To construct a 90% confidence interval for $\mu_1 - \mu_2$, we will use the given statistics and the provided formula. The confidence interval formula for the difference in means when variances are not equal is given by:

$(\bar{x}_1 - \bar{x}_2) \pm t_c \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

First, we need to determine the degrees of freedom (d.f.) for the t-distribution. According to the problem, d.f. is the smaller of $n_1 - 1$ or $n_2 - 1$.

Given:

• $\bar{x}_1 = 133$ mg
• $s_1 = 3.98$ mg
• $n_1 = 13$
• $\bar{x}_2 = 121$ mg
• $s_2 = 2.23$ mg
• $n_2 = 18$

The degrees of freedom is: $\text{d.f.} = \min(n_1 - 1, n_2 - 1) = \min(13 - 1, 18 - 1) = \min(12, 17) = 12$

Next, we look up the critical value $t_c$ for a 90% confidence level and 12 degrees of freedom. From the t-distribution table, $t_c \approx 1.782$.

Now we can calculate the standard error (SE):

$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{(3.98)^2}{13} + \frac{(2.23)^2}{18}}$

Calculate each component:

$\frac{(3.98)^2}{13} = \frac{15.8404}{13} \approx 1.2185$ $\frac{(2.23)^2}{18} = \frac{4.9729}{18} \approx 0.2763$

Thus,

$SE = \sqrt{1.2185 + 0.2763} = \sqrt{1.4948} \approx 1.222$

Now calculate the confidence interval:

$(\bar{x}_1 - \bar{x}_2) \pm t_c \cdot SE = (133 - 121) \pm 1.782 \cdot 1.222$

Calculate the difference in means and the margin of error:

$\bar{x}_1 - \bar{x}_2 = 133 - 121 = 12$ [ \text{Margin of Error} = 1.