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 36 cars of this model is chosen and found to have a mean gas mileage of 28.4 mi/gallon. Construct a 98% confidence interval for the mean gas mileage for this car model.

Answer 1

The 98% confidence interval for the mean gas mileage for this car model is (26.07,30.73).

Confidence intervals are defined as a range of values with a known chance that a parameter's value falls inside them.

The confidence interval of statistical data is computed using the formula:

$(\overline{x} - Z\frac{\sigma }{n},\overline{x} + Z\frac{\sigma }{n})$

where $\overline{x}$ is the mean, Z is the Z-score corresponding to the confidence interval, σ is the standard deviation, and n is the sample size.

In the question, the sample size (n) = 36, the mean of the sample ($\overline{x}$) = 28.4 mi/gallon, the standard deviation (σ) = 6 mi/gallon.

The confidence interval given to us is 98%.

Z-score corresponding to this (Z) = 2.33.

Thus, the confidence interval can be calculated as:

(28.4 - 2.33{6/√36},28.4 + 2.33{6/√36})

= (28.4 - 2.33,28.4 + 2.33)

= (26.07,30.73).

Thus, the 98% confidence interval for the mean gas mileage for this car model is (26.07,30.73).

Learn more about constructing confidence intervals at

brainly.com/question/17030704

#SPJ4