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

Answer:

The answer is "(27.030,29.770)"

Step-by-step explanation:

$(\bar{x}) = 28.4$

$(n)= 49$

$(\sigma) = 6$

$(CI) = \bar{x}\pm z^*_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}$

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

$CI = 28.4 \pm 1.598 \times \frac{6}{\sqrt{49}}=(27.030,29.770)$