Question

gasoline use a random sample of 25 drivers used on average 756 gallons of gasoline per year. the standard deviation of the population is 35 gallons. part: 0 / 2 part 1 of 2 (a) find the 99% confidence interval of the mean for all drivers. round intermediate answers to at least three decimal places. round your final answers to the nearest whole number. \\(\square < \mu < \square\\)

Explanation:

Step1: Identify given values

Sample size $n=25$, sample mean $\bar{x}=756$, population standard deviation $\sigma=35$, confidence level 99%.

Step2: Find z-score for 99% confidence

Two-tailed z-score for 99% confidence is $z=2.576$.

Step3: Calculate standard error

Standard error $SE=\frac{\sigma}{\sqrt{n}}=\frac{35}{\sqrt{25}}=7$.

Step4: Compute margin of error

Margin of error $E=z\times SE=2.576\times7\approx18.032$.

Step5: Find confidence interval

Lower bound: $756 - 18.032\approx738$; Upper bound: $756 + 18.032\approx774$.

Answer:

738 < μ < 774