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

Answer:

D 88.9 mm and 93.1 mm

Step-by-step explanation:

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

$\alpha = \frac{1-0.85}{2} = 0.075$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = 1.44*\frac{8}{\sqrt{30}} = 2.1$

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