Step 1 :Given that the sample mean \(\bar{x}\) is 744 gallons, the standard deviation \(\sigma\) is 33 gallons, and the sample size \(n\) is 49 drivers.
Step 2 :The formula for the confidence interval is \(CI = \bar{x} \pm Z \times \left(\frac{\sigma}{\sqrt{n}}\right)\), where \(Z\) is the Z-score, which is 1.645 for a 90% confidence interval.
Step 3 :Substitute the given values into the formula: \(CI = 744 \pm 1.645 \times \left(\frac{33}{\sqrt{49}}\right)\).
Step 4 :Simplify the expression inside the parentheses: \(CI = 744 \pm 1.645 \times \left(\frac{33}{7}\right)\).
Step 5 :Calculate the multiplication: \(CI = 744 \pm 1.645 \times 4.714\).
Step 6 :Calculate the addition and subtraction: \(CI = 744 \pm 7.75\).
Step 7 :So, the 90% confidence interval for the mean is \(\boxed{736 \text{ to } 752}\) gallons.
Step 8 :If a driver said that he used 804 gallons per year, it would be outside our confidence interval of 736 to 752 gallons. Therefore, it would be unlikely and we would not readily believe that.