Step 1 :Given values are: mean of group 1 (x1) = 119, standard deviation of group 1 (s1) = 13, sample size of group 1 (n1) = 14, mean of group 2 (x2) = 81, standard deviation of group 2 (s2) = 15, sample size of group 2 (n2) = 10.
Step 2 :Calculate the t-score for a 90% confidence interval with 22 degrees of freedom. The t-score is approximately 1.717.
Step 3 :Calculate the standard error (se) using the formula \(\sqrt{\frac{s1^2}{n1} + \frac{s2^2}{n2}}\). The standard error is approximately 5.880.
Step 4 :Calculate the lower limit of the confidence interval using the formula \((x1 - x2) - t_{score} \times se\). The lower limit is approximately 27.90.
Step 5 :Calculate the upper limit of the confidence interval using the formula \((x1 - x2) + t_{score} \times se\). The upper limit is approximately 48.10.
Step 6 :The lower limit of the 90% confidence interval for the difference between the mean IQ scores of group 1 and group 2 is approximately 27.90, and the upper limit is approximately 48.10. Therefore, we are 90% confident that the true difference between the mean IQ scores of the two groups is between 27.90 and 48.10. So, the final answer is \(\boxed{27.90}\) and \(\boxed{48.10}\).