Question

The gas mileage for a certain model of car is known to have a standard deviation of 6 mi/gallon. A simple random sample of 49 cars of this model is chosen and found to have a mean gas mileage of 28.4 mi/gallon. Construct a 89% confidence interval for the mean gas mileage for this car model.

Answer 1

The answer is "(27.030,29.770)"

Explanation:

`(\bar{x}) = 28.4`

`(n)= 49`

`(\sigma) = 6`

`(CI) = \bar{x}\pm z^*_{(\alpha)/(2)} (\sigma)/(√(n))`

`z^*_{(\alpha)/(2)}\ for \ 89\%\ confidence = 1.598 \\\\\text{Using z table }\\\\ NORM.S.INV(1-((1-0.89))/(2))`

`CI = 28.4 \pm 1.598 * (6)/(√(49))=(27.030,29.770)`