(4.8 - 5.6 \pm 2.101\times 1.29 \) which simplifies to -0.8 \pm 2.73.

Final Result

Thus, the 95% confidence interval for the difference between the means \(\mu_{1}-\mu_{2}\) is (-3.53, 1.93)