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To calculate the z-score for a 99% confidence level, find the z-value such that the area under the standard normal curve is 0.99. This can be done using a standard normal distribution table or a z-score calculator. The z-score for a 99% confidence level is approximately 2.576.

The standard error (SE) is equaled to the standard deviation divided by the square root of the sample size. Given the standard deviation as \(2.3\) miles per gallon and sample size of \(36\), the standard error becomes \(SE = \frac{2.3}{\sqrt{36}} = 0.383\) miles per gallon.

Now construct the confidence interval using the formula \( \mu = M \pm (Z*SE) \) where \( \mu \) is the population mean, \( M \) is the sample mean, \( Z \) is the z-score and \( SE \) is the standard error. Substituting for \( M = 26.4 \), \( Z = 2.576 \) and \( SE = 0.383 \), we obtain the confidence interval \( \mu = 26.4 \pm (2.576*0.383) \) yielding to \( \mu \) between \( 25.41 \) and \( 27.39 \) miles per gallon.

To reduce the width of the confidence interval, one could: \n1. Decrease the confidence level: A lower confidence level will result in a smaller z-score and thus, a narrower confidence interval. However, this also implies taking more risk of being wrong.\n2. Increase the sample size: A larger sample size reduces the standard error, resulting in a more narrow interval. This can be more costly and time-consuming but doesn't compromise on the precision.\n3. Decrease the standard deviation: More consistency in data collection can lead to a lower standard deviation, thereby resulting in a more narrow confidence interval. This might not, however, always be in one's control. \nThe best alternative would be to increase the sample size because it enhances the precision of the estimate without compromising the level of confidence.