Question

he gas mileage for a certain model of car is known to have a standard deviation of 4 mi/gallon. A simple random sample of 49 cars of this model is chosen and found to have a mean gas mileage of 27.5 mi/gallon. Construct a 96.5% confidence interval for the mean gas mileage for this car model. a) (27.328, 27.672) b) (19.068, 35.932) c) (26.295, 28.705) d) (20.252, 34.748) e) (26.465, 28.535) f) None of the above

Answer 1

c) (26.295, 28.705)

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.965)/(2) = 0.0175`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.0175 = 0.9825`, so
`z = 2.11`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 2.11(4)/(√(49)) = 1.205`

The lower end of the interval is the sample mean subtracted by M. So it is 27.5 - 1.205 = 26.295 mi/gallon

The upper end of the interval is the sample mean added to M. So it is 27.5 + 1.205 = 28.705 mi/gallon

So the correct answer is:

c) (26.295, 28.705)