Question

A sample of 30 randomly selected oranges was taken from a large population, and their diameters were measured.The mean diameter of the sample was 91 mm and the standard deviation was 8 mm.Assuming a Normal distribution, calculate (correct to one decimal place) 85% confidence limits for the mean diameter of the whole population of oranges.

A 87.9 mm and 94.1 mm

B 89.9 mm and 93.1 mm

C 88.1 mm and 93.9 mm

D 88.9 mm and 93.1 mm

Answer 1

D 88.9 mm and 93.1 mm

Explanation:

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

`\alpha = (1-0.85)/(2) = 0.075`

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
`1-0.075 = 0.925`, so
`z = 1.44`

Now, find 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 = 1.44*(8)/(√(30)) = 2.1`

The lower end of the interval is the sample mean subtracted by M. So it is 91 - 2.1 = 88.9mm

The upper end of the interval is the sample mean added to M. So it is 91 + 2.1 = 93.1 mm

So the correct answer is:

D 88.9 mm and 93.1 mm