A car company claims that its new SUV gets better gas mileage than its competitor's SUV. A random sample of 49 of its SUVs has a mean gas mileage of 14.3 miles per gallon (mpg). The population standard deviation is known to be $0.5 \mathrm{mpg}$. A random sample of 38 competitor's SUVs has a mean gas mileage of $13.8 \mathrm{mpg}$. The population standard deviation for the competitor is known to be $1.3 \mathrm{mpg}$. Test the company's claim at the 0.05 level of significance. Let the car company's SUVs be Population 1 and let the competitor's SUVs be Population 2.

Step 1 ：Define the null hypothesis as the mean gas mileage of the company's SUVs being equal to the mean gas mileage of the competitor's SUVs.

Step 2 ：Define the alternative hypothesis as the mean gas mileage of the company's SUVs not being equal to the mean gas mileage of the competitor's SUVs.

Step 3 ：Use the formula for the z-score for the difference between two means to calculate the test statistic: \[z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]

Step 4 ：Substitute the given values into the formula: \[z = \frac{(14.3 - 13.8)}{\sqrt{\frac{0.5^2}{49} + \frac{1.3^2}{38}}}\]

Step 5 ：Calculate the z-score: \[z = 2.245615861384521\]

Step 6 ：Compare the calculated z-score to the critical z-score for a 0.05 level of significance. The critical z-score is 1.959963984540054.

Step 7 ：Since the calculated z-score is greater than the critical z-score, reject the null hypothesis. This means that the difference between the mean gas mileage of the company's SUVs and the mean gas mileage of the competitor's SUVs is statistically significant at the 0.05 level of significance.

Step 8 ：This supports the company's claim that its new SUV gets better gas mileage than its competitor's SUV.

Step 9 ：Final Answer: \(\boxed{\text{Reject the null hypothesis}}\)