A random sample of 24 adult male wolves from the Canadian Northwest Territories gave an average weight [tex]x_1 = 97.8[/tex] pounds with an estimated sample standard deviation [tex]s_1 = 7.5[/tex] pounds. Another sample of 21 adult male wolves from Alaska gave an average weight [tex]x_2 = 89.2[/tex] pounds with an estimated sample standard deviation [tex]s_2 = 7.1[/tex] pounds. Let [tex]\mu_1[/tex] represent the population mean weight of adult male wolves from the Northwest Territories, and let [tex]\mu_2[/tex] represent the population mean weight of adult male wolves from Alaska.

Find a 75% confidence interval for [tex]\mu_1 - \mu_2[/tex] (in pounds). Enter your answer in the form: lower limit to upper limit. Include the word "to." Round your answer to one decimal place.

To find a 75% confidence interval for the difference between the population means [tex]\mu_1 - \mu_2[/tex] of adult male wolves from the Northwest Territories and Alaska, we need the sample means, standard deviations, and sample sizes for both groups:

Sample 1 (Northwest Territories):
- Average weight [tex]\bar{x}_1 = 97.8[/tex] pounds
- Sample standard deviation [tex]s_1 = 7.5[/tex] pounds
- Sample size [tex]n_1 = 24[/tex]
Sample 2 (Alaska):
- Average weight [tex]\bar{x}_2 = 89.2[/tex] pounds
- Sample standard deviation [tex]s_2 = 7.1[/tex] pounds
- Sample size [tex]n_2 = 21[/tex]

We use the formula for the confidence interval of the difference between two means:

[tex](\bar{x}_1 - \bar{x}_2) \pm t^* \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}[/tex]

Where [tex]t^*[/tex] is the critical value from the t-distribution for 75% confidence level. Since this is a two-tailed distribution, 75% confidence corresponds to 25% on one side (or 12.5% in each tail of the distribution), therefore we look for [tex]t^*[/tex] with [tex]\text{degrees of freedom}[/tex] near the smaller of the two sample sizes minus one, which can be calculated or approximated.

Using the above setup:

[tex]\text{Estimated standard error} = \sqrt{\frac{7.5^2}{24} + \frac{7.1^2}{21}} = \sqrt{\frac{56.25}{24} + \frac{50.41}{21}} = \sqrt{2.34375 + 2.40048} = \sqrt{4.74423} \approx 2.178[/tex]

For a 75% confidence level, we find the critical t-value. Since sample sizes are not very large, we'll use an approximate degree of freedom based on the smaller sample, [tex]n_2 - 1 = 20[/tex]. The value of [tex]t^*[/tex] can be found using statistical tables or software. Let's approximate here:

Assuming [tex]t^* \approx 1.318[/tex] for 20 degrees of freedom and 75% confidence:

[tex](97.8 - 89.2) \pm 1.318 \cdot 2.178 = 8.6 \pm 2.868[/tex]

Therefore, the 75% confidence interval for [tex]\mu_1 - \mu_2[/tex] is approximately:

[tex]5.7 \text{ to } 11.5\, \text{pounds}[/tex]

This interval suggests that, with 75% confidence, the average weight of adult male wolves from the Northwest Territories is between 5.7 to 11.5 pounds more than those from Alaska.