Two brands of car batteries, both carrying 6-year warranties, were sampled and tested under controlled conditions. Five of each brand failed after the number of months shown. Calculate a) both sample means and b) both sample standard deviations. Decide c) which brand battery lasts longer and d) which brand has the more consistent lifetime. Brand A: $73,65,68,70,65$ Brand B: $60,67,72,70,67$ a) What are the sample means? \[ \bar{x}_{A}=68.2 \] \[ \bar{x}_{B}=67.2 \] b) What are the sample standard deviations? \[ \mathrm{s}_{\mathrm{A}}=\square \quad \mathrm{s}_{\mathrm{B}}=4.55 \] (Round to the nearest hundredth as needed.)

Step 1 ：Given the lifetimes of two brands of car batteries, Brand A: $73,65,68,70,65$ and Brand B: $60,67,72,70,67$, we are asked to calculate the sample means and sample standard deviations for both brands.

Step 2 ：To calculate the sample mean, we sum all the data points and divide by the number of data points. For Brand A, the sample mean is calculated as \(\bar{x}_{A}=\frac{73+65+68+70+65}{5}=68.2\). Similarly, for Brand B, the sample mean is calculated as \(\bar{x}_{B}=\frac{60+67+72+70+67}{5}=67.2\).

Step 3 ：To calculate the sample standard deviation, we first calculate the variance, which is the average of the squared differences from the mean. Then, we take the square root of the variance to get the standard deviation. For Brand A, the sample standard deviation is calculated as \(s_{A}=\sqrt{\frac{(73-68.2)^2+(65-68.2)^2+(68-68.2)^2+(70-68.2)^2+(65-68.2)^2}{5-1}}=3.42\). Similarly, for Brand B, the sample standard deviation is calculated as \(s_{B}=\sqrt{\frac{(60-67.2)^2+(67-67.2)^2+(72-67.2)^2+(70-67.2)^2+(67-67.2)^2}{5-1}}=4.55\).

Step 4 ：From the calculated sample means and sample standard deviations, we can see that Brand A batteries last longer on average and have a more consistent lifetime. Therefore, the final answer is \(\boxed{\bar{x}_{A}=68.2, \bar{x}_{B}=67.2, s_{A}=3.42, s_{B}=4.55}\).