K.Brew sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet. Random samples of sales receipts were studied for mail-order sales and internet sales, with the total purchase being recorded for each sale. A random sample of 6 sales receipts for mail-order sales results in a mean sale amount of $\$ 71.60$ with a standard deviation of $\$ 15.75$. A random sample of 10 sales receipts for internet sales results in a mean sale amount of $\$ 85.30$ with a standard deviation of $\$ 16.25$. Using this data, find the $90 \%$ confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases. Assume that the population variances are not equal and that the two populations are normally distributed. Step 1 of 3: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 1 ：We are given two samples of sales receipts from K.Brew, one from mail-order sales and the other from internet sales. The mail-order sales sample has a size of 6, a mean of $71.60, and a standard deviation of $15.75. The internet sales sample has a size of 10, a mean of $85.30, and a standard deviation of $16.25. We are asked to find the 90% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases. We are told to assume that the population variances are not equal and that the two populations are normally distributed.

Step 2 ：The first step in constructing the confidence interval is to find the critical value. This is done by looking up the value in a t-distribution table. However, since we are dealing with two samples, we need to calculate the degrees of freedom first. The formula for degrees of freedom for two samples with unequal variances is given by: \[df = \frac{(s1^2/n1 + s2^2/n2)^2}{(s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1)}\] where s1 and s2 are the standard deviations of the two samples, and n1 and n2 are the sizes of the two samples.

Step 3 ：Substituting the given values into the formula, we find that the degrees of freedom for this problem is 11.

Step 4 ：Now that we have the degrees of freedom, we can find the critical value from the t-distribution table. The critical value for a 90% confidence interval with 11 degrees of freedom is approximately 1.796.

Step 5 ：Final Answer: The critical value that should be used in constructing the 90% confidence interval is approximately \(\boxed{1.796}\).