**Statistical Analysis: Confidence Intervals****Problem Statement:**8. A sample of 16 gas stations shows that the mean price for regular gasoline is $ \bar{x} = \$3.25 $. It is known that the standard deviation is $ \sigma = \$0.08 $. a. Find a 95% confidence interval. b. If we had used a 99% confidence interval, what effect, if any, would this have on the length of the confidence interval?**Solution:****Part a: Finding a 95% Confidence Interval**To find the 95% confidence interval for the mean price of regular gasoline, we use the formula for the confidence interval of the mean for a sample:\[ CI = \bar{x} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right) \]Where:- $ \bar{x} $ is the sample mean- $ Z $ is the Z-value that corresponds to the desired confidence level- $ \sigma $ is the population standard deviation- $ n $ is the sample sizeGiven:- $ \bar{x} = \$3.25 $- $ \sigma = \$0.08 $- $ n = 16 $For a 95% confidence level, the Z-value is 1.96.So, we calculate:\[ CI = 3.25 \pm 1.96 \left( \frac{0.08}{\sqrt{16}} \right) \]\[ CI = 3.25 \pm 1.96 \left( \frac{0.08}{4} \right) \]\[ CI = 3.25 \pm 1.96 \left( 0.02 \right) \]\[ CI = 3.25 \pm 0.0392 \]Thus, the 95% confidence interval is:\[ (\$3.2108, \$3.2892) \]**Part b: Effect of Using a 99% Confidence Interval**If we use a 99% confidence interval, the Z-value would change. For a 99% confidence interval, the Z-value is approximately 2.576.The formula to recalculate the confidence interval with a 99% confidence level would be:\[ CI = 3.25 \pm 2.576 \

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8. A sample of 16 gas stations shows that the mean price for regular gasoline is x= $3.25. It is known that the standard deviation is o = $.08. a. Find a 95% confidence interval. b. If we had used a 99% confidence interval, what effect if any would this have on the length of the confidence interval.